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I have a dictionary of values read from two fields in a database: a string field and a numeric field. The string field is unique, so that is the key of the dictionary.
I can sort on the keys, but how can I sort based on the values?
Note: I have read Stack Overflow question here How do I sort a list of dictionaries by a value of the dictionary? and probably could change my code to have a list of dictionaries, but since I do not really need a list of dictionaries I wanted to know if there is a simpler solution to sort either in ascending or descending order.
Python 3.7+ or CPython 3.6
Dicts preserve insertion order in Python 3.7+. Same in CPython 3.6, but it's an implementation detail.
>>> x = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}
>>> {k: v for k, v in sorted(x.items(), key=lambda item: item[1])}
{0: 0, 2: 1, 1: 2, 4: 3, 3: 4}
or
>>> dict(sorted(x.items(), key=lambda item: item[1]))
{0: 0, 2: 1, 1: 2, 4: 3, 3: 4}
Older Python
It is not possible to sort a dictionary, only to get a representation of a dictionary that is sorted. Dictionaries are inherently orderless, but other types, such as lists and tuples, are not. So you need an ordered data type to represent sorted values, which will be a list—probably a list of tuples.
For instance,
import operator
x = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}
sorted_x = sorted(x.items(), key=operator.itemgetter(1))
sorted_x will be a list of tuples sorted by the second element in each tuple. dict(sorted_x) == x.
And for those wishing to sort on keys instead of values:
import operator
x = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}
sorted_x = sorted(x.items(), key=operator.itemgetter(0))
In Python3 since unpacking is not allowed we can use
x = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}
sorted_x = sorted(x.items(), key=lambda kv: kv[1])
If you want the output as a dict, you can use collections.OrderedDict:
import collections
sorted_dict = collections.OrderedDict(sorted_x)
As simple as: sorted(dict1, key=dict1.get)
Well, it is actually possible to do a "sort by dictionary values". Recently I had to do that in a Code Golf (Stack Overflow question Code golf: Word frequency chart). Abridged, the problem was of the kind: given a text, count how often each word is encountered and display a list of the top words, sorted by decreasing frequency.
If you construct a dictionary with the words as keys and the number of occurrences of each word as value, simplified here as:
from collections import defaultdict
d = defaultdict(int)
for w in text.split():
d[w] += 1
then you can get a list of the words, ordered by frequency of use with sorted(d, key=d.get) - the sort iterates over the dictionary keys, using the number of word occurrences as a sort key .
for w in sorted(d, key=d.get, reverse=True):
print(w, d[w])
I am writing this detailed explanation to illustrate what people often mean by "I can easily sort a dictionary by key, but how do I sort by value" - and I think the original post was trying to address such an issue. And the solution is to do sort of list of the keys, based on the values, as shown above.
You could use:
sorted(d.items(), key=lambda x: x[1])
This will sort the dictionary by the values of each entry within the dictionary from smallest to largest.
To sort it in descending order just add reverse=True:
sorted(d.items(), key=lambda x: x[1], reverse=True)
Input:
d = {'one':1,'three':3,'five':5,'two':2,'four':4}
a = sorted(d.items(), key=lambda x: x[1])
print(a)
Output:
[('one', 1), ('two', 2), ('three', 3), ('four', 4), ('five', 5)]
Dicts can't be sorted, but you can build a sorted list from them.
A sorted list of dict values:
sorted(d.values())
A list of (key, value) pairs, sorted by value:
from operator import itemgetter
sorted(d.items(), key=itemgetter(1))
In recent Python 2.7, we have the new OrderedDict type, which remembers the order in which the items were added.
>>> d = {"third": 3, "first": 1, "fourth": 4, "second": 2}
>>> for k, v in d.items():
... print "%s: %s" % (k, v)
...
second: 2
fourth: 4
third: 3
first: 1
>>> d
{'second': 2, 'fourth': 4, 'third': 3, 'first': 1}
To make a new ordered dictionary from the original, sorting by the values:
>>> from collections import OrderedDict
>>> d_sorted_by_value = OrderedDict(sorted(d.items(), key=lambda x: x[1]))
The OrderedDict behaves like a normal dict:
>>> for k, v in d_sorted_by_value.items():
... print "%s: %s" % (k, v)
...
first: 1
second: 2
third: 3
fourth: 4
>>> d_sorted_by_value
OrderedDict([('first': 1), ('second': 2), ('third': 3), ('fourth': 4)])
Using Python 3.5
Whilst I found the accepted answer useful, I was also surprised that it hasn't been updated to reference OrderedDict from the standard library collections module as a viable, modern alternative - designed to solve exactly this type of problem.
from operator import itemgetter
from collections import OrderedDict
x = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}
sorted_x = OrderedDict(sorted(x.items(), key=itemgetter(1)))
# OrderedDict([(0, 0), (2, 1), (1, 2), (4, 3), (3, 4)])
The official OrderedDict documentation offers a very similar example too, but using a lambda for the sort function:
# regular unsorted dictionary
d = {'banana': 3, 'apple':4, 'pear': 1, 'orange': 2}
# dictionary sorted by value
OrderedDict(sorted(d.items(), key=lambda t: t[1]))
# OrderedDict([('pear', 1), ('orange', 2), ('banana', 3), ('apple', 4)])
Pretty much the same as Hank Gay's answer:
sorted([(value,key) for (key,value) in mydict.items()])
Or optimized slightly as suggested by John Fouhy:
sorted((value,key) for (key,value) in mydict.items())
As of Python 3.6 the built-in dict will be ordered
Good news, so the OP's original use case of mapping pairs retrieved from a database with unique string ids as keys and numeric values as values into a built-in Python v3.6+ dict, should now respect the insert order.
If say the resulting two column table expressions from a database query like:
SELECT a_key, a_value FROM a_table ORDER BY a_value;
would be stored in two Python tuples, k_seq and v_seq (aligned by numerical index and with the same length of course), then:
k_seq = ('foo', 'bar', 'baz')
v_seq = (0, 1, 42)
ordered_map = dict(zip(k_seq, v_seq))
Allow to output later as:
for k, v in ordered_map.items():
print(k, v)
yielding in this case (for the new Python 3.6+ built-in dict!):
foo 0
bar 1
baz 42
in the same ordering per value of v.
Where in the Python 3.5 install on my machine it currently yields:
bar 1
foo 0
baz 42
Details:
As proposed in 2012 by Raymond Hettinger (cf. mail on python-dev with subject "More compact dictionaries with faster iteration") and now (in 2016) announced in a mail by Victor Stinner to python-dev with subject "Python 3.6 dict becomes compact and gets a private version; and keywords become ordered" due to the fix/implementation of issue 27350 "Compact and ordered dict" in Python 3.6 we will now be able, to use a built-in dict to maintain insert order!!
Hopefully this will lead to a thin layer OrderedDict implementation as a first step. As #JimFasarakis-Hilliard indicated, some see use cases for the OrderedDict type also in the future. I think the Python community at large will carefully inspect, if this will stand the test of time, and what the next steps will be.
Time to rethink our coding habits to not miss the possibilities opened by stable ordering of:
Keyword arguments and
(intermediate) dict storage
The first because it eases dispatch in the implementation of functions and methods in some cases.
The second as it encourages to more easily use dicts as intermediate storage in processing pipelines.
Raymond Hettinger kindly provided documentation explaining "The Tech Behind Python 3.6 Dictionaries" - from his San Francisco Python Meetup Group presentation 2016-DEC-08.
And maybe quite some Stack Overflow high decorated question and answer pages will receive variants of this information and many high quality answers will require a per version update too.
Caveat Emptor (but also see below update 2017-12-15):
As #ajcr rightfully notes: "The order-preserving aspect of this new implementation is considered an implementation detail and should not be relied upon." (from the whatsnew36) not nit picking, but the citation was cut a bit pessimistic ;-). It continues as " (this may change in the future, but it is desired to have this new dict implementation in the language for a few releases before changing the language spec to mandate order-preserving semantics for all current and future Python implementations; this also helps preserve backwards-compatibility with older versions of the language where random iteration order is still in effect, e.g. Python 3.5)."
So as in some human languages (e.g. German), usage shapes the language, and the will now has been declared ... in whatsnew36.
Update 2017-12-15:
In a mail to the python-dev list, Guido van Rossum declared:
Make it so. "Dict keeps insertion order" is the ruling. Thanks!
So, the version 3.6 CPython side-effect of dict insertion ordering is now becoming part of the language spec (and not anymore only an implementation detail). That mail thread also surfaced some distinguishing design goals for collections.OrderedDict as reminded by Raymond Hettinger during discussion.
It can often be very handy to use namedtuple. For example, you have a dictionary of 'name' as keys and 'score' as values and you want to sort on 'score':
import collections
Player = collections.namedtuple('Player', 'score name')
d = {'John':5, 'Alex':10, 'Richard': 7}
sorting with lowest score first:
worst = sorted(Player(v,k) for (k,v) in d.items())
sorting with highest score first:
best = sorted([Player(v,k) for (k,v) in d.items()], reverse=True)
Now you can get the name and score of, let's say the second-best player (index=1) very Pythonically like this:
player = best[1]
player.name
'Richard'
player.score
7
I had the same problem, and I solved it like this:
WantedOutput = sorted(MyDict, key=lambda x : MyDict[x])
(People who answer "It is not possible to sort a dict" did not read the question! In fact, "I can sort on the keys, but how can I sort based on the values?" clearly means that he wants a list of the keys sorted according to the value of their values.)
Please notice that the order is not well defined (keys with the same value will be in an arbitrary order in the output list).
If values are numeric you may also use Counter from collections.
from collections import Counter
x = {'hello': 1, 'python': 5, 'world': 3}
c = Counter(x)
print(c.most_common())
>> [('python', 5), ('world', 3), ('hello', 1)]
Starting from Python 3.6, dict objects are now ordered by insertion order. It's officially in the specifications of Python 3.7.
>>> words = {"python": 2, "blah": 4, "alice": 3}
>>> dict(sorted(words.items(), key=lambda x: x[1]))
{'python': 2, 'alice': 3, 'blah': 4}
Before that, you had to use OrderedDict.
Python 3.7 documentation says:
Changed in version 3.7: Dictionary order is guaranteed to be insertion
order. This behavior was implementation detail of CPython from 3.6.
In Python 2.7, simply do:
from collections import OrderedDict
# regular unsorted dictionary
d = {'banana': 3, 'apple':4, 'pear': 1, 'orange': 2}
# dictionary sorted by key
OrderedDict(sorted(d.items(), key=lambda t: t[0]))
OrderedDict([('apple', 4), ('banana', 3), ('orange', 2), ('pear', 1)])
# dictionary sorted by value
OrderedDict(sorted(d.items(), key=lambda t: t[1]))
OrderedDict([('pear', 1), ('orange', 2), ('banana', 3), ('apple', 4)])
copy-paste from : http://docs.python.org/dev/library/collections.html#ordereddict-examples-and-recipes
Enjoy ;-)
This is the code:
import operator
origin_list = [
{"name": "foo", "rank": 0, "rofl": 20000},
{"name": "Silly", "rank": 15, "rofl": 1000},
{"name": "Baa", "rank": 300, "rofl": 20},
{"name": "Zoo", "rank": 10, "rofl": 200},
{"name": "Penguin", "rank": -1, "rofl": 10000}
]
print ">> Original >>"
for foo in origin_list:
print foo
print "\n>> Rofl sort >>"
for foo in sorted(origin_list, key=operator.itemgetter("rofl")):
print foo
print "\n>> Rank sort >>"
for foo in sorted(origin_list, key=operator.itemgetter("rank")):
print foo
Here are the results:
Original
{'name': 'foo', 'rank': 0, 'rofl': 20000}
{'name': 'Silly', 'rank': 15, 'rofl': 1000}
{'name': 'Baa', 'rank': 300, 'rofl': 20}
{'name': 'Zoo', 'rank': 10, 'rofl': 200}
{'name': 'Penguin', 'rank': -1, 'rofl': 10000}
Rofl
{'name': 'Baa', 'rank': 300, 'rofl': 20}
{'name': 'Zoo', 'rank': 10, 'rofl': 200}
{'name': 'Silly', 'rank': 15, 'rofl': 1000}
{'name': 'Penguin', 'rank': -1, 'rofl': 10000}
{'name': 'foo', 'rank': 0, 'rofl': 20000}
Rank
{'name': 'Penguin', 'rank': -1, 'rofl': 10000}
{'name': 'foo', 'rank': 0, 'rofl': 20000}
{'name': 'Zoo', 'rank': 10, 'rofl': 200}
{'name': 'Silly', 'rank': 15, 'rofl': 1000}
{'name': 'Baa', 'rank': 300, 'rofl': 20}
Try the following approach. Let us define a dictionary called mydict with the following data:
mydict = {'carl':40,
'alan':2,
'bob':1,
'danny':3}
If one wanted to sort the dictionary by keys, one could do something like:
for key in sorted(mydict.iterkeys()):
print "%s: %s" % (key, mydict[key])
This should return the following output:
alan: 2
bob: 1
carl: 40
danny: 3
On the other hand, if one wanted to sort a dictionary by value (as is asked in the question), one could do the following:
for key, value in sorted(mydict.iteritems(), key=lambda (k,v): (v,k)):
print "%s: %s" % (key, value)
The result of this command (sorting the dictionary by value) should return the following:
bob: 1
alan: 2
danny: 3
carl: 40
You can create an "inverted index", also
from collections import defaultdict
inverse= defaultdict( list )
for k, v in originalDict.items():
inverse[v].append( k )
Now your inverse has the values; each value has a list of applicable keys.
for k in sorted(inverse):
print k, inverse[k]
You can use the collections.Counter. Note, this will work for both numeric and non-numeric values.
>>> x = {1: 2, 3: 4, 4:3, 2:1, 0:0}
>>> from collections import Counter
>>> #To sort in reverse order
>>> Counter(x).most_common()
[(3, 4), (4, 3), (1, 2), (2, 1), (0, 0)]
>>> #To sort in ascending order
>>> Counter(x).most_common()[::-1]
[(0, 0), (2, 1), (1, 2), (4, 3), (3, 4)]
>>> #To get a dictionary sorted by values
>>> from collections import OrderedDict
>>> OrderedDict(Counter(x).most_common()[::-1])
OrderedDict([(0, 0), (2, 1), (1, 2), (4, 3), (3, 4)])
The collections solution mentioned in another answer is absolutely superb, because you retain a connection between the key and value which in the case of dictionaries is extremely important.
I don't agree with the number one choice presented in another answer, because it throws away the keys.
I used the solution mentioned above (code shown below) and retained access to both keys and values and in my case the ordering was on the values, but the importance was the ordering of the keys after ordering the values.
from collections import Counter
x = {'hello':1, 'python':5, 'world':3}
c=Counter(x)
print( c.most_common() )
>> [('python', 5), ('world', 3), ('hello', 1)]
You can also use a custom function that can be passed to parameter key.
def dict_val(x):
return x[1]
x = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}
sorted_x = sorted(x.items(), key=dict_val)
You can use a skip dict which is a dictionary that's permanently sorted by value.
>>> data = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}
>>> SkipDict(data)
{0: 0.0, 2: 1.0, 1: 2.0, 4: 3.0, 3: 4.0}
If you use keys(), values() or items() then you'll iterate in sorted order by value.
It's implemented using the skip list datastructure.
Of course, remember, you need to use OrderedDict because regular Python dictionaries don't keep the original order.
from collections import OrderedDict
a = OrderedDict(sorted(originalDict.items(), key=lambda x: x[1]))
If you do not have Python 2.7 or higher, the best you can do is iterate over the values in a generator function. (There is an OrderedDict for 2.4 and 2.6 here, but
a) I don't know about how well it works
and
b) You have to download and install it of course. If you do not have administrative access, then I'm afraid the option's out.)
def gen(originalDict):
for x, y in sorted(zip(originalDict.keys(), originalDict.values()), key=lambda z: z[1]):
yield (x, y)
#Yields as a tuple with (key, value). You can iterate with conditional clauses to get what you want.
for bleh, meh in gen(myDict):
if bleh == "foo":
print(myDict[bleh])
You can also print out every value
for bleh, meh in gen(myDict):
print(bleh, meh)
Please remember to remove the parentheses after print if not using Python 3.0 or above
from django.utils.datastructures import SortedDict
def sortedDictByKey(self,data):
"""Sorted dictionary order by key"""
sortedDict = SortedDict()
if data:
if isinstance(data, dict):
sortedKey = sorted(data.keys())
for k in sortedKey:
sortedDict[k] = data[k]
return sortedDict
Here is a solution using zip on d.values() and d.keys(). A few lines down this link (on Dictionary view objects) is:
This allows the creation of (value, key) pairs using zip(): pairs = zip(d.values(), d.keys()).
So we can do the following:
d = {'key1': 874.7, 'key2': 5, 'key3': 8.1}
d_sorted = sorted(zip(d.values(), d.keys()))
print d_sorted
# prints: [(5, 'key2'), (8.1, 'key3'), (874.7, 'key1')]
As pointed out by Dilettant, Python 3.6 will now keep the order! I thought I'd share a function I wrote that eases the sorting of an iterable (tuple, list, dict). In the latter case, you can sort either on keys or values, and it can take numeric comparison into account. Only for >= 3.6!
When you try using sorted on an iterable that holds e.g. strings as well as ints, sorted() will fail. Of course you can force string comparison with str(). However, in some cases you want to do actual numeric comparison where 12 is smaller than 20 (which is not the case in string comparison). So I came up with the following. When you want explicit numeric comparison you can use the flag num_as_num which will try to do explicit numeric sorting by trying to convert all values to floats. If that succeeds, it will do numeric sorting, otherwise it'll resort to string comparison.
Comments for improvement welcome.
def sort_iterable(iterable, sort_on=None, reverse=False, num_as_num=False):
def _sort(i):
# sort by 0 = keys, 1 values, None for lists and tuples
try:
if num_as_num:
if i is None:
_sorted = sorted(iterable, key=lambda v: float(v), reverse=reverse)
else:
_sorted = dict(sorted(iterable.items(), key=lambda v: float(v[i]), reverse=reverse))
else:
raise TypeError
except (TypeError, ValueError):
if i is None:
_sorted = sorted(iterable, key=lambda v: str(v), reverse=reverse)
else:
_sorted = dict(sorted(iterable.items(), key=lambda v: str(v[i]), reverse=reverse))
return _sorted
if isinstance(iterable, list):
sorted_list = _sort(None)
return sorted_list
elif isinstance(iterable, tuple):
sorted_list = tuple(_sort(None))
return sorted_list
elif isinstance(iterable, dict):
if sort_on == 'keys':
sorted_dict = _sort(0)
return sorted_dict
elif sort_on == 'values':
sorted_dict = _sort(1)
return sorted_dict
elif sort_on is not None:
raise ValueError(f"Unexpected value {sort_on} for sort_on. When sorting a dict, use key or values")
else:
raise TypeError(f"Unexpected type {type(iterable)} for iterable. Expected a list, tuple, or dict")
I just learned a relevant skill from Python for Everybody.
You may use a temporary list to help you to sort the dictionary:
# Assume dictionary to be:
d = {'apple': 500.1, 'banana': 1500.2, 'orange': 1.0, 'pineapple': 789.0}
# Create a temporary list
tmp = []
# Iterate through the dictionary and append each tuple into the temporary list
for key, value in d.items():
tmptuple = (value, key)
tmp.append(tmptuple)
# Sort the list in ascending order
tmp = sorted(tmp)
print (tmp)
If you want to sort the list in descending order, simply change the original sorting line to:
tmp = sorted(tmp, reverse=True)
Using list comprehension, the one-liner would be:
# Assuming the dictionary looks like
d = {'apple': 500.1, 'banana': 1500.2, 'orange': 1.0, 'pineapple': 789.0}
# One-liner for sorting in ascending order
print (sorted([(v, k) for k, v in d.items()]))
# One-liner for sorting in descending order
print (sorted([(v, k) for k, v in d.items()], reverse=True))
Sample Output:
# Ascending order
[(1.0, 'orange'), (500.1, 'apple'), (789.0, 'pineapple'), (1500.2, 'banana')]
# Descending order
[(1500.2, 'banana'), (789.0, 'pineapple'), (500.1, 'apple'), (1.0, 'orange')]
Use ValueSortedDict from dicts:
from dicts.sorteddict import ValueSortedDict
d = {1: 2, 3: 4, 4:3, 2:1, 0:0}
sorted_dict = ValueSortedDict(d)
print sorted_dict.items()
[(0, 0), (2, 1), (1, 2), (4, 3), (3, 4)]
Iterate through a dict and sort it by its values in descending order:
$ python --version
Python 3.2.2
$ cat sort_dict_by_val_desc.py
dictionary = dict(siis = 1, sana = 2, joka = 3, tuli = 4, aina = 5)
for word in sorted(dictionary, key=dictionary.get, reverse=True):
print(word, dictionary[word])
$ python sort_dict_by_val_desc.py
aina 5
tuli 4
joka 3
sana 2
siis 1
If your values are integers, and you use Python 2.7 or newer, you can use collections.Counter instead of dict. The most_common method will give you all items, sorted by the value.
This works in 3.1.x:
import operator
slovar_sorted=sorted(slovar.items(), key=operator.itemgetter(1), reverse=True)
print(slovar_sorted)
For the sake of completeness, I am posting a solution using heapq. Note, this method will work for both numeric and non-numeric values
>>> x = {1: 2, 3: 4, 4:3, 2:1, 0:0}
>>> x_items = x.items()
>>> heapq.heapify(x_items)
>>> #To sort in reverse order
>>> heapq.nlargest(len(x_items),x_items, operator.itemgetter(1))
[(3, 4), (4, 3), (1, 2), (2, 1), (0, 0)]
>>> #To sort in ascending order
>>> heapq.nsmallest(len(x_items),x_items, operator.itemgetter(1))
[(0, 0), (2, 1), (1, 2), (4, 3), (3, 4)]
Related
I have 2 lists that I want to convert them into a dict with key and values. I managed to do so but there are too many steps so I would like to know if there's a simpler way of achieving this. Basically I would like to create the dict directly in the loop without having the extra steps bellow. I just started working with python and I don't quite understand all the datatypes that it provides.
The jName form can be modified if needed.
jName=["Nose", "Neck", "RShoulder", "RElbow", "RWrist", "LShoulder", "LElbow", "LWrist", "RHip",
"RKnee","RAnkle","LHip", "LKnee", "LAnkle", "REye", "LEye", "REar", "LEar"]
def get_joints(subset, candidate):
joints_per_skeleton = [[] for i in range(len(subset))]
# for each detected skeleton
for n in range(len(subset)):
# for each joint
for i in range(18):
cidx = subset[n][i]
if cidx != -1:
y = candidate[cidx.astype(int), 0]
x = candidate[cidx.astype(int), 1]
joints_per_skeleton[n].append((y, x))
else:
joints_per_skeleton[n].append(None)
return joints_per_skeleton
joints = get_joints(subset,candidate)
print joints
Here is the output of the joints list of list
[[None, (48.0, 52.0), (72.0, 50.0), None, None, (24.0, 55.0), (5.0, 105.0), None, (63.0, 159.0), (57.0, 221.0), (55.0, 281.0), (28.0, 154.0), (23.0, 219.0), (23.0, 285.0), None, (25.0, 17.0), (55.0, 18.0), (30.0, 21.0)]]
Here I defined a function to create the dictionary from the 2 lists
def create_dict(keys, values):
return dict(zip(keys, values))
my_dict = create_dict(jointsName, joints[0])
Here is the result:
{'LAnkle': (23.0, 285.0),
'LEar': (30.0, 21.0),
'LElbow': (5.0, 105.0),
'LEye': (25.0, 17.0),
'LHip': (28.0, 154.0),
'LKnee': (23.0, 219.0),
'LShoulder': (24.0, 55.0),
'LWrist': None,
'Neck': (48.0, 52.0),
'Nose': None,
'RAnkle': (55.0, 281.0),
'REar': (55.0, 18.0),
'RElbow': None,
'REye': None,
'RHip': (63.0, 159.0),
'RKnee': (57.0, 221.0),
'RShoulder': (72.0, 50.0),
'RWrist': None}
I think defaultdict could help you. I made my own example to show that you could predefine the keys and then go through a double for loop and have the values of the dict be lists of potentially different sizes. Please let me know if this answers your question:
from collections import defaultdict
import random
joint_names = ['hip','knee','wrist']
num_skeletons = 10
d = defaultdict(list)
for skeleton in range(num_skeletons):
for joint_name in joint_names:
r1 = random.randint(0,10)
r2 = random.randint(0,10)
if r1 > 4:
d[joint_name].append(r1*r2)
print d
Output:
defaultdict(<type 'list'>, {'hip': [0, 5, 30, 36, 56], 'knee': [35, 50, 10], 'wrist': [27, 5, 15, 64, 30]})
As a note I found it very difficult to read through your code since there were some variables that were defined before the snippet you posted.
I currently have a list of lists that looks like this:
My_List = [[This, Is, A, Sample, Text, Sentence] [This, too, is, a, sample, text] [finally, so, is, this, one]]
Now what I need to do is "tag" each of these words with one of 3, in this case arbitrary, tags such as "EE", "FF", or "GG" based on which list the word is in and then reassemble them into the same order they came in. My final code would need to look like:
GG_List = [This, Sentence]
FF_List = [Is, A, Text]
EE_List = [Sample]
My_List = [[(This, GG), (Is, FF), (A, FF), (Sample, "EE), (Text, FF), (Sentence, GG)] [*same with this sentence*] [*and this one*]]
I tried this by using for loops to turn each item into a dict but the dicts then got rearranged by their tags which sadly can't happen because of the nature of this thing... the experiment needs everything to stay in the same order because eventually I need to measure the proximity of tags relative to others but only in the same sentence (list).
I thought about doing this with NLTK (which I have little experience with) but it looks like that is much more sophisticated then what I need and the tags aren't easily customized by a novice like myself.
I think this could be done by iterating through each of these items, using an if statement as I have to determine what tag they should have, and then making a tuple out of the word and its associated tag so it doesn't shift around within its list.
I've devised this.. but I can't figure out how to rebuild my list-of-lists and keep them in order :(.
for i in My_List: #For each list in the list of lists
for h in i: #For each item in each list
if h in GG_List: # Check for the tag
MyDicts = {"GG":h for h in i} #Make Dict from tag + word
Thank you so much for your help!
Putting the tags in a dictionary would work:
My_List = [['This', 'Is', 'A', 'Sample', 'Text', 'Sentence'],
['This', 'too', 'is', 'a', 'sample', 'text'],
['finally', 'so', 'is', 'this', 'one']]
GG_List = ['This', 'Sentence']
FF_List = ['Is', 'A', 'Text']
EE_List = ['Sample']
zipped = zip((GG_List, FF_List, EE_List), ('GG', 'FF', 'EE'))
tags = {item: tag for tag_list, tag in zipped for item in tag_list}
res = [[(word, tags[word]) for word in entry if word in tags] for entry in My_List]
Now:
>>> res
[[('This', 'GG'),
('Is', 'FF'),
('A', 'FF'),
('Sample', 'EE'),
('Text', 'FF'),
('Sentence', 'GG')],
[('This', 'GG')],
[]]
Dictionary works by key-value pairs. Each key is assigned a value. To search the dictionary, you search the index by the key, e.g.
>>> d = {1:'a', 2:'b', 3:'c'}
>>> d[1]
'a'
In the above case, we always search the dictionary by its keys, i.e. the integers.
In the case that you want to assign the tag/label to each word, you are searching by the key word and finding the "value", i.e. the tag/label, so your dictionary would have to look something like this (assuming that the strings are words and numbers as tag/label):
>>> d = {'a':1, 'b':1, 'c':3}
>>> d['a']
1
>>> sent = 'a b c a b'.split()
>>> sent
['a', 'b', 'c', 'a', 'b']
>>> [d[word] for word in sent]
[1, 1, 3, 1, 1]
This way the order of the tags follows the order of the words when you use a list comprehension to iterate through the words and find the appropriate tags.
So the problem comes when you have the initial dictionary indexed with the wrong way, i.e. key -> labels, value -> words, e.g.:
>>> d = {1:['a', 'd'], 2:['b', 'h'], 3:['c', 'x']}
>>> [d[word] for word in sent]
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
KeyError: 'a'
Then you would have to reverse your dictionary, assuming that all elements in your value lists are unique, you can do this:
>>> from collections import ChainMap
>>> d = {1:['a', 'd'], 2:['b', 'h'], 3:['c', 'x']}
>>> d_inv = dict(ChainMap(*[{value:key for value in values} for key, values in d.items()]))
>>> d_inv
{'h': 2, 'c': 3, 'a': 1, 'x': 3, 'b': 2, 'd': 1}
But the caveat is that ChainMap is only available in Python3.5 (yet another reason to upgrade your Python ;P). For Python <3.5, solutions, see How do I merge a list of dicts into a single dict?.
So going back to the problem of assigning labels/tags to words, let's say we have these input:
>>> d = {1:['a', 'd'], 2:['b', 'h'], 3:['c', 'x']}
>>> sent = 'a b c a b'.split()
First, we invert the dictionary (assuming that there're one to one mapping for every word and its tag/label:
>>> d_inv = dict(ChainMap(*[{value:key for value in values} for key, values in d.items()]))
Then, we apply the tags to the words through a list comprehension:
>>> [d_inv[word] for word in sent]
[1, 2, 3, 1, 2]
And for multiple sentences:
>>> sentences = ['a b c'.split(), 'h a x'.split()]
>>> [[d_inv[word] for word in sent] for sent in sentences]
[[1, 2, 3], [2, 1, 3]]
I have a list - let's call it 'catalog' - of object properties. Different objects can have a different number of properties; for example items 1 and 2 have "colour" and "sizeid", but item 3 has an additional "onsale" property.
[{'item_id': 1, 'property_name': 'colour', 'property_value': 'blk'},
{'item_id': 1, 'property_name': 'sizeid', 'property_value': '1'},
{'item_id': 2, 'property_name': 'colour', 'property_value': 'grn'},
{'item_id': 2, 'property_name': 'sizeid', 'property_value': '2'},
{'item_id': 3, 'property_name': 'colour', 'property_value': 'blue'},
{'item_id': 3, 'property_name': 'sizeid', 'property_value': '2'},
{'item_id': 3, 'property_name': 'onsale', 'property_value': True}]
I would like to be able to access property_value based on item_id and property_name, like so:
>>> print catalog[1]['sizeid']
>>> '1'
>>> print catalog[3]['onsale']
>>> True
>>> print catalog[2]['onsale']
>>> KeyError: 'onsale'
I had a cunning plan to do this procedurally with something like
for i, val in enumerate(catalog):
tidy_catalog[val['item_id']][val['property_name']] = val['property_value']
But this hasn't worked - probably for many reasons, but I think primarily because dictionaries won't allow duplicate values.
Appreciate any suggestions, or even helpful man pages - I am stuck in PHP ways of thinking and while I think I get Python data structures at a conceptual level, using them in practice is proving a bit challenging.
The reason this hasn't worked is assigning values to keys which do not exist.
To get around this probem, you can use a defaultdict:
>>> tidy_catalog = collections.defaultdict(dict)
>>> for val in catalog:
... tidy_catalog[val['item_id']][val['property_name']] = val['property_value']
...
>>> pp(tidy_catalog)
defaultdict(<type 'dict'>, {
1: {'colour': 'blk', 'sizeid': '1'},
2: {'colour': 'grn', 'sizeid': '2'},
3: {'colour': 'blue', 'onsale': True, 'sizeid': '2'},
})
Consider adjusting your catalog array to array of proper dictionaries, where you store these informations:
catalog = [{'item_d' : 1, 'colour' : 'blk', 'sizeid' : '1'}, ...]
I want to map a function that takes a dictionary as the input and returns a list of the keys.
The keys in the list must be of only the unique values present in the dictionary.
So, this is what I have done.
bDict={}
for key,value in aDict.items():
if bDict.has_key(value) == False:
bDict[value]=key
else:
bDict.pop(value,None)
This is the output :
>>> aDict.keys()
Out[4]: [1, 3, 6, 7, 8, 10]
>>> aDict.values()
Out[5]: [1, 2, 0, 0, 4, 0]
>>> bDict.keys()
Out[6]: [0, 1, 2, 4]
>>> bDict.values()
Out[7]: [10, 1, 3, 8]
But, the expected output should be for bDict.values() : [*1,3,8*]
This may help.
CODE
aDict = { 1:1, 3:2, 6:0, 7:0, 8:4, 10:0, 11:0}
bDict = {}
for i,j in aDict.items():
if j not in bDict:
bDict[j] = [i]
else:
bDict[j].append(i)
print map(lambda x: x[0],filter(lambda x: len(x) == 1,bDict.values()))
OUTPUT
[1, 3, 8]
So it appears you're creating a new dictionary with the keys and values inverted, keeping pairs where the value is unique. You can figure out which of the items are unique first then build a dictionary off of that.
def distinct_values(d):
from collections import Counter
counts = Counter(d.itervalues())
return { v: k for k, v in d.iteritems() if counts[v] == 1 }
This yields the following result:
>>> distinct_values({ 1:1, 3:2, 6:0, 7:0, 8:4, 10:0 })
{1: 1, 2: 3, 4: 8}
Here is a solution (with two versions of the aDict to test a rand case which failed in another solution):
#aDict = { 1:1, 3:2, 6:0, 7:0, 8:4, 10:0}
aDict = { 1:1, 3:2, 6:0, 7:0, 8:4, 10:0, 11:2}
seenValues = {}
uniqueKeys = set()
for aKey, aValue in aDict.items():
if aValue not in seenValues:
# Store the key of the value, and assume it is unique
seenValues[aValue] = aKey
uniqueKeys.add(aKey)
elif seenValues[aValue] in uniqueKeys:
# The value has been seen before, and the assumption of
# it being unique was wrong, so remove it
uniqueKeys.remove(seenValues[aValue])
print "Remove non-unique key/value pair: {%d, %d}" % (aKey, aValue)
else:
print "Non-unique key/value pair: {%d, %d}" % (aKey, aValue)
print "Unique keys: ", sorted(uniqueKeys)
And this produces the output:
Remove non-unique key/value pair: {7, 0}
Non-unique key/value pair: {10, 0}
Remove non-unique key/value pair: {11, 2}
Unique keys: [1, 8]
Or with original version of aDict:
Remove non-unique key/value pair: {7, 0}
Non-unique key/value pair: {10, 0}
Unique keys: [1, 3, 8]
As a python 2.7 one-liner,
[k for k,v in aDict.iteritems() if aDict.values().count(v) == 1]
Note that the above
Calls aDict.values() many times, once for each entry in the dictionary, and
Calls aDict.values().count(v) multiple times for each replicated value.
This is not a problem if the dictionary is small. If the dictionary isn't small, the creation and destruction of those duplicative lists and the duplicative calls to count() may be costly. It may help to cache the value of adict.values(), and it may also help to create a dictionary that maps the values in the dictionary to the number of occurrences as a dictionary entry value.
Say I have a dict of country -> [cities] (potentially an ordered dict):
{'UK': ['Bristol', 'Manchester' 'London', 'Glasgow'],
'France': ['Paris', 'Calais', 'Nice', 'Cannes'],
'Germany': ['Munich', 'Berlin', 'Cologne']
}
The number of keys (countries) is variable: and the number of elements cities in the array, also variable. The resultset comes from a 'search' on city name so, for example, a search on "San%" could potentially meet with 50k results (on a worldwide search)
The data is to be used to populate a select2 widget --- and I'd like to use its paging functionality...
Is there a smart way to slice this such that [3:8] would yield:
{'UK': ['Glasgow'],
'France': ['Paris', 'Calais', 'Nice', 'Cannes'],
'Germany': ['Munich']
}
(apologies for the way this question was posed earlier -- I wasn't sure that the real usage would clarify the issue...)
If I understand your problem correctly, as talked about in the comments, this should do it
from pprint import pprint
def slice_dict(d,a, b):
big_list = []
ret_dict = {}
# Make one big list of all numbers, tagging each number with the key
# of the dict they came from.
for k, v in d.iteritems():
for n in v:
big_list.append({k:n})
# Slice it
sliced = big_list[a:b]
# Put everything back in order
for k, v in d.iteritems():
for subd in sliced:
for subk, subv in subd.iteritems():
if k == subk:
if k in ret_dict:
ret_dict[k].append(subv)
else:
ret_dict[k] = [subv]
return ret_dict
d = {
'a': [1, 2, 3, 4],
'b': [5, 6, 7, 8, 9],
'c': [10, 11, 12, 13, 14]
}
x = slice_dict(d, 3, 11)
pprint(x)
$ python slice.py
{'a': [4], 'b': [5, 6], 'c': [10, 11, 12, 13, 14]}
The output is a little different from your example output, but that's because the dict was not ordered when it was passed to the function. It was a-c-b, that's why b is cut off at 6 and c is not cut off