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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)]
I have a list of integers which I need to parse into a string of ranges.
For example:
[0, 1, 2, 3] -> "0-3"
[0, 1, 2, 4, 8] -> "0-2,4,8"
And so on.
I'm still learning more pythonic ways of handling lists, and this one is a bit difficult for me. My latest thought was to create a list of lists which keeps track of paired numbers:
[ [0, 3], [4, 4], [5, 9], [20, 20] ]
I could then iterate across this structure, printing each sub-list as either a range, or a single value.
I don't like doing this in two iterations, but I can't seem to keep track of each number within each iteration. My thought would be to do something like this:
Here's my most recent attempt. It works, but I'm not fully satisfied; I keep thinking there's a more elegant solution which completely escapes me. The string-handling iteration isn't the nicest, I know -- it's pretty early in the morning for me :)
def createRangeString(zones):
rangeIdx = 0
ranges = [[zones[0], zones[0]]]
for zone in list(zones):
if ranges[rangeIdx][1] in (zone, zone-1):
ranges[rangeIdx][1] = zone
else:
ranges.append([zone, zone])
rangeIdx += 1
rangeStr = ""
for range in ranges:
if range[0] != range[1]:
rangeStr = "%s,%d-%d" % (rangeStr, range[0], range[1])
else:
rangeStr = "%s,%d" % (rangeStr, range[0])
return rangeStr[1:]
Is there a straightforward way I can merge this into a single iteration? What else could I do to make it more Pythonic?
>>> from itertools import count, groupby
>>> L=[1, 2, 3, 4, 6, 7, 8, 9, 12, 13, 19, 20, 22, 23, 40, 44]
>>> G=(list(x) for _,x in groupby(L, lambda x,c=count(): next(c)-x))
>>> print ",".join("-".join(map(str,(g[0],g[-1])[:len(g)])) for g in G)
1-4,6-9,12-13,19-20,22-23,40,44
The idea here is to pair each element with count(). Then the difference between the value and count() is constant for consecutive values. groupby() does the rest of the work
As Jeff suggests, an alternative to count() is to use enumerate(). This adds some extra cruft that needs to be stripped out in the print statement
G=(list(x) for _,x in groupby(enumerate(L), lambda (i,x):i-x))
print ",".join("-".join(map(str,(g[0][1],g[-1][1])[:len(g)])) for g in G)
Update: for the sample list given here, the version with enumerate runs about 5% slower than the version using count() on my computer
Whether this is pythonic is up for debate. But it is very compact. The real meat is in the Rangify() function. There's still room for improvement if you want efficiency or Pythonism.
def CreateRangeString(zones):
#assuming sorted and distinct
deltas = [a-b for a, b in zip(zones[1:], zones[:-1])]
deltas.append(-1)
def Rangify((b, p), (z, d)):
if p is not None:
if d == 1: return (b, p)
b.append('%d-%d'%(p,z))
return (b, None)
else:
if d == 1: return (b, z)
b.append(str(z))
return (b, None)
return ','.join(reduce(Rangify, zip(zones, deltas), ([], None))[0])
To describe the parameters:
deltas is the distance to the next value (inspired from an answer here on SO)
Rangify() does the reduction on these parameters
b - base or accumulator
p - previous start range
z - zone number
d - delta
To concatenate strings you should use ','.join. This removes the 2nd loop.
def createRangeString(zones):
rangeIdx = 0
ranges = [[zones[0], zones[0]]]
for zone in list(zones):
if ranges[rangeIdx][1] in (zone, zone-1):
ranges[rangeIdx][1] = zone
else:
ranges.append([zone, zone])
rangeIdx += 1
return ','.join(
map(
lambda p: '%s-%s'%tuple(p) if p[0] != p[1] else str(p[0]),
ranges
)
)
Although I prefer a more generic approach:
from itertools import groupby
# auxiliary functor to allow groupby to compare by adjacent elements.
class cmp_to_groupby_key(object):
def __init__(self, f):
self.f = f
self.uninitialized = True
def __call__(self, newv):
if self.uninitialized or not self.f(self.oldv, newv):
self.curkey = newv
self.uninitialized = False
self.oldv = newv
return self.curkey
# returns the first and last element of an iterable with O(1) memory.
def first_and_last(iterable):
first = next(iterable)
last = first
for i in iterable:
last = i
return (first, last)
# convert groups into list of range strings
def create_range_string_from_groups(groups):
for _, g in groups:
first, last = first_and_last(g)
if first != last:
yield "{0}-{1}".format(first, last)
else:
yield str(first)
def create_range_string(zones):
groups = groupby(zones, cmp_to_groupby_key(lambda a,b: b-a<=1))
return ','.join(create_range_string_from_groups(groups))
assert create_range_string([0,1,2,3]) == '0-3'
assert create_range_string([0, 1, 2, 4, 8]) == '0-2,4,8'
assert create_range_string([1,2,3,4,6,7,8,9,12,13,19,20,22,22,22,23,40,44]) == '1-4,6-9,12-13,19-20,22-23,40,44'
This is more verbose, mainly because I have used generic functions that I have and that are minor variations of itertools functions and recipes:
from itertools import tee, izip_longest
def pairwise_longest(iterable):
"variation of pairwise in http://docs.python.org/library/itertools.html#recipes"
a, b = tee(iterable)
next(b, None)
return izip_longest(a, b)
def takeuntil(predicate, iterable):
"""returns all elements before and including the one for which the predicate is true
variation of http://docs.python.org/library/itertools.html#itertools.takewhile"""
for x in iterable:
yield x
if predicate(x):
break
def get_range(it):
"gets a range from a pairwise iterator"
rng = list(takeuntil(lambda (a,b): (b is None) or (b-a>1), it))
if rng:
b, e = rng[0][0], rng[-1][0]
return "%d-%d" % (b,e) if b != e else "%d" % b
def create_ranges(zones):
it = pairwise_longest(zones)
return ",".join(iter(lambda:get_range(it),None))
k=[0,1,2,4,5,7,9,12,13,14,15]
print create_ranges(k) #0-2,4-5,7,9,12-15
def createRangeString(zones):
"""Create a string with integer ranges in the format of '%d-%d'
>>> createRangeString([0, 1, 2, 4, 8])
"0-2,4,8"
>>> createRangeString([1,2,3,4,6,7,8,9,12,13,19,20,22,22,22,23,40,44])
"1-4,6-9,12-13,19-20,22-23,40,44"
"""
buffer = []
try:
st = ed = zones[0]
for i in zones[1:]:
delta = i - ed
if delta == 1: ed = i
elif not (delta == 0):
buffer.append((st, ed))
st = ed = i
else: buffer.append((st, ed))
except IndexError:
pass
return ','.join(
"%d" % st if st==ed else "%d-%d" % (st, ed)
for st, ed in buffer)
Here is my solution. You need to keep track of various pieces of information while you iterate through the list and create the result - this screams generator to me. So here goes:
def rangeStr(start, end):
'''convert two integers into a range start-end, or a single value if they are the same'''
return str(start) if start == end else "%s-%s" %(start, end)
def makeRange(seq):
'''take a sequence of ints and return a sequence
of strings with the ranges
'''
# make sure that seq is an iterator
seq = iter(seq)
start = seq.next()
current = start
for val in seq:
current += 1
if val != current:
yield rangeStr(start, current-1)
start = current = val
# make sure the last range is included in the output
yield rangeStr(start, current)
def stringifyRanges(seq):
return ','.join(makeRange(seq))
>>> l = [1,2,3, 7,8,9, 11, 20,21,22,23]
>>> l2 = [1,2,3, 7,8,9, 11, 20,21,22,23, 30]
>>> stringifyRanges(l)
'1-3,7-9,11,20-23'
>>> stringifyRanges(l2)
'1-3,7-9,11,20-23,30'
My version will work correctly if given an empty list, which I think some of the others will not.
>>> stringifyRanges( [] )
''
makeRanges will work on any iterator that returns integers and lazily returns a sequence of strings so can be used on infinite sequences.
edit: I have updated the code to handle single numbers that are not part of a range.
edit2: refactored out rangeStr to remove duplication.
how about this mess...
def rangefy(mylist):
mylist, mystr, start = mylist + [None], "", 0
for i, v in enumerate(mylist[:-1]):
if mylist[i+1] != v + 1:
mystr += ["%d,"%v,"%d-%d,"%(start,v)][start!=v]
start = mylist[i+1]
return mystr[:-1]