I am new to python and my coding experience so far is with MATLAB.
I am trying to understand more about lists and dictionaries as i am using a library about DOEs that takes an dictionary as a passing argument.
But my trouble so far is that this dictionary assumes the form of ex.
DOE={'Elastic Modulus':[10,20,30], 'Density':[1,2,3], 'Thickness':[2,3,5]}
But i need this dictionary to be user defined, for example:
Have an input to define how many variables are needed (in this example are 3: Elastic Modulus','Density'and 'Thickness)
as the variables are defined, it should be able to store values in the dictionary over a for loop.
Is this possible using dictionaries?
Or is it better to use a list and convert in a dicionary later?
Thank you in advance
One can add keys and the corresponding values to a dict one at a time like so:
my_dict = {}
num_entries = int(input("How many entries "))
for _ in range(num_entries):
key = input("Enter the key: ")
value = input("Enter the value: ")
my_dict[key] = value
Presumably you would have a loop to do the entry of key and value for the number of values you wish to enter. Also if you are in python 2 it needs to be raw_input rather than input function. [Edit: Showing how to do the loop, since I noticed that was part of your question]
I have function that returns just one list of key-value pair. How can I convert this to an actual key value or an object type so I can get each attribute from the list. For example I would like to be able to just get the time or price or any other property and not the whole list as one item.
{'time': 1512858529643, 'price': '0.00524096', 'origQty': '530.00000000'
I know it doesn't look like a list but it actually is! The function that I am calling returns this as a list. I am simply storing it to a variable and nothign else.
open_order=client.get_open_orders(symbol="BNBETH",recvWindow=1234567)
If you still have doubts. When I try to print a dictionary item like this print(open_order['time'])
I get the following error.
Traceback (most recent call last):
File "C:\Python27\python-binance-master\main.py", line 63, in <module>
print(open_order['time'])
TypeError: list indices must be integers, not str
Also If I show type it shows as list.
print(type(open_order))
So, I was able to come up with a solution, sort of... by converting the list to string and splitting at the "," character. Now I have list of items that I can actually print by selecting one print(split_order_items[5]) There has to be a better solution.
open_order=client.get_open_orders(symbol="BNBETH",recvWindow=1234567)
y=''.join(str(e)for e in open_order)
split_order_items =([x.strip() for x in y.split(',')])
print(split_order_items[5])
I was able to create a multiple list items using the above code. I just can't seem to convert it to dictionary object!
Thanks!
What you have posted is a dict, not a list. You can do something like this:
data = {'time': 1512858529643, 'price': '0.00524096', 'orderId': 7848174, 'origQty': '530.00000000'}
print(data['time']) # this gets just the time and prints it
print(data['price']) # this gets just the price and prints it
I strongly suggest reading up on the Python dict: https://docs.python.org/3/tutorial/datastructures.html#dictionaries
I have a dictionary that I declared in a particular order and want to keep it in that order all the time. The keys/values can't really be kept in order based on their value, I just want it in the order that I declared it.
So if I have the dictionary:
d = {'ac': 33, 'gw': 20, 'ap': 102, 'za': 321, 'bs': 10}
It isn't in that order if I view it or iterate through it. Is there any way to make sure Python will keep the explicit order that I declared the keys/values in?
From Python 3.6 onwards, the standard dict type maintains insertion order by default.
Defining
d = {'ac':33, 'gw':20, 'ap':102, 'za':321, 'bs':10}
will result in a dictionary with the keys in the order listed in the source code.
This was achieved by using a simple array with integers for the sparse hash table, where those integers index into another array that stores the key-value pairs (plus the calculated hash). That latter array just happens to store the items in insertion order, and the whole combination actually uses less memory than the implementation used in Python 3.5 and before. See the original idea post by Raymond Hettinger for details.
In 3.6 this was still considered an implementation detail; see the What's New in Python 3.6 documentation:
The order-preserving aspect of this new implementation is considered an implementation detail and should not be relied upon (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).
Python 3.7 elevates this implementation detail to a language specification, so it is now mandatory that dict preserves order in all Python implementations compatible with that version or newer. See the pronouncement by the BDFL. As of Python 3.8, dictionaries also support iteration in reverse.
You may still want to use the collections.OrderedDict() class in certain cases, as it offers some additional functionality on top of the standard dict type. Such as as being reversible (this extends to the view objects), and supporting reordering (via the move_to_end() method).
from collections import OrderedDict
OrderedDict((word, True) for word in words)
contains
OrderedDict([('He', True), ('will', True), ('be', True), ('the', True), ('winner', True)])
If the values are True (or any other immutable object), you can also use:
OrderedDict.fromkeys(words, True)
Rather than explaining the theoretical part, I'll give a simple example.
>>> from collections import OrderedDict
>>> my_dictionary=OrderedDict()
>>> my_dictionary['foo']=3
>>> my_dictionary['aol']=1
>>> my_dictionary
OrderedDict([('foo', 3), ('aol', 1)])
>>> dict(my_dictionary)
{'foo': 3, 'aol': 1}
Note that this answer applies to python versions prior to python3.7. CPython 3.6 maintains insertion order under most circumstances as an implementation detail. Starting from Python3.7 onward, it has been declared that implementations MUST maintain insertion order to be compliant.
python dictionaries are unordered. If you want an ordered dictionary, try collections.OrderedDict.
Note that OrderedDict was introduced into the standard library in python 2.7. If you have an older version of python, you can find recipes for ordered dictionaries on ActiveState.
Dictionaries will use an order that makes searching efficient, and you cant change that,
You could just use a list of objects (a 2 element tuple in a simple case, or even a class), and append items to the end. You can then use linear search to find items in it.
Alternatively you could create or use a different data structure created with the intention of maintaining order.
I came across this post while trying to figure out how to get OrderedDict to work. PyDev for Eclipse couldn't find OrderedDict at all, so I ended up deciding to make a tuple of my dictionary's key values as I would like them to be ordered. When I needed to output my list, I just iterated through the tuple's values and plugged the iterated 'key' from the tuple into the dictionary to retrieve my values in the order I needed them.
example:
test_dict = dict( val1 = "hi", val2 = "bye", val3 = "huh?", val4 = "what....")
test_tuple = ( 'val1', 'val2', 'val3', 'val4')
for key in test_tuple: print(test_dict[key])
It's a tad cumbersome, but I'm pressed for time and it's the workaround I came up with.
note: the list of lists approach that somebody else suggested does not really make sense to me, because lists are ordered and indexed (and are also a different structure than dictionaries).
You can't really do what you want with a dictionary. You already have the dictionary d = {'ac':33, 'gw':20, 'ap':102, 'za':321, 'bs':10}created. I found there was no way to keep in order once it is already created. What I did was make a json file instead with the object:
{"ac":33,"gw":20,"ap":102,"za":321,"bs":10}
I used:
r = json.load(open('file.json'), object_pairs_hook=OrderedDict)
then used:
print json.dumps(r)
to verify.
from collections import OrderedDict
list1 = ['k1', 'k2']
list2 = ['v1', 'v2']
new_ordered_dict = OrderedDict(zip(list1, list2))
print new_ordered_dict
# OrderedDict([('k1', 'v1'), ('k2', 'v2')])
Another alternative is to use Pandas dataframe as it guarantees the order and the index locations of the items in a dict-like structure.
I had a similar problem when developing a Django project. I couldn't use OrderedDict, because I was running an old version of python, so the solution was to use Django's SortedDict class:
https://code.djangoproject.com/wiki/SortedDict
e.g.,
from django.utils.datastructures import SortedDict
d2 = SortedDict()
d2['b'] = 1
d2['a'] = 2
d2['c'] = 3
Note: This answer is originally from 2011. If you have access to Python version 2.7 or higher, then you should have access to the now standard collections.OrderedDict, of which many examples have been provided by others in this thread.
Generally, you can design a class that behaves like a dictionary, mainly be implementing the methods __contains__, __getitem__, __delitem__, __setitem__ and some more. That class can have any behaviour you like, for example prividing a sorted iterator over the keys ...
if you would like to have a dictionary in a specific order, you can also create a list of lists, where the first item will be the key, and the second item will be the value
and will look like this
example
>>> list =[[1,2],[2,3]]
>>> for i in list:
... print i[0]
... print i[1]
1
2
2
3
You can do the same thing which i did for dictionary.
Create a list and empty dictionary:
dictionary_items = {}
fields = [['Name', 'Himanshu Kanojiya'], ['email id', 'hima#gmail.com']]
l = fields[0][0]
m = fields[0][1]
n = fields[1][0]
q = fields[1][1]
dictionary_items[l] = m
dictionary_items[n] = q
print dictionary_items
I have a file in the format:
0000 | a1_1,a3_2 | b2_1, b3_2
0001 | a1_3 | b4_1
and I'm trying to create a dictionary which has
{ 'a1' : set(['b2', 'b3', 'b4']), 'a3': set(['b2', 'b3']) }
and this is how my code looks like:
def get_ids(row, col):
ids = set()
x = row.strip().split('|')
for a in x[col].split(','):
ids.add(a.split('_')[0])
return ids
def add_to_dictionary(funky_dictionary,key, values):
if key in funky_dictionary:
funky_dictionary[key].update(values)
else:
funky_dictionary[key] = values
def get_dict(input_file):
funky_dictionary = {}
with open(input_file,'r') as ip:
for row in ip:
a_ids = get_ids(row,1)
b_ids = get_ids(row,2)
for key in a_ids:
add_to_dictionary(funky_dictionary,key,b_ids)
return funky_dictionary
So my problem is this when I lookup values for certain key in the dictionary, it returns me with way more values than expected. E.g.
For the above example the expected value of a3 would be set(['b2', ' b3'])
However with the code, I'm getting set(['b2', ' b3', 'b4'])
I cant figure out whats wrong with the code. Any help?
The issue you have is that many of your dictionary's values are in fact references to the same set instances. In your example data, when the first line is processed, 'a1' and 'a3' both get mapped to the same set object (containing 'b2' and 'b3'). When you process the second line and call update on that set via the key 'a1', you'll see the added value through 'a3' too, since both values are references to the same set.
You need to change the code so that each value is a separate set object. I'd suggest getting rid of add_to_dictionary and just using the dictionary's own setdefault method, like this:
for key in a_ids:
funky_dictionary.setdefault(key, set()).update(b_ids)
This code always starts with a new empty set for a new key, and always updates it with new values (rather than adding a reference to the b_ids set to the dictionary directly).
How can I create a series of variables automatically with python?
Like this:
tmp1=1;
tmp2=1;
tmp3=1;
tmp4=1;
tmp5=1；
Look at this SO question. you have several ways to do that:
Use dict or collection instead
Using globals
Using the exec() method
About the first solution (dict or collection) - is not actually what you asked for, which is to create a variable in the global scope.. but I would go with that anytime. I don't see really any reason why I'd need to create variables dynamically instead of using some datatype.
I would say that using both globals and exec() method for this is a bad practice.
Store them in a dictionary
d = {}
value = ['a','b','c']
for key in range(1, 3):
d[key]= value[key]
print d
> {0:'a',1:'b',2:'c'}
print d[0]
> 'a'
(Comments? I am new to python too!)
I think I have got an answer somewhere:
for i in range(100):
locals()['tmp%d'%i]=1
or:
>>> for i in range(1, 10):
... exec 'tmp' + str(i) + '=1'
I don't know if I have a ambiguity describe, the above two is exactly I want.