create a sub-dict with first element in dict for every key-CodePudding

I have a dictionary containing different lists, I would need a fast & efective way to create a new dictionary containing same keys (better if I can be independent of key names) and one value of the list let's say the i-th

one example

 someDict = {'key1': [1,2,3,4,5],'otherkey': [.256,.221,.487,.454,.555]}

then the first elements (i=0) should return

subdict = {'key1':1,'otherkey':0.256}

the third element would be:

subdict = {'key1':3,'otherkey':0.487}

I could do some for key,val in someDict... to create new dict but i'm not sure it's the best option

CodePudding user response：

Depending on what you need the new dict objects for, a pandas dataframe may be the best choice. It will generally be more performant for table-like operations. For instance, if your plan is to put those dicts in a list, the datatrame could be a good fit.

>>> import pandas as pd
>>> someDict = {'key1': [1,2,3,4,5],'otherkey': [.256,.221,.487,.454,.555]} 
>>> df = pd.DataFrame(someDict)
>>> df
   key1  otherkey
0     1     0.256
1     2     0.221
2     3     0.487
3     4     0.454
4     5     0.555

CodePudding user response：

Try this:

def ith_val_subdict(input_dict, i):
    return {k: v[i] for k, v in input_dict.items()}

Example:

someDict = {'key1': [1,2,3,4,5],'otherkey': [.256,.221,.487,.454,.555]}

subdict_0 = ith_val_subdict(someDict, 0)
print(subdict_0)
# {'key1': 1, 'otherkey': 0.256}

subdict_2 = ith_val_subdict(someDict, 2)
print(subdict_2)
# {'key1': 3, 'otherkey': 0.487}

CodePudding user response：

You can use two zip in a list comprehension:

[dict(zip(someDict, vals)) for vals in zip(*someDict.values())]

output:

[{'key1': 1, 'otherkey': 0.256},
 {'key1': 2, 'otherkey': 0.221},
 {'key1': 3, 'otherkey': 0.487},
 {'key1': 4, 'otherkey': 0.454},
 {'key1': 5, 'otherkey': 0.555}]

NB. this will truncate the values to the length of the shortest sublist, if there is an uneven length and padding is needed, use itertools.zip_longest