Pythonic将两个不同长度的列表压缩成字典

Question

I have two lists of strings with different lengths:

k = ['a', 'b', 'c', 'd', 'e', 'f', 'g']
v = ['z', 'y', 'x', 'w',]
dict = {}

I want to have the following output:

{'a': ['z', 'y', 'x', 'w',], 'b': ['z', 'y', 'x', 'w',], 'c': ['z', 'y', 'x', 'w',], 'd': ['z', 'y', 'x', 'w',], 'e': ['z', 'y', 'x', 'w',], 'f': ['z', 'y', 'x', 'w',], 'g': ['z', 'y', 'x', 'w',]}    

The closest I have gotten is this:

{key: [key, value] for key, value in zip(k, v)}

and

(kvalue,vvalue) for kkey, kvalue in enumerate(k) for vkey, vvalue in enumerate(v)

The ultimate goal with this is to use the key, value pairs to insert the contained strings into a string in a separate for loop. I'm using Python 3.6.

Pseudocode:

for filenames in directory:
    var = 'some string' + each key/value pairing
    print(var)

>>> some string a z
>>> some string a y
>>> some string a x
>>> some string a w
>>> some string b z
>>> some string b y
>>> some string b x
>>> some string b w
>>> some string c z
>>> some string c y
>>> some string c x
>>> some string c w
>>> some string d z
>>> some string d y
>>> some string d x
>>> some string d w
>>> some string e z
>>> some string e y
>>> some string e x
>>> some string e w
>>> some string f z
>>> some string f y
>>> some string f x
>>> some string f w
>>> some string g z
>>> some string g y
>>> some string g x
>>> some string g w

Answer 1

For the first part of your question, use dict.fromkeys() :

k = ['a', 'b', 'c', 'd', 'e', 'f', 'g']
v = ['z', 'y', 'x', 'w',]
d = dict.fromkeys(k, v)

output:

{'a': ['z', 'y', 'x', 'w'],
 'b': ['z', 'y', 'x', 'w'],
 'c': ['z', 'y', 'x', 'w'],
 'd': ['z', 'y', 'x', 'w'],
 'e': ['z', 'y', 'x', 'w'],
 'f': ['z', 'y', 'x', 'w'],
 'g': ['z', 'y', 'x', 'w']}

For the second part, you actually don't need the first, use itertools.product :

from itertools import product

k = ['a', 'b', 'c', 'd', 'e', 'f', 'g']
v = ['z', 'y', 'x', 'w',]

out = ['some string {} {}'.format(key, val) for key, val in product(k, v)]

output:

['some string a z',
 'some string a y',
 'some string a x',
 'some string a w',
 'some string b z',
 'some string b y',
 'some string b x',
 'some string b w',
 'some string c z',
 'some string c y',
 'some string c x',
 'some string c w',
 'some string d z',
 'some string d y',
 'some string d x',
 'some string d w',
...

Answer 2

You can use a dictionary-comprehension:

result = {x: v for x in k}
print(result)
{'a': ['z', 'y', 'x', 'w'], 'b': ['z', 'y', 'x', 'w'], 'c': ['z', 'y', 'x', 'w'], 'd': ['z', 'y', 'x', 'w'], 'e': ['z', 'y', 'x', 'w'], 'f': ['z', 'y', 'x', 'w'], 'g': ['z', 'y', 'x', 'w']}

This creates a dictionary with keys x from k and values v .

Answer 3

A dictionary comprehension would do:

k = ['a', 'b', 'c', 'd', 'e', 'f', 'g']
v = ['z', 'y', 'x', 'w',]
print({key: v for key in k})

# {'a': ['z', 'y', 'x', 'w',], 'b': ['z', 'y', 'x', 'w',], 'c': ['z', 'y', 'x', 'w',], 'd': ['z', 'y', 'x', 'w',], 'e': ['z', 'y', 'x', 'w',], 'f': ['z', 'y', 'x', 'w',], 'g': ['z', 'y', 'x', 'w',]} 

Answer 4

You're using the wrong method altogether: zip is for corresponding pairs only, giving you (az) (by) (cx) and so on, without providing the other pairings.

Is there some reason you need to generate this more than once? If not, this is a job for itertools.product , not zip .

from itertools import product

k = ['a', 'b', 'c', 'd', 'e', 'f', 'g']
v = ['z', 'y', 'x', 'w',]

for pair in product(k, v):
    print("some string " + ' '.join(pair))

Output:

some string a z
some string a y
some string a x
some string a w
some string b z
some string b y
some string b x
some string b w
...
some string g x
some string g w