Extract keys from a list of dicts with single key-value pairs

Question

I have a list which is like this:

[{0: 26},
 {0: 36},
 {1: 1},
 {0: 215},
 {1: 63},
 {0: 215}]

How can I extract a list of keys from it?

[0, 0, 1, 0, 1, 0]

Sunitha · Answer 1 · 2018-08-15T08:46:54.807

4

Use dict.keys to extract the keys of each of the dict, convert to a list, and then extract the first element

>>> lst = [{0: 26}, {0: 36}, {1: 1}, {0: 215}, {1: 63}, {0: 215}]
>>> [list(d.keys())[0] for d in lst]
[0, 0, 1, 0, 1, 0]

Alternatively, you can use list comprehension as below

>>> [k for d in lst for k in d.keys()]
[0, 0, 1, 0, 1, 0]

edited Aug 15 '18 at 08:46

answered Aug 15 '18 at 08:32

Sunitha

11,777
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3

Or `[next(iter(i)) for i in L]`. Please can you rename your list, in many fonts `l` looks the same as `I`. In addition, you don't need to use `d.keys()`. `d` should suffice when iterating or using `list(d)`. – jpp Aug 15 '18 at 08:43

Austin · Accepted Answer · 2018-08-15T08:39:03.533

Use dict.keys() to get keys out of a dictionary and use it in a list-comprehension like below:

lst = [{0: 26},
       {0: 36},
       {1: 1},
       {0: 215},
       {1: 63},
       {0: 215}]

print([y for x in lst for y in x.keys()])
# [0, 0, 1, 0, 1, 0]

Or, this should be further simplified as:

print([y for x in lst for y in x])

Because, when you simply iterate through dictionary like for y in x, you are actually iterating through keys of the dictionary.

score 2 · Answer 3 · answered Aug 15 '18 at 08:36

I suggest you to simply iterate over the list of dictionaries, and to iterate over the keys of each dictionary, using list comprehension as follows:

myList = [{0: 26}, {0: 36}, {1: 1}, {0: 215}, {1: 63}, {0: 215}]
newList = [k for d in myList for k in d]
print(newList) # [0, 0, 1, 0, 1, 0]

score 1 · Answer 4 · answered Aug 15 '18 at 08:32

1

Those are the keys so you would use the dict.keys() method:

L = [{0: 26},
     {0: 36},
     {1: 1},
     {0: 215},
     {1: 63},
     {0: 215}]

L2 = [list(d.keys())[0] for d in L]

answered Aug 15 '18 at 08:32

Mathieu

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SanthoshSolomon · Answer 5 · 2018-08-15T08:34:00.120

1

use keys() function of dict().

a = [{0: 26}, {0: 36}, {1: 1}, {0: 215}, {1: 63}, {0: 215}]
keys = list(a.keys()[0])
vaues = (a.values()[0])

Cheers!

edited Aug 15 '18 at 08:34

answered Aug 15 '18 at 08:32

SanthoshSolomon

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Good luck applying `.keys()` and `.values()` on a list. – Mathieu Aug 15 '18 at 08:33
I did not notice that properly. Thanks though! – SanthoshSolomon Aug 15 '18 at 08:34

score 1 · Answer 6 · answered Aug 15 '18 at 08:37

1

you can do it this way :

list = [list(d.keys())[0] for d in originalList] #  originalList is the first list you posted

Here's the output : [0, 0, 1, 0, 1, 0]

answered Aug 15 '18 at 08:37

Ines Tlili

790
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score 1 · Answer 7 · answered Aug 15 '18 at 08:47

1

You can use dict.keys and itertools.chain to build a list of all keys:

from itertools import chain

w = [{0: 26},
   {0: 36},
   {1: 1},
   {0: 215},
   {1: 63},
   {0: 215}]

keys = list(chain.from_iterable(map(dict.keys, w)))

answered Aug 15 '18 at 08:47

Daniel

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score 1 · Answer 8 · answered Aug 15 '18 at 09:32

1

OR try the below code, using map:

lod=[{0: 26},
 {0: 36},
 {1: 1},
 {0: 215},
 {1: 63},
 {0: 215}]
print(list(map(lambda x: list(x.keys())[0])))

answered Aug 15 '18 at 09:32

U13-Forward

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Extract keys from a list of dicts with single key-value pairs

8 Answers8