Modifying list of dictionary using list comprehension

Question

So I have the following list of dictionary

myList = [{'one':1, 'two':2,'three':3},
          {'one':4, 'two':5,'three':6},
          {'one':7, 'two':8,'three':9}]

This is just an example of a dictionary that I have. My question is, it possible to somehow modify say key two in all the dictionary to become twice their value, using list comprehension ?

I know how to use list comprehension to create new list of dictionary, but don't know how to modify them, I have come up with something like this

new_list = { <some if condiftion> for (k,v) in x.iteritems() for x in myList  }

I am not sure how to specify a condition in the <some if condiftion>, also is the nested list comprehension format I am thinking of correct ?

I want the final output as per my example like this

[ {'one':1, 'two':4,'three':3},{'one':4, 'two':10,'three':6},{'one':7, 'two':16,'three':9}  ]

List comprehension is used to build new lists, not modify old ones. — Scott Hunter
– Scott Hunter, Commented Jun 30, 2018 at 14:07
What you created is a dictionary comprehension, not a list comprehension. Well, so far as { and [ are concerned — roganjosh
– roganjosh, Commented Jun 30, 2018 at 14:08

jezrael · Accepted Answer · 2018-06-30 14:10:29Z

8

Use list comprehension with nested dict comprehension:

new_list = [{ k: v * 2 if k == 'two' else v for k,v in x.items()} for x in myList]
print (new_list)
[{'one': 1, 'two': 4, 'three': 3}, 
 {'one': 4, 'two': 10, 'three': 6}, 
 {'one': 7, 'two': 16, 'three': 9}]

edited Jun 30, 2018 at 14:10

answered Jun 30, 2018 at 14:07

jezrael

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Comments

Aran-Fey · Accepted Answer · 2018-06-30 14:51:43Z

3

In python 3.5+ you can use the new unpacking syntax in dict literals introduced in PEP 448. This creates a copy of each dict and then overwrites the value for the key two:

new_list = [{**d, 'two': d['two']*2} for d in myList]
# result:
# [{'one': 1, 'two': 4, 'three': 3},
#  {'one': 4, 'two': 10, 'three': 6},
#  {'one': 7, 'two': 16, 'three': 9}]

answered Jun 30, 2018 at 14:51

Aran-Fey

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lenik · Accepted Answer · 2018-06-30 14:12:52Z

1

myList = [ {'one':1, 'two':2,'three':3},{'one':4, 'two':5,'three':6},{'one':7, 'two':8,'three':9}  ]

[ { k: 2*i[k] if k == 'two' else i[k] for k in i } for i in myList ]

[{'one': 1, 'three': 3, 'two': 4}, {'one': 4, 'three': 6, 'two': 10}, {'one': 7, 'three': 9, 'two': 16}]

answered Jun 30, 2018 at 14:12

lenik

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Comments

jpp · Accepted Answer · 2018-06-30 14:17:28Z

1

A simple for loop should be sufficient. However, if you want to use a dictionary comprehension, I find defining a mapping dictionary more readable and extendable than ternary statements:

factor = {'two': 2}

res = [{k: v*factor.get(k, 1) for k, v in d.items()} for d in myList]

print(res)

[{'one': 1, 'two': 4, 'three': 3},
 {'one': 4, 'two': 10, 'three': 6},
 {'one': 7, 'two': 16, 'three': 9}]

answered Jun 30, 2018 at 14:17

jpp

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2 Comments

Gambit1614 Over a year ago

Thanks! I didn't knew we could use mapping like this!

jpp Over a year ago

@MohammedKashif, Yep, the main point is for one condition an if / else ternary is fine. But it will become (more) unreadable if you have further conditions.

Black Lion · Accepted Answer · 2018-06-30 14:15:59Z

0

Hello did you tried this :

for d in myList:
  d.update((k, v*2) for k, v in d.iteritems() if k == "two")

Thanks

answered Jun 30, 2018 at 14:15

Black Lion

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1 Comment

Aran-Fey Over a year ago

Why not use d['two'] = d['two'] * 2 instead of that d.update(...) thing?

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