Applying a function to values in dict

PythonDictionary

Python Problem Overview


I want to apply a function to all values in dict and store that in a separate dict. I am just trying to see how I can play with python and want to see how I can rewrite something like this

for i in d:
    d2[i] = f(d[i])

to something like

d2[i] = f(d[i]) for i in d

The first way of writing it is of course fine, but I am trying to figure how python syntax can be changed

Python Solutions


Solution 1 - Python

If you're using Python 2.7 or 3.x:

d2 = {k: f(v) for k, v in d1.items()}

Which is equivalent to:

d2 = {}
for k, v in d1.items():
    d2[k] = f(v)

Otherwise:

d2 = dict((k, f(v)) for k, v in d1.items())

Solution 2 - Python

You could use map:

d2 = dict(d, map(f, d.values()))

If you don't mind using an extension. You can also use valmap in the toolz library which is functionally equivalent to using the map solution:

from toolz.dicttoolz import valmap

d2 = valmap(f, d)

If not for the clean presentation of the method, you also have the option of supplying a default return class as well, for people that need something other than a dict.

Solution 3 - Python

d2 = dict((k, f(v)) for k,v in d.items())

Solution 4 - Python

Dictionaries can be nested in Python and in this case the solution d2 = {k: f(v) for k, v in d1.items()} will not work.

For nested dictionaries one needs some function to transverse the whole data structure. For instance if values are allowed to be themselves dictionaries, one can define a function like:

def myfun(d):
  for k, v in d.iteritems():
    if isinstance(v, dict):
      d[k] = myfun(v)
    else:
      d[k] = f(v)
  return d

And then

d2 = myfun(d)

Attributions

All content for this solution is sourced from the original question on Stackoverflow.

The content on this page is licensed under the Attribution-ShareAlike 4.0 International (CC BY-SA 4.0) license.

Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionchriseView Question on Stackoverflow
Solution 1 - PythonJoel CornettView Answer on Stackoverflow
Solution 2 - PythonJabView Answer on Stackoverflow
Solution 3 - PythonAlexey KachayevView Answer on Stackoverflow
Solution 4 - Pythonuser2314737View Answer on Stackoverflow