Btukfyl

So let's say i have a pandas data-frame as below:

df=pd.DataFrame({'a':[1,2,3,0]})

So my goal is to replace 0 value with (empty list) in this data-frame, but i did:

print(df.replace(0,))

But it gives me an error:

TypeError: Invalid "to_replace" type: 'int'

I tried everything that's possible i.e:

df[df==0]=

etc...

But nothing works.

Desired output (in case for confusion):

   a

0  1

1  2

2  3

3 

It is possible by list comprehension, but because mixed content - numeric with list it is not recommended:

df['a'] = [ if x == 0 else x for x in df.a]



print (df)



    a

0   1

1   2

2   3

3  

And replace all values in all columns:

df = df.applymap(lambda x:  if x == 0 else x)

print (df)

    a

0   1

1   2

2   3

3  

There are two issues here. First is pandas' quirkiness when dealing with lists. To replace values in a DataFrame with list you need to do something like this;

df.loc[df.a == 0, "a"] = [ for _ in df[df.a == 0]]

This creates n empty list based on the number of items that matches the criteria (df == 0)

The second issue is that your column is of integer type, and you can't store a list in an integer column. So before you can assign the list, you would first need to convert the column type to object first.

df = df.astype(object)

df.loc[df.a == 0, "a"] = [ for _ in df[df.a == 0]]

This works for me:

df=pd.DataFrame({'a':[1,8,3,0]})

print(df)



  a

0  1

1  8

2  3

3  0



df['a'] = df['a'].apply(lambda d: d if d!=0 else )

print(df)



   a

0   1

1   8

2   3

3