Lambda function in pd df with if statement result in Syntax Error

Question

I have a df, where in one column contains values in the form of strings which includes numbers & nan values. That column is represented in the form of string, where each values in the string seperated using :(colan)instead of , (comma)(see below column no n represeneted as cln)

index cl1   cl2    .... cln
0      1.1  6.4    .    '12.30 : 14.27 : nan : 15.87, nan .....'
1      2    5.3         '12.8 : 13.73 : nan : 15.87, 12.8 .....'
2      5    6.6
.     .     . 
.     .     .
.     .     .
.    
m     cl1m cl2m          '12.8 : 13.73 : nan : 15.87, 12.8 .....'

I want to replace : instead of comma using a single line of code using lambda fn.

df['cln'] = df['cln'].apply(lambda x : x.replace(':', ',') if x != np.nan)

But i received the following error

SyntaxError: invalid syntax

Even i tried the below code but it shows the same error

df['cln'] = df['cln'].apply(lambda x : x.replace(':', ',') if type(x) = str)
# since nan is float

i know how to use for loop and solve this problem, but i just want to know why it doesn't work using Lambda fn

Expecting pythonic answers for this problem

You need to have else clause to if.

Austin
– Austin

2020-03-23 12:21:49 +00:00
Commented Mar 23, 2020 at 12:21 — Austin
– Austin, Commented Mar 23, 2020 at 12:21

C.Nivs · Accepted Answer · 2020-03-23 12:28:50Z

3

You need an else:

df['cln'] = df['cln'].apply(lambda x : x.replace(':', ',') if x != np.nan else x)

Or better yet, just filter out the null:

df[df['cln'].notnull()].apply(lambda x: x.replace(':', ','))

Otherwise, you could see this result:

np.nan == np.nan
False
np.nan != np.nan
True

edited Mar 23, 2020 at 12:28

answered Mar 23, 2020 at 12:22

C.Nivs

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