How can I check if a value is nan or a list?

Question

I'm currently working with pandas DataFrames. While iterating over it I want to check if a value is numpy.nan or a list.

for i, row in df.iterrows(): value = row["Name"] if pd.isnull(value): dosomething()

This works just fine, except if

type(value) == list

Then I thought of maybe putting any() around:

for i, row in df.iterrows(): value = row["Name"] if any(pd.isnull(value)): dosomething()

But now I get a exception everytime a NaN is in value because it's obviously not iterable.

Is there a better solution then checking the type of value?

jezrael · Accepted Answer · 2019-03-12 09:17:59Z

4

Use or:

for i, row in df.iterrows(): value = row["Name"] if pd.isnull(value) or (type(value) == list): dosomething()

Another way for check is isinstance:

for i, row in df.iterrows(): value = row["Name"] if pd.isnull(value) or isinstance(value, list): dosomething()

answered Mar 12, 2019 at 9:17

jezrael

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3 Comments

aAnnAa Over a year ago

But I still get an errormessage if type(value)==list (ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all())

jezrael Over a year ago

@aAnnAa - Maybe duplicated columns, then need any(pd.isnull(value) or (type(value) == list))

aAnnAa Over a year ago

It works now with "if not isinstance(value, list) and pd.isnull(value)", thanks a lot for your help :)

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