updating column value as loop over dataframe rows

Question

I am trying to take links and push them via TinyURL.. I have a dataframe containing Login Links column and I would like to take those individually and add to column 'Tiny URL', the tiny_urlize link.

for index, row in df.iterrows():
    df.loc[index, 'Tiny Url'] = tiny_urlize(row['Login Link'])

The error I am getting looks like this:

A value is trying to be set on a copy of a slice from a DataFrame.
Try using .loc[row_indexer,col_indexer] = value instead

but this is exactly what I am doing. Am I missing something?

Is df itself a copy (e.g. did you do at some point before df = other_df[some_slice] ? — rafaelc
– rafaelc, Commented Jan 31, 2019 at 19:02
It can also be the result of a few methods that you wouldn't necessarily expect to return a copy. For instance df = df.drop_duplicates() could be the culprit. — ALollz
– ALollz, Commented Jan 31, 2019 at 19:13

Sven Harris · Accepted Answer · 2019-01-31 19:05:53Z

2

This error can pop up for a number of different reasons, a common case for example is that df is a slice of another dataframe. I suspect if you write df = df.copy() before your iterrows you won't get the error.

You can also simplify and speed up your expression by writing it as an apply:

df['Tiny Url'] = df['Login Link'].apply(tiny_urlize)

which I believe will also prevent the warning

answered Jan 31, 2019 at 19:05

Sven Harris

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Raunaq Jain · Accepted Answer · 2019-01-31 19:05:33Z

0

The issue is that you are updating the value of df using loc while looping (which is essentially a slice). Try,

for index, row in df.iterrows():
    row['Tiny Url'] = tiny_urlize(row['Login Link'])

Moreover, pandas has in-built functions which are very efficient for such things, one of them being apply.

df['Tiny Url'] = df['Login Link'].apply(tiny_urlize)

answered Jan 31, 2019 at 19:05

Raunaq Jain

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