Using .loc on DataFrame while ignoring several rows

Question

I have a DataFrame that I need to modify based on one of column values. In particular, when the value in column a is above 110, I want the column b to be assigned value of -99. The only issue is that first 3 rows of the dataframe contain a mix of string and numerical data types so when I try:

df.loc[df['a'] >= 110, 'b'] = -99

I get a TypeError because comparison between str and int is not allowed.

So my question is: how do I do this assignment while ignoring the first 3 rows of the dataframe?

So far I came up with this rather dodgy way:

try:
    df.loc[df['a'] >= 110, 'b'] = -99
except TypeError:
    pass

This does seem to work, but it obviously doesn't seem like the proper way to do it.

EDIT: And also this method just skips first 3 rows, but I really need to keep them as is.

Kenan · Accepted Answer · 2020-01-09 01:56:42Z

1

Try:

df.loc[df['a'].apply(pd.to_numeric, errors='coerce').ge(110), 'b'] = -99

or use errors='ignore'

edited Jan 9, 2020 at 1:56

answered Jan 9, 2020 at 1:04

Kenan

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3 Comments

NotAName Over a year ago

I want to keep the data in the top 3 rows as is. Wouldn't apply(pd.to_numeric,errors='coerse') replace the data with a bunch of NaN?

Kenan Over a year ago

make a copy of the first 3 rows and re append after that operation

NotAName Over a year ago

When I try to do it the way you suggested, I get a weird error: ValueError: invalid error value specified. Although it works if I do errors='ignore'

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