Change value based on condition on slice of dataframe

Question

I have a dataframe like this:

df = pd.DataFrame(columns=['Dog', 'Small', 'Adult'])
df.Dog = ['Poodle', 'Shepard', 'Bird dog','St.Bernard']
df.Small = [1,1,0,0]
df.Adult = 0

That will look like this:

       Dog         Small   Adult
0      Poodle          1   0
1      Shepard         1   0
2      Bird dog        0   0
3      St.Bernard      0   0

Then I would like to change one column based on another. I can do that:

df.loc[df.Small == 0, 'Adult'] = 1

However, I just want to do so for the 3 first rows.

I can select the first three rows:

df.iloc[0:2]

But if I try to change values on the first three rows:

df.iloc[0:2, df.Small == 0, 'Adult'] = 1

I get an error.

I also get an error if I merge the two:

df.iloc[0:2].loc[df.Small == 0, 'Adult'] = 1

It tells me that I am trying to set a value on a copy of a slice.

How should I do this correctly?

Asked By: Oeyvind

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Answer 1

You could include the range as another condition in your .loc selection (for the general case, I’ll explicitly include the 0):

df.loc[(df.Small == 0) & (0 <= df.index) & (df.index <= 2), 'Adult'] = 1

Another option is to transform the index into a series to use pd.Series.between:

df.loc[(df.Small == 0) & (df.index.to_series().between(0, 2)), 'Adult'] = 1

Answered By: aaossa

Answer 2

adding conditionals based on index works only if the index is already sorted. Alternatively, you can do the following:

ind = df[df.Small == 0].index[:2]
df.loc[ind, 'Adult'] = 1

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