CODEWITHSUNDEEP
pythonpandasdataframe
SchwarzbrotMitHummus
SchwarzbrotMitHummus

Reputation: 79

Pandas Dataframe Set value in cell if condition (Element is in list)

I have a dataframe containing a column called "column_B" with only nan-values. It contains also "column_A", which is a list of strings for every cell in the dataframe.

index   column_A    column_B  
1       ['A','B']    np.nan
2       ['B']        np.nan
3       ['C']        np.nan

Now I want that column_B is set to 0 if any element in a list of column_A contains 'B'.

Desired result:

   index   column_A    column_B  
    1       ['A','B']    0
    2       ['B']        0
    3       ['C']        np.nan

I tried

df['column_B'][any([x == 'B' for x in df['column_A']])] = 0

that doesn't work. Any ideas? :)

Thanks

Upvotes: 2

Views: 2638

Answers (4)

ashkangh
ashkangh

Reputation: 1624

Another solution is:

df['column_B'] = np.where(df['column_A'].astype(str).str.contains(r'\bB\b'), 0, np.nan)

Output:

    index   column_A    column_B
0   1      ['A','B']      0.0
1   2       ['B']         0.0
2   3       ['C']         NaN

Upvotes: 1

piRSquared
piRSquared

Reputation: 294488

Comprehension and mask

df['column_B'] = df.column_B.mask(['B' in x for x in df.column_A], 0)

df

  column_A  column_B
0   [A, B]       0.0
1      [B]       0.0
2      [C]       NaN

Without mask

df['column_B'] = [0 if 'B' in x else np.nan for x in df.column_A]

df

  column_A  column_B
0   [A, B]       0.0
1      [B]       0.0
2      [C]       NaN

Explode

df['column_B'] = df.column_B.mask(df.column_A.explode().eq('B').any(level=0), 0)

df

  column_A  column_B
0   [A, B]       0.0
1      [B]       0.0
2      [C]       NaN

Upvotes: 2

rhug123
rhug123

Reputation: 8768

using explode():

df2 = df.explode('column_A')
df.loc[df2['column_A'].isin(['B']).groupby(level=0).any(),'column_B'] = 0

Upvotes: 0

jezrael
jezrael

Reputation: 863226

Use Series.map or Series.apply for test value in list by in and set value in Series.mask:

df['column_B'] = df['column_B'].mask(df['column_A'].map(lambda x: 'B' in x), 0)
#df['column_B'] = df['column_B'].mask(df['column_A'].apply(lambda x: 'B' in x), 0)

Or by DataFrame.loc:

df.loc[df['column_A'].map(lambda x: 'B' in x), 'column_B'] = 0

Upvotes: 3

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