CODEWITHSUNDEEP
pythonpandas
rer49
rer49

Reputation: 235

Comparing column values within a row within a Pandas Dataframe

I have a specific and a general problem I'm trying to solve.

Specific Problem: I want to create a new column in a data frame that gives a 1 if column C1 is 8 and all other values in the row are less than 8.How do I logically negate all of the other columns at the same time? Here is the code from my flawed attempt:

df["C1is_8"] = df.apply(lambda row:(row['C1']==8)& ~(row['C1']<8) ,axis=1).astype(int)

The code below produces the dataframe for the code above.

dict = { 'C1':[4,3,0,0,2,3,4,5,8,8,8,8],
         'C2':[8,3,3,7,6,5,3,5,6,8,8,8],
         'C3':[2,3,6,4,5,0,0,4,6,7,8,8],
         'C4':[8,5,4,4,4,3,2,1,4,2,6,8]
       }
columns = ['C1','C2','C3','C4']
Index = [1,2,3,4,5,6,7,8,9,10,11,12]
df = pd.DataFrame(dict,index = Index,columns = columns)
df = OGdf[::-1]
df

General Problem: How do I rewrite some version of the code above so that I can generalize it (i.e. row[ i ] ) so that it could apply to any column not just 'C1'?

Upvotes: 2

Views: 1209

Answers (1)

sacuL
sacuL

Reputation: 51335

I think this answer satisfies both your specific and your general problems, using filter and all:

# define the column you want to apply your first condition to
col = 'C1'

# Python 3.6 or above, with f-strings:
df['new_col'] = ((df[col] == 8) & (df.filter(regex=f'[^{col}]') < 8).all(1)).astype(int)
# Otherwise:
df['new_col'] = ((df[col] == 8) & (df.filter(regex='[^{}]'.format(col)) < 8).all(1)).astype(int)

>>> df
    C1  C2  C3  C4  new_col
1    4   8   2   8        0
2    3   3   3   5        0
3    0   3   6   4        0
4    0   7   4   4        0
5    2   6   5   4        0
6    3   5   0   3        0
7    4   3   0   2        0
8    5   5   4   1        0
9    8   6   6   4        1
10   8   8   7   2        0
11   8   8   8   6        0
12   8   8   8   8        0

Upvotes: 2

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