Create a column filled with 0 or 1 based on other columns in Pandas

Question

I have a data frame and need to fill a column with 0 or 1 based on the values of another 29 columns. If you have 1 in any of the columns 0 to 28 the column with Dor_leve receives 1, otherwise receives 0, for the column Dor_media is if you have 2 in any of the columns 0 to 28 and to Dor_grave if it has 3 in any column from 0 to 28.

I have the columns Col_0 to Col_28 and I need to generate the columns Dor_leve, Dor_media and Dor_grave as below. The values contained in the columns Col_0 ... Col_28 are 0 or 1 or 2 or 3.

Example of what I need to do:

Col_0 Col_1 Col_2 ... Col_28 Dor_leve Dor_media Dor_grave
0      1     1    ...   0         1         0        0
1      0     2    ...   0         1         1        0  
1      0     3    ...   1         1         0        1
2      3     0    ...   0         0         1        1 
0      0     0    ...   0         0         0        0
3      0     0    ...   1         1         0        1

The columns are different parts of the body, so the person can have 1, 2 or 3 in each column or 0 in all. Attempt was:

import pandas as pd
df = pd.read_excel('arquivo.xlsx', sheet_name=0, header=0)

# Crio as colunas com 0 para todos

df['Dor_leve']= 0
df['Dor_media']= 0
df['Dor_grave']= 0

# Atualizar os valores, com 1 para sim, se tem o problema

for index, value in df.items():
    if(value==1):
        df.loc[index,'Dor_leve']=1
    elif(value==2):
        df.loc[index, 'Dor_media']=1
    elif(value==3):
        df.loc[index, 'Dor_grave']=1
    insira o código aqui

Returned this error:

> ValueError: The truth value of a Series is ambiguous. Use a.empty,
> a.bool(), a.item(), a.any() or a.all().

I tried to:

if(df.items()==1):
   df['Dor_leve']=1

if(df.items()==2):
   df['Dor_media']=1
    
if(df.items()==3):
  df['Dor_grave']=1

It didn’t work either. With the if (df.items()==valor) the columns were only 0.

Answer 1

First establish the columns you want to use to check the condition:

# definindo colunas manualmente
cols = ['Col_0', 'Col_1', 'Col_2', ...]  # preencha com as colunas aqui

# ou então usando uma compreensão de lista, dada uma lógica no nome das colunas
cols = [f'Col_{x}' for x in range(29)]

Then, "ask" for each row of the columns of interest if there is the presence of 1, 2, or 3. The result of this question (True/False) is converted to the integer 1 or 0. We can ask this question using df.apply and applying a function to each line (with the argument axis=1):

df['Dor_leve'] = df[cols].apply(lambda row: int(1 in row.values), axis=1)
df['Dor_media'] = df[cols].apply(lambda row: int(2 in row.values), axis=1)
df['Dor_grave'] = df[cols].apply(lambda row: int(3 in row.values), axis=1)