how to find consecutive 1 from last column of data frame and count how many time-CodePudding

there are 5 column in 1st data frame . by using this find consecutive 1 from last D_4 to D_1, if find 0 in between then break and till that how many ones and that will be output

CodePudding user response：

You can concatenate each row as string from D_4 to D_1, split strings once at first '0' then get the length of the first part:

df['lco'] = (df.iloc[:, :0:-1].astype(str).apply(''.join, axis=1)
                              .str.split('0', n=1).str[0].str.len())
print(df)

# Output
  Code  D_1  D_2  D_3  D_4  lco
0    A    0    1    0    1    1
1    B    1    1    0    1    1
2    C    0    0    1    1    2
3    D    1    1    1    1    4
4    E    0    0    0    1    1
5    F    0    0    0    0    0
6    G    1    1    1    0    0
7    H    1    0    1    1    2

CodePudding user response：

You can melt, use a reverse cummin per group to get rid of the trailing 1s, then count the 1s:

df.merge(df.melt('Code', value_name='num')
           .groupby('Code')['num']
           .apply(lambda s: s[::-1].cummin().sum()),
         on='Code'
         )

or, in place, with stack:

df['num'] = (df
   .iloc[:,1:].stack()
   .groupby(level=0)
   .apply(lambda s: s[::-1].cummin().sum())
)

output:

  Code  D_1  D_2  D_3  D_4  num
0    A    0    1    0    1    1
1    B    1    1    0    1    1
2    C    0    0    1    1    2
3    D    1    1    1    1    4
4    E    0    0    0    1    1
5    F    0    0    0    0    0
6    G    1    1    1    0    0
7    H    1    0    1    1    2