Compare last element of a row to the rest of a row in pandas

chrys :

I am working with a dataframe in pandas which holds numeric data.

E.g:

d = {'col1': [1, 2,3,2], 'col2': [3, 4,1,2],'col3':[1,3,4,1}
df = pd.DataFrame(data=d)

What I want to do is compare the elements in the third column with the other elements in their respective row in terms of something each element in row n < last element of row n return true / false or 1 / 0.

#Desired Output:
resDf = {'col1':[False,True,True,False],'col2':[False,False,True,False]}

What I have done so far is use apply like this:

resultBoolDf = df.iloc[:,:-1].apply(lambda x: np.where(x < df.col3,1,0),axis = 0)

So this does not seem to work since I assume that the comparison is not iterating correctly. Could somebody give me a tip on how to solve this? Thanks!

jezrael :

Use DataFrame.lt for compare with last column selected by position:

df1 = df.iloc[:,:-1].lt(df.iloc[:, -1], axis=0)
#if want specify last column by label
#df1 = df.iloc[:,:-1].lt(df.col3, axis=0)
print (df1)
    col1   col2
0  False  False
1   True  False
2   True   True
3  False  False

Last if need 0,1 convert to integers by DataFrame.astype:

df1 = df.iloc[:,:-1].lt(df.iloc[:, -1], axis=0).astype(int)
#if want specify last column by label
#df1 = df.iloc[:,:-1].lt(df.col3, axis=0).astype(int)
print (df1)
   col1  col2
0     0     0
1     1     0
2     1     1
3     0     0

Your solution with numpy.where is possible use with DataFrame constructor:

arr = np.where(df.iloc[:,:-1].lt(df.col3, axis=0),1,0)
df1 = pd.DataFrame(arr, index=df.index, columns = df.columns[:-1])
print (df1)
   col1  col2
0     0     0
1     1     0
2     1     1
3     0     0