using index value greater than as a condition for np where-CodePudding

I'm trying to look for true conditions where "this" is greater than "that" starting with row 2 (index is greater than 1). For instance if the df is 5 rows long and "this" is greater than "that" in rows 1, 4, and 5 then the return would be 0,0,0,1,1. The index itself is a datetime string. However, someinteger is the actual index row value (a value from zero to length-1 of the df) It seems pretty easy and trivial but I'm having difficulty. I have tried this, but it's not working.

import panda as pd
import numpy as np

df = pd.DataFrame({
'this': [5,2,2,5,5], 
'that': [3,3,3,3,3]})

someintegervalue = 1

df['blah blah blah'] = np.where(df.index > someintegervalue & df['this'] > df['that'], 0, 1)

CodePudding user response：

IIUC, you could just do:

import numpy as np
df['blah blah blah']= np.where(np.arange(len(df)) > someintegervalue & df['this']>df['that'], 0, 1)

CodePudding user response：

You were close. You just need to enclose your conditions in () and flip the last two np.where() arguments.

import pandas as pd
import numpy as np

df = pd.DataFrame({
    'this': [5,2,2,5,5], 
    'that': [3,3,3,3,3]})

someintegervalue = 1

df['blah blah blah']= np.where((df.index > someintegervalue) & (df['this'] > df['that']), 1, 0)

Output:

    this    that    blah blah blah
0   5       3       0
1   2       3       0
2   2       3       0
3   5       3       1
4   5       3       1