np.where 检查 True 为第一列然后在另一列

Question

i have below dataframe with np.where condition as:

df=pd.DataFrame(data = {'First':[1,2,4,6,2,7,8,9],'Second':[4,6,7,3,1,3,9,3]})
df['First_check']=np.where(df['First']==2,'T','F')

df

   First    Second  First_check
0       1       4       F
1       2       6       T
2       4       7       F
3       6       3       F
4       2       1       T
5       7       3       F
6       8       9       F
7       9       3       F

Now i want to check for df['Second']==3 but only after df['First_check']=='T' also i want only first occurance of condition.

Below is My desired output:

   First    Second  First_check Second_check
0       1       4       F           F
1       2       6       T           F
2       4       7       F           F
3       6       3       F           T
4       2       1       T           F
5       7       3       F           T
6       8       9       F           F
7       9       3       F           F

Edit: i want df['Second']==3 to become True But 1st df['First_check']=='T' should become True then it may or may not be on same row. say for row 2 df['First_check']=T then it should check next rows 2,3,4... for df['Second']==3 it matched at row 4th

Answer 1

I think this is what you are looking for.

Criteria:

• Step 1: Look for value in 'First_check' . If the value is 'T', reset the flag to check for 3 in Second
• Step 2: Now check for 3 in Second . Turn the value of the first 3 to 'T' . All subsequent 3 should be 'F' until you get a new T in 'First_check'
• Step 3: When you get a new 'T' , return to Step 1 and continue.

To do this, you need to look back both 'First_check' and 'Second'

Here's code to solve for it.

import pandas as pd
import numpy as np
df=pd.DataFrame(data = {'First':[1,2,4,6,2,7,8,9],'Second':[4,6,7,3,1,3,9,3]})

df['First_check']=np.where(df['First']==2,'T','F')
print (df)

df['tempF'] = df.groupby((df['First_check'].eq('T')).cumsum()).cumcount()+1
df['tempS'] = df.groupby((df['Second'].eq(3)).cumsum()).cumcount()+1
df['Second_check'] = np.where((df['tempS'] == 1) & (df['tempF'] == df['tempS'].shift(1)),'T','F')
df.drop(['tempF','tempS'],axis=1,inplace=True)
print (df)

The output is as per your required output:

   First  Second First_check Second_check
0      1       4           F            F
1      2       6           T            F
2      4       7           F            F
3      6       3           F            T
4      2       1           T            F
5      7       3           F            T
6      8       9           F            F
7      9       3           F            F

Answer 2

Alternatively you can create a new Series where 2 s and 3 s are filled in while other places have nan s, and then you can do a forward fill on this new Series, which will give a Series with 2 s and 3 s only (with possibly nan s at the beginning). Finally you can check if 3 s in the Second has a preceding value of 2 in the preceding column:

# firstly merge two and three into a single Series and do a forward fill
preceding = df.First.shift(-1).where(
    df.First.shift(-1).eq(2), 
    df.Second.where(df.Second.eq(3))
).ffill()

preceding
#0    2.0
#1    2.0
#2    2.0
#3    2.0
#4    2.0
#5    3.0
#6    3.0
#7    3.0
#Name: First, dtype: float64

# after the forward fill if a 3 is preceded by a 2, then it should be True
df.Second.eq(3) & preceding.shift().eq(2)
#0    False
#1    False
#2    False
#3     True
#4    False
#5     True
#6    False
#7    False
#Name: First, dtype: bool