如何基于另一列的值（其中两个不同的列中的值在pandas中相等）找到一列的行索引？

Question

I have some pandas output:

seq X1      X2   
0   0.59    NaN  
1   -1.28   NaN  
2   -1.26   NaN 
3   -0.79   NaN 
4   1.03    NaN
5   -1.43   NaN 
6   0.03    1.03    
7   0.92    1.03
8   -2.21   1.03    

How do I get a third column X3, that takes the 1.03 in X2 and finds the seq number associated with the same number in column X1? In my example starting from row7(row index 6), X3 should return a 4, since seq = 4 when X1 is 1.03.

I desire:

seq  X1     X2  X3
0   0.59    NaN  NaN
1   -1.28   NaN  NaN
2   -1.26   NaN  NaN
3   -0.79   NaN  NaN
4   1.03    NaN  NaN
5   -1.43   NaN  NaN
6   0.03    1.03 4  
7   0.92    1.03 4
8   -2.21   1.03 4 

First ever Stack question. Pardon my folly!

Answer 1

Can you explain why you want the number 4 to appear in all rows of X3 ?

You can get the seq value ( 4 ) where X1 == 1.03 by typing:

df.loc[df['X1']==1.03, 'seq'].values[0]

But that only gives you one 4. Note that I've taken the first seq value (by typing [0] ) because if you have more than one place where X1 == 1.03 a list of numbers will be returned (as a data frame) and you haven't explained how you'd like to deal with multiple seq matches.

The following code will run and return the data frame you requested, but I suggest you spend a little time thinking about whether you need X2 and X3 to be part of the data frame at all...

# Import what you need
import pandas as pd
import numpy as np

# Define the data
x1 = np.array([0.59, -1.28, -1.27, -0.79, \
                1.03, -1.43, 0.03, 0.92, -2.21])
x2 = np.array([np.nan, np.nan, np.nan, np.nan, \
                np.nan, np.nan, 1.03, 1.03, 1.03])

# Create a pandas dataframe
df = pd.DataFrame( { 'seq' : range(9),
                     'X1'  : x1,
                     'X2'  : x2 } )

# Figure out where the first instance of X1==1.03
# occurs and grab that seq value
s_first = df.loc[df['X1']==1.03,'seq'].values[0]

# Fill in X3 according to the values in X2
df.loc[df['X2'].isnull(), 'X3'] = np.nan
df.loc[df['X2'].notnull(), 'X3'] = s_first

# Show the 9 rows in the data frame
df.head(9)