Pandas dataframe，在一行中，查找所选列中的最大值，并根据该值查找另一列的值

Question

I have a dataframe like this:

import pandas as pd
df = pd.DataFrame({'x1':[20,25],'y1':[5,8],'x2':[22,27],'y2':[10,2]})

df

       x1  y1  x2  y2
    0  20   5  22  10
    1  25   8  27   2

X and Y pair together. I need to compare y1 and y2 and get the max in every row. And find the corresponding x. Hence the max of row [0] is y2 (=10), and the corresponding x is x2 (=22). The second row will be y1 (=8) and x1(=25). Expected result, new columns x and y:

   x1  y1  x2  y2   x   y
0  20   5  22  10  22  10
1  25   8  27   2  25   8

This is a simple dataframe I made to elaborate on the question. X and Y pairs, in my case, can be 30 pairs.

Answer 1

# get a hold on "y*" columns
y_cols = df.filter(like="y")

# get the maximal y-values' suffixes, and then add from front "x" to them
max_x_vals = y_cols.idxmax(axis=1).str.extract(r"(\d+)$", expand=False).radd("x")
# get the locations of those x* values
max_x_ids = df.columns.get_indexer(max_x_vals)

# now we have the indexes of x*'s in the columns; NumPy's indexing
# helps to get a cross section
df["max_xs"] = df.to_numpy()[np.arange(len(df)), max_x_ids]

# for y*'s, it's directly the maximum per row
df["max_ys"] = y_cols.max(axis=1)

to get

>>> df

   x1  y1  x2  y2  max_xs  max_ys
0  20   5  22  10      22      10
1  25   8  27   2      25       8

Answer 2

You can do it with the help of.apply function.

import pandas as pd
import numpy as np

df = pd.DataFrame({'x1':[20,25],'y1':[5,8],'x2':[22,27],'y2':[10,2]})
y_cols = [col for col in df.columns if col[0] == 'y'] 
x_cols = [col for col in df.columns if col[0] == 'x'] 

def find_corresponding_x(row):
    max_y_index = np.argmax(row[y_cols])
    return row[f'{x_cols[max_y_index]}']

df['corresponding_x'] = df.apply(find_corresponding_x, axis = 1)

Answer 3

this is one solution:

a = df[df['y1'] < df['y2']].drop(columns=['y1','x1']).rename(columns={'y2':'y', 'x2':'x'})
b = df[df['y1'] >= df['y2']].drop(columns=['y2','x2']).rename(columns={'y1':'y', 'x1':'x'})

result = pd.concat([a,b])

if you need to keep order then maybe add another column with original index and sort by it after concatenation

Answer 4

you can use the function below. remember to import pandas and numpy like I did in this code. import your data set and use Max_number function.

import pandas as pd
import numpy as np
df = pd.DataFrame({'x1':[20,25],'y1':[5,8],'x2':[22,27],'y2':[10,2]})

def Max_number (df):
    columns = list(df.columns)
    rows = df.shape[0]
    max_value = []
    column_name = []

    for i in range(rows):
        row_array = list(np.array(df[i:i+1])[0])
        maximum = max(row_array)
        max_value.append(maximum)
        index=row_array.index(maximum)
        column_name.append(columns[index])
    
    return pd.DataFrame({"column":column_name,"max_value":max_value})

returns this:

row index	column	max_value
0	x2	22
1	x2	27

Answer 5

if x1 column comes first and then y1, then x2, y2 and so on, you can just try:

a = df.columns.get_indexer(y_cols.idxmax(axis=1))
df[['y', 'x']] = df.to_numpy()[np.arange(len(df)), [a, a - 1]].T

Answer 6

I hope it works for your solution,

import pandas as pd
df = pd.DataFrame({'x1':[20,25],'y1':[5,8],'x2':[22,27],'y2':[10,2]})
df['x_max'] = df[['x1', 'x2']].max(axis=1)
df['y_max'] = df[['y1', 'y2']].max(axis=1)
df