如何创建包含每行倒数第二个值的列？

Question

I have a DataFrame and I need to create a new column which contains the second largest value of each row in the original Dataframe.

Sample:

 df = pd.DataFrame(np.random.randint(1,100, 80).reshape(8, -1))

Desired output:

 0   1   2   3   4   5   6   7   8   9  penultimate
0  52  69  62   7  20  69  38  10  57  17           62
1  52  94  49  63   1  90  14  76  20  84           90
2  78  37  58   7  27  41  27  26  48  51           58
3   6  39  99  36  62  90  47  25  60  84           90
4  37  36  91  93  76  69  86  95  69   6           93
5   5  54  73  61  22  29  99  27  46  24           73
6  71  65  45   9  63  46   4  93  36  18           71
7  85   7  76  46  65  97  64  52  28  80           85

How can this be done in as little code as possible?

Answer 1

You could use NumPy for this:

import numpy as np

df = pd.DataFrame(np.random.randint(1,100, 80).reshape(8, -1))
df['penultimate'] = np.sort(df.values, 1)[:, -2]
print(df)

Using NumPy is faster.

Answer 2

Here is a simple lambda function!

# Input
df = pd.DataFrame(np.random.randint(1,100, 80).reshape(8, -1))

# Output
out = df.apply(lambda x: x.sort_values().unique()[-2], axis=1)
df['penultimate'] = out
print(df)

Cheers!