Python Dataframe find minimum among multiple set of columns

Question

I am having a data frame of four columns. I want to find the minimum among the first two columns and the last two columns for each row. Code:

np.random.seed(0)
xdf = pd.DataFrame({'a':np.random.rand(1,10)[0]*10,'b':np.random.rand(1,10)[0]*10,'c':np.random.rand(1,10)[0]*10,'d':np.random.rand(1,10)[0]*10,},index=np.arange(0,10,1))

xdf['ab_min'] = xdf[['a','b']].min(axis=1)
xdf['cd_min'] = xdf[['c','d']].min(axis=1)
xdf['minimum'] = xdf['ab_min'].list()+xdf['cd_min'].list()

Expected answer:

xdf['minimum'] 

0   [ab_min,cd_min]
1   [ab_min,cd_min]
2   [ab_min,cd_min]
3   [ab_min,cd_min]

Present answer:

AttributeError: 'Series' object has no attribute 'list'

Shubham Sharma · Accepted Answer · 2022-02-17 04:22:58Z

2

Select the columns ab_min and cd_min then use to_numpy to convert it to numpy array and assign the result to minimum column

xdf['minimum'] = xdf[['ab_min', 'cd_min']].to_numpy().tolist()

>>> xdf['minimum']

0      [3.23307959607905, 1.9836323494587338]
1     [6.189440334168731, 1.0578078219990983]
2    [3.1194570407645217, 1.2816570607783184]
3     [1.9170068676155894, 7.158027504597937]
4     [0.6244579166416464, 8.568849995324166]
5     [4.108986697339397, 0.6201685780268684]
6     [4.170639127277155, 2.3385281968695693]
7      [2.0831140755567814, 5.94063873401418]
8     [0.4887113296319978, 6.380570614449363]
9     [2.844815261473105, 0.9146457613970793]
Name: minimum, dtype: object

answered Feb 17, 2022 at 4:22

Shubham Sharma

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Comments

Tikkaty · Accepted Answer · 2022-02-17 04:33:39Z

2

try this:

import pandas as pd
import numpy as np

xdf = pd.DataFrame({'a':np.random.rand(1,10)[0]*10,'b':np.random.rand(1,10)[0]*10,'c':np.random.rand(1,10)[0]*10,'d':np.random.rand(1,10)[0]*10,},index=np.arange(0,10,1))

print(xdf)

ab = xdf['ab_min'] = xdf[['a','b']].min(axis=1)
cd = xdf['cd_min'] = xdf[['c','d']].min(axis=1)
blah = pd.concat([ab, cd], axis=1)

print(blah)

results:

answered Feb 17, 2022 at 4:33

Tikkaty

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Comments

Derek O · Accepted Answer · 2022-02-17 04:20:41Z

1

You can use .apply with a lambda function along axis=1:

xdf['minimum'] = xdf.apply(lambda x: [x[['a','b']].min(),x[['c','d']].min()], axis=1)

Result:

>>> xdf
          a         b         c         d                                    minimum
0  0.662634  4.166338  8.864823  9.004818    [0.6626341544146663, 8.864822751494284]
1  6.854054  6.163417  6.510728  0.049498   [6.163416966676091, 0.04949754019059838]
2  6.389760  4.462319  2.435369  3.732534    [4.462318678134215, 2.4353686460846893]
3  4.628735  7.571098  1.900726  9.046384    [4.628735362058981, 1.9007255361271058]
4  3.203285  4.364302  2.473973  2.911911    [3.203285015796596, 2.4739732602476727]
5  5.357440  3.166420  9.908758  0.910704      [3.166420385020304, 0.91070444348338]
6  8.120486  6.395869  0.970977  5.278279    [6.395868901095546, 0.9709769503958143]
7  1.574765  7.184971  3.835641  4.495135     [1.574765093192545, 3.835640598199231]
8  8.688497  0.069061  0.771772  8.971878  [0.06906065557899743, 0.7717717844423222]
9  5.455920  2.630342  1.966357  7.374366    [2.6303421168291843, 1.966357159086991]

answered Feb 17, 2022 at 4:20

Derek O

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2 Comments

Mainland Over a year ago

You nailed it. Thanks. I tried .apply but couldn't know how to iterate through two columns only. Also, my actual dataframe has 12 columns. I don't want to mention the column names but considered xdf.apply(lambda x: [x[xdf.columns[0:2]].min(),x[xdf.columns[2:4]].min()], axis=1)

Derek O Over a year ago

@Mainland yeah that would work too but has the possible drawback of hardcoding the location of the columns. glad my answer was helpful!

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