在 python 中，如何计算 3 个或更多数据帧的 MAX 和 MIN

Question

How to calculate MAX and MIN for 3 or more dataframes
I can do calculate the difference between prices simply by adding this line of code

diff['list2-list1'] = diff['price2'] - diff['price1']

but it not work to calculate MIN with

diff['min'] =  (df1,df2.df3).min()

or

diff['min'] = (diff['price2'],diff['price1'],diff['price3']).min()

or

diff['min'] = (diff['price2'],diff['price1'],diff['price3']).idxmin()

and do not print result of if in new column when latest list (list3) have minimum value

if diff['min'] == diff['price3']
     diff['Lowest now?'] = "yes"

The python code I have

import pandas 
import numpy as np
import csv
from csv_diff import load_csv, compare



df1 = pandas.read_csv('list1.csv')
df1['version'] = 'list1'

df2 = pandas.read_csv('list2.csv')
df2['version'] = 'list2'

df3 = pandas.read_csv('list3.csv')
df3['version'] = 'list3'

# keep only columns 'version', 'ean', 'price'
diff = df1.append([df2,df3])[['version', 'ean','price']]


# keep only duplicated eans, which will only occur
# for eans in both original lists
diff = diff[diff['ean'].duplicated(keep=False)]
# perform a pivot https://pandas.pydata.org/pandas-docs/stable/user_guide/reshaping.html
diff = diff.pivot_table(index='ean', columns='version', values='price', aggfunc='first')

# back to a normal dataframe
diff = diff.reset_index()
diff.columns.name = None

# rename columns and keep only what we want
diff = diff.rename(columns={'list1': 'price1', 'list2': 'price2', 'list3': 'price3'})[['ean', 'price1', 'price2','price3']]
diff['list2-list1'] = diff['price2'] - diff['price1']
diff['list3-list2'] = diff['price3'] - diff['price2']

diff['min'] =  (df1,df2).min()
if diff['min'] == diff['price3']
     diff['Lowest now?'] = "yes"

diff.to_csv('diff.csv')

more information

headers of list1,lsit2,list3 are the same

price,ean,unit 

example of list1

price,ean,unit 
143.80,2724316972629,0
125.00,2724456127521,0
158.00,2724280705919,0
19.99,2724342954019,0
20.00,2724321942662,0
212.00,2724559841560,0
1322.98,2724829673686

example of list2

price,ean,unit 
55.80,2724316972629,0
15.00,2724456127521,0
66.00,2724559841560,0
1622.98,2724829673686,0

example of list3

price,ean,unit 
139.99,2724342954019,0
240.00,2724321942662,0
252.00,2724559841560,0
1422.98,2724829673686,0

Answer 1

There you go:

data = pd.concat([df1, df2, df3], axis=1).fillna(0).astype('float')
data['minimum_price'] = data['price'].min(1)
data['maximum_price'] = data['price'].max(1)

Out:

     price           ean  units    price           ean  units    price           ean  units  minimum_price  maximum_price
0   143.80  2.724317e+12    0.0    55.80  2.724317e+12    0.0   139.99  2.724343e+12    0.0          55.80         143.80
1   125.00  2.724456e+12    0.0    15.00  2.724456e+12    0.0   240.00  2.724322e+12    0.0          15.00         240.00
2   158.00  2.724281e+12    0.0    66.00  2.724560e+12    0.0   252.00  2.724560e+12    0.0          66.00         252.00
3    19.99  2.724343e+12    0.0  1622.98  2.724830e+12    0.0  1422.98  2.724830e+12    0.0          19.99        1622.98
4    20.00  2.724322e+12    0.0     0.00  0.000000e+00    0.0     0.00  0.000000e+00    0.0           0.00          20.00
5   212.00  2.724560e+12    0.0     0.00  0.000000e+00    0.0     0.00  0.000000e+00    0.0           0.00         212.00
6  1322.98  2.724830e+12    0.0     0.00  0.000000e+00    0.0     0.00  0.000000e+00    0.0           0.00        1322.98

Answer 2

Assuming the dataframes have the same columns, you can use pd.concat .

min = pd.concat(df1, df2,df3).min() .