使用pandas在多个工作表中查找最小值

How can I find the minimum values among multiple worksheets for each index across total worksheet

如何在整个工作表中找到每个索引的多个工作表中的最小值

suppose,

假设，

  worksheet 1

    index    A   B   C
       0     2   3   4.28
       1     3   4   5.23
    worksheet 2

    index    A   B   C
        0    9   6   5.9
        1    1   3   4.1

    worksheet 3

    index    A   B   C
        0    9   6   6.0
        1    1   3   4.3
 ...................(Worksheet 4,Worksheet 5)...........
by comparing C column, I want an answer, where dataframe looks like

index      min(c)
    0       4.28
    1       4.1

2 个解决方案

#1

from functools import reduce

reduce(np.fmin, [ws1.C, ws2.C, ws3.C])

index
0    4.28
1    4.10
Name: C, dtype: float64

This generalizes nicely with a comprehension

这很好地概括了理解

reduce(np.fmin, [w.C for w in [ws1, ws2, ws3, ws4, ws5]])

If you must insist on your column name

如果您必须坚持您的列名称

from functools import reduce

reduce(np.fmin, [ws1.C, ws2.C, ws3.C]).to_frame('min(C)')

       min(C)
index        
0        4.28
1        4.10

You can also use pd.concat on a dictionary and use pd.Series.min with the level=1 parameter

您还可以在字典上使用pd.concat，并使用pd.Series.min和level = 1参数

pd.concat(dict(enumerate([w.C for w in [ws1, ws2, ws3]]))).min(level=1)
# equivalently
# pd.concat(dict(enumerate([w.C for w in [ws1, ws2, ws3]])), axis=1).min(1)

index
0    4.28
1    4.10
Name: C, dtype: float64

Note:

注意：

dict(enumerate([w.C for w in [ws1, ws2, ws3]]))

is another way of saying

是另一种说法

{0: ws1.C, 1: ws2.C, 2: ws3.C}

#2

You need read_excel with parameter sheetname=None for OrderedDicts from all sheetnames and then list comprehension with reduce with numpy.fmin:

对于所有工作表名称中的OrderedDicts，您需要带参数sheetname = None的read_excel，然后使用numpy.fmin的reduce列表理解：

dfs = pd.read_excel('file.xlsx', sheetname=None)
print (dfs)
OrderedDict([('Sheet1',    A  B     C
0  2  3  4.28
1  3  4  5.23), ('Sheet2',    A  B    C
0  9  6  5.9
1  1  3  4.1), ('Sheet3',    A  B    C
0  9  6  6.0
1  1  3  4.3)])

from functools import reduce

df = reduce(np.fmin, [v['C'] for k,v in dfs.items()])
print (df)
0    4.28
1    4.10
Name: C, dtype: float64

Solution with concat:

concat的解决方案：

df = pd.concat([v['C'] for k,v in dfs.items()],axis=1).min(axis=1)
print (df)
0    4.28
1    4.10
dtype: float64

If need define index in read_excel:

如果需要在read_excel中定义索引：

dfs = pd.read_excel('file.xlsx', sheetname=None, index_col='index')
print (dfs)
OrderedDict([('Sheet1',        A  B     C
index            
0      2  3  4.28
1      3  4  5.23), ('Sheet2',        A  B    C
index           
0      9  6  5.9
1      1  3  4.1), ('Sheet3',        A  B    C
index           
0      9  6  6.0
1      1  3  4.3)])


df = pd.concat([v['C'] for k,v in dfs.items()], axis=1).min(axis=1)
print (df)
index
0    4.28
1    4.10
dtype: float64

#1

from functools import reduce

reduce(np.fmin, [ws1.C, ws2.C, ws3.C])

index
0    4.28
1    4.10
Name: C, dtype: float64