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
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
解决方法:
from functools import reduce
reduce(np.fmin, [ws1.C, ws2.C, ws3.C])
index
0 4.28
1 4.10
Name: C, dtype: float64
这很好地概括了理解
reduce(np.fmin, [w.C for w in [ws1, ws2, ws3, ws4, ws5]])
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
您还可以在字典上使用pd.concat,并将pd.Series.min与level = 1参数一起使用
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
注意:
dict(enumerate([w.C for w in [ws1, ws2, ws3]]))
是另一种说法
{0: ws1.C, 1: ws2.C, 2: ws3.C}