从python中的列表中提取数据

Question

我有以下清单：

[N               12.000000
 mean             2.011608
 median           2.021611
 std              0.572034
 relative std     0.284350
 dtype: float64,
 N               12.000000
 mean             2.011608
 median           2.021611
 std              0.571815
 relative std     0.284262
 dtype: float64,
 N               12.000000
 mean             2.011608
 median           2.021611
 std              0.572101
 relative std     0.284412
 dtype: float64,
 N               12.000000
 mean             2.011608
 median           2.021611
 std              0.572115
 relative std     0.284440
 dtype: float64,
 N               12.000000
 mean             2.011608
 median           2.021611
 std              0.571872
 relative std     0.284313
 dtype: float64]

我想从具有最小相对标准值的列表中提取数据（N、均值、标准、相对标准）。上面列表的输出应该如下：

 N               12.000000
 mean             2.011608
 median           2.021611
 std              0.571815
 relative std     0.284262
 dtype: float64

到目前为止我尝试了什么？

min(list)

但是抛出以下错误ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().

Answer 1

使用 min 和 lambda 函数通过标签 std 选择 Series 的值：

s1 = pd.Series([0,1,2], index=['std','min','max'])
s2 = pd.Series([4,1,2], index=['std','min','max'])
s3 = pd.Series([0.8,1,2], index=['std','min','max'])

L = [s1, s2, s3]

s = min(L, key=lambda x:x.loc['std'])
print (s)
std    0
min    1
max    2
dtype: int64

测试所有最小值：

print ([x.loc['std'] for x in L])
[0, 4, 0.8]

对于索引使用np.argmin:

import numpy as np
print (np.argmin([x.loc['std'] for x in L]))
0