[英]Pandas series mean and standard deviation
I have a list: 我有一个清单:
data = [
{'A': [2.0, 3.0, 4.0, 5.0, 6.0], 'B':[27.0, 28.0, 29.0, 30.0], 'C': ['lic1'],
'D': ['soy1'], 'E': ['foo1']},
{'A': [7.0, 11.0, 90.0, 43.0, 87.0], 'B':[27.0, 28.0, 29.0, 30.0], 'C': ['lic1'],
'D': ['soy1'], 'E': ['foo1']},
# ... etc
] ]
The data on 'A' is a Pandas Series. “A”的数据是熊猫系列。 I would like to compute the average and standard deviation for the data in 'A' (there are several records for A) for example: (mean=(2.0+3.0+4.0+5.0+6.0+7.0+11.0+90.0+43.0+87.0)/len(A)=25.8) 我想计算'A'中数据的平均值和标准差(例如,A有几个记录):(均值=(2.0 + 3.0 + 4.0 + 5.0 + 6.0 + 7.0 + 11.0 + 90.0 + 43.0 + 87.0)/ LEN(A)= 25.8)
You can use list comprehension
with concat
and then mean
or std
. 您可以将list comprehension
与concat
,然后使用mean
或std
。
For converting to float
( int
) add astype
, if still problem need to_numeric
with parameter errors='coerce'
. 为了转换为float
( int
)添加astype
,如果仍然有问题需要带参数errors='coerce'
to_numeric
。
s = pd.concat([pd.Series(x['A']) for x in data]).astype(float)
print (s)
0 2.0
1 3.0
2 4.0
3 5.0
4 6.0
0 7.0
1 11.0
2 90.0
3 43.0
4 87.0
dtype: float64
print (s.mean())
25.8
print (s.std())
35.15299892375234
Another solution: 另一种方案:
from itertools import chain
s = pd.Series(list(chain.from_iterable([x['A'] for x in data]))).astype(float)
print (s)
0 2.0
1 3.0
2 4.0
3 5.0
4 6.0
5 7.0
6 11.0
7 90.0
8 43.0
9 87.0
dtype: float64
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