pandas 熊猫系列均值和标准差
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Pandas series mean and standard deviation
提问by magicsword
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. 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)
'A' 上的数据是 Pandas 系列。我想计算“A”中数据的平均值和标准偏差（A 有几条记录），例如：(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)
采纳答案by jezrael
You can use list comprehensionwith concatand then meanor std.
您可以使用list comprehensionwithconcat和 thenmean或std。
For converting to float(int) add astype, if still problem need to_numericwith parameter errors='coerce'.
对于转换为float( int) add astype，如果仍然有问题需要to_numeric参数errors='coerce'。
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:
另一种解决方案：
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
