pandas 将系列设置为索引
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Setting Series as index
提问by zthomas.nc
I'm using python 2.7 to take a numerical column of my dataframe dataand make it an individual object (series) with an index of dates that is another column from data.
我正在使用 python 2.7 来获取我的数据框的数字列，data并使其成为一个单独的对象（系列），其日期索引是data.
new_series = pd.Series(data['numerical_column'] , index=data['dates'])
However, when I do this, I get a bunch of NaNvalues in the Series:
但是，当我这样做时，我NaN在系列中得到了一堆值：
dates
1980-01-31 NaN
1980-02-29 NaN
1980-03-31 NaN
1980-04-30 NaN
1980-05-31 NaN
1980-06-30 NaN
...
Why does my numerical_datavalues just disappear?
为什么我的numerical_data价值观消失了？
I realize that I can apparently achieve this goal by doing the following, although I'm curious why my initial approach failed.
我意识到我显然可以通过执行以下操作来实现这个目标，尽管我很好奇为什么我最初的方法失败了。
new_series = data.set_index('dates')['numerical_column']
回答by jezrael
I think there is problem with not align index of column data['numerical_column'].
我认为不对齐 column 索引存在问题data['numerical_column']。
So need convert it to numpy arrayby values:
因此需要将其转换为numpy array通过values：
new_series = pd.Series(data['numerical_column'].values , index=data['dates'])
Sample:
样本：
import pandas as pd
import datetime
data = pd.DataFrame({
'dates': {0: datetime.date(1980, 1, 31), 1: datetime.date(1980, 2, 29),
2: datetime.date(1980, 3, 31), 3: datetime.date(1980, 4, 30),
4: datetime.date(1980, 5, 31), 5: datetime.date(1980, 6, 30)},
'numerical_column': {0: 1, 1: 4, 2: 5, 3: 3, 4: 1, 5: 0}})
print (data)
dates numerical_column
0 1980-01-31 1
1 1980-02-29 4
2 1980-03-31 5
3 1980-04-30 3
4 1980-05-31 1
5 1980-06-30 0
new_series = pd.Series(data['numerical_column'].values , index=data['dates'])
print (new_series)
dates
1980-01-31 1
1980-02-29 4
1980-03-31 5
1980-04-30 3
1980-05-31 1
1980-06-30 0
dtype: int64
But method with set_indexis nicer, but slowier:
但方法 withset_index更好，但更慢：
#[60000 rows x 2 columns]
data = pd.concat([data]*10000).reset_index(drop=True)
In [65]: %timeit pd.Series(data['numerical_column'].values , index=data['dates'])
1000 loops, best of 3: 308 μs per loop
In [66]: %timeit data.set_index('dates')['numerical_column']
1000 loops, best of 3: 1.28 ms per loop
Verification:
验证：
If index of column has same index, it works nice:
如果列的索引具有相同的索引，则效果很好：
s = data.set_index('dates')['numerical_column']
df = s.to_frame()
print (df)
numerical_column
dates
1980-01-31 1
1980-02-29 4
1980-03-31 5
1980-04-30 3
1980-05-31 1
1980-06-30 0
new_series = pd.Series(df['numerical_column'] , index=data['dates'])
print (new_series)
dates
1980-01-31 1
1980-02-29 4
1980-03-31 5
1980-04-30 3
1980-05-31 1
1980-06-30 0
Name: numerical_column, dtype: int64
