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python pandas字符串列时间滚动不重复计数

[英]python pandas string column time rolling distinct count

As I want to count the unique number of column A in a moving time window(60 seconds): 由于我想计算移动时间窗口(60秒)中列A的唯一数量:

fn = lambda x: len(np.unique(x)) 
df = pd.DataFrame({'A':['a', 'b', 'a', 'b', 'e'], 'B': [0, 1, 2, 3, 4]},
                index = [pd.Timestamp('20130101 09:01:00'),
                         pd.Timestamp('20130101 09:01:32'),
                         pd.Timestamp('20130101 09:02:03'),
                         pd.Timestamp('20130101 09:02:25'),
                         pd.Timestamp('20130101 09:03:06')])


df[['A']].rolling('60s').apply(fn)

I expect the result as 我期望结果为

2013-01-01 09:01:00 1
2013-01-01 09:01:32 2
2013-01-01 09:02:03 2
2013-01-01 09:02:25 2
2013-01-01 09:03:06 2

however, the result is: 但是,结果是:

2013-01-01 09:01:00 a
2013-01-01 09:01:32 b
2013-01-01 09:02:03 a
2013-01-01 09:02:25 b
2013-01-01 09:03:06 e

what's the problem? 有什么问题?

You can use column B instead A : 您可以使用B列而不是A列:

a = df[['B']].rolling('60s').apply(fn)
print (a)
                       B
2013-01-01 09:01:00  1.0
2013-01-01 09:01:32  2.0
2013-01-01 09:02:03  2.0
2013-01-01 09:02:25  3.0
2013-01-01 09:03:06  2.0

And if need convert to int : 如果需要转换为int

a = df[['B']].rolling('60s').apply(fn).astype(int)
print (a)
                     B
2013-01-01 09:01:00  1
2013-01-01 09:01:32  2
2013-01-01 09:02:03  2
2013-01-01 09:02:25  3
2013-01-01 09:03:06  2

If no column you can create it: 如果没有列,则可以创建它:

a = df.assign(B=np.arange(len(df.index)))[['B']].rolling('60s').apply(fn).astype(int)
print (a)
                     B
2013-01-01 09:01:00  1
2013-01-01 09:01:32  2
2013-01-01 09:02:03  2
2013-01-01 09:02:25  3
2013-01-01 09:03:06  2

df['B'] = np.arange(len(df.index))
a = df[['B']].rolling('60s').apply(fn).astype(int)
print (a)
                     B
2013-01-01 09:01:00  1
2013-01-01 09:01:32  2
2013-01-01 09:02:03  2
2013-01-01 09:02:25  3
2013-01-01 09:03:06  2

EDIT1: EDIT1:

df['B'] = np.arange(len(df.index))
a = df.groupby('A')[['B']].rolling('60s').apply(fn).astype(int)
print (a)
                       B
A                       
a 2013-01-01 09:01:00  1
  2013-01-01 09:02:03  1
b 2013-01-01 09:01:32  1
  2013-01-01 09:02:25  2
e 2013-01-01 09:03:06  1

Simply try this way : 只需尝试这种方式:

In [40]: import pandas as pd

In [41]: fn = lambda x: len(np.unique(x)) 
    ...: df = pd.DataFrame({'A':['a', 'b', 'c', 'd', 'e'], 'B': [0, 1, 2, 3, 4]},
    ...:                 index = [pd.Timestamp('20130101 09:01:00'),
    ...:                          pd.Timestamp('20130101 09:01:32'),
    ...:                          pd.Timestamp('20130101 09:02:03'),
    ...:                          pd.Timestamp('20130101 09:02:25'),
    ...:                          pd.Timestamp('20130101 09:03:06')])

In [42]: df[['B']] = df[['B']].rolling('60s').apply(fn).astype(int)

In [43]: df[['']] = df[['B']]

In [44]: df[['']]
Out[44]: 

2013-01-01 09:01:00  1
2013-01-01 09:01:32  2
2013-01-01 09:02:03  2
2013-01-01 09:02:25  3
2013-01-01 09:03:06  2

In [45]: 

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