在滚动窗口中取第一个和最后一个值

本文介绍了在滚动窗口中取第一个和最后一个值的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

使用pandas，我想应用可用于resample() 但不适用于rolling() 的函数.

Using pandas, I would like to apply function available for resample() but not for rolling().

这有效:

df1 = df.resample(to_freq, closed='left', kind='period', ).agg(OrderedDict([('Open', 'first'), ('Close', 'last'), ]))

这不会:

df2 = df.rolling(my_indexer).agg( OrderedDict([('Open', 'first'), ('Close', 'last') ])) >>> AttributeError: 'first' is not a valid function for 'Rolling' object df3 = df.rolling(my_indexer).agg( OrderedDict([ ('Close', 'last') ])) >>> AttributeError: 'last' is not a valid function for 'Rolling' object

对于将滚动窗口的第一个和最后一个值保留在两个不同的列中，您有什么建议?

What would be your advice to keep first and last value of a rolling windows to be put into two different columns?

import pandas as pd from random import seed from random import randint from collections import OrderedDict # DataFrame ts_1h = pd.date_range(start='2020-01-01 00:00+00:00', end='2020-01-02 00:00+00:00', freq='1h') seed(1) values = [randint(0,10) for ts in ts_1h] df = pd.DataFrame({'Values' : values}, index=ts_1h) # First & last work with resample resampled_first = df.resample('3H', closed='left', kind='period', ).agg(OrderedDict([('Values', 'first')])) resampled_last = df.resample('3H', closed='left', kind='period', ).agg(OrderedDict([('Values', 'last')])) # They don't with rolling rolling_first = df.rolling(3).agg(OrderedDict([('Values', 'first')])) rolling_first = df.rolling(3).agg(OrderedDict([('Values', 'last')]))

感谢您的帮助！最好的，

Thanks for your help! Bests,

推荐答案

你可以使用自己的函数获取滚动窗口中的第一个或最后一个元素

You can use own function to get first or last element in rolling window

rolling_first = df.rolling(3).agg(lambda rows: rows[0]) rolling_last = df.rolling(3).agg(lambda rows: rows[-1])

示例

import pandas as pd from random import seed, randint # DataFrame ts_1h = pd.date_range(start='2020-01-01 00:00+00:00', end='2020-01-02 00:00+00:00', freq='1h') seed(1) values = [randint(0, 10) for ts in ts_1h] df = pd.DataFrame({'Values' : values}, index=ts_1h) df['first'] = df['Values'].rolling(3).agg(lambda rows: rows[0]) df['last'] = df['Values'].rolling(3).agg(lambda rows: rows[-1]) print(df)

结果

Values first last 2020-01-01 00:00:00+00:00 2 NaN NaN 2020-01-01 01:00:00+00:00 9 NaN NaN 2020-01-01 02:00:00+00:00 1 2.0 1.0 2020-01-01 03:00:00+00:00 4 9.0 4.0 2020-01-01 04:00:00+00:00 1 1.0 1.0 2020-01-01 05:00:00+00:00 7 4.0 7.0 2020-01-01 06:00:00+00:00 7 1.0 7.0 2020-01-01 07:00:00+00:00 7 7.0 7.0 2020-01-01 08:00:00+00:00 10 7.0 10.0 2020-01-01 09:00:00+00:00 6 7.0 6.0 2020-01-01 10:00:00+00:00 3 10.0 3.0 2020-01-01 11:00:00+00:00 1 6.0 1.0 2020-01-01 12:00:00+00:00 7 3.0 7.0 2020-01-01 13:00:00+00:00 0 1.0 0.0 2020-01-01 14:00:00+00:00 6 7.0 6.0 2020-01-01 15:00:00+00:00 6 0.0 6.0 2020-01-01 16:00:00+00:00 9 6.0 9.0 2020-01-01 17:00:00+00:00 0 6.0 0.0 2020-01-01 18:00:00+00:00 7 9.0 7.0 2020-01-01 19:00:00+00:00 4 0.0 4.0 2020-01-01 20:00:00+00:00 3 7.0 3.0 2020-01-01 21:00:00+00:00 9 4.0 9.0 2020-01-01 22:00:00+00:00 1 3.0 1.0 2020-01-01 23:00:00+00:00 5 9.0 5.0 2020-01-02 00:00:00+00:00 0 1.0 0.0

使用字典你必须直接输入lambda，而不是字符串

Using dictionary you have to put directly lambda, not string

result = df['Values'].rolling(3).agg({'first': lambda rows: rows[0], 'last': lambda rows: rows[-1]}) print(result)

和自己的函数一样——你必须输入它的名字，而不是带有名字的字符串

The same with own function - you have to put its name, not string with name

def first(rows): return rows[0] def last(rows): return rows[-1] result = df['Values'].rolling(3).agg({'first': first, 'last': last}) print(result)

示例

import pandas as pd from random import seed, randint # DataFrame ts_1h = pd.date_range(start='2020-01-01 00:00+00:00', end='2020-01-02 00:00+00:00', freq='1h') seed(1) values = [randint(0, 10) for ts in ts_1h] df = pd.DataFrame({'Values' : values}, index=ts_1h) result = df['Values'].rolling(3).agg({'first': lambda rows: rows[0], 'last': lambda rows: rows[-1]}) print(result) def first(rows): return rows[0] def mylast(rows): return rows[-1] result = df['Values'].rolling(3).agg({'first': first, 'last': last}) print(result)