本文介绍了使用Pandas GroupBy和size()/ count()来生成聚合的DataFrame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

所以我现在有一个名为 df 的DataFrame:

So I currently have a DataFrame called df that goes:

date                       tag
2011-02-18 12:57:00-07:00  A
2011-02-19 12:57:00-07:00  A
2011-03-18 12:57:00-07:00  B
2011-04-01 12:57:00-07:00  C
2011-05-19 12:57:00-07:00  Z
2011-06-03 12:57:00-07:00  A
2011-06-05 12:57:00-07:00  A
...

我正在尝试使用GroupBy标签,日期(年/月),所以它看起来像:

I'm trying to do a GroupBy the tag, and the date (yr/month), so it looks like:

date     A  B  C  Z
2011-02  2  0  0  0
2011-03  0  1  0  0
2011-04  0  0  1  0
2011-05  0  0  0  1
2011-06  2  0  0  0
...

我尝试了以下操作,但并不能给我所需的。

I've tried the following, but it doesn't quite give me what I want.

grouped_series = df.groupby([["%s-%s" % (d.year, d.month) for d in df.date], df.tag]).size()

我知道哪个标签存在,任何帮助将不胜感激。

I know which tag exists etc. Any help will be greatly appreciated.

更新(适用于未来的人):

的字符串格式。相信我,在绘图时会更好:

Ended up keeping the datetime, instead of string format. Trust me, this will be better when plotting:

grouped_df = df.groupby([[ datetime.datetime(d.year, d.month, 1, 0, 0) for d in df.date], df.name]).size()
grouped_df = grouped_df.unstack().fillna(0)


推荐答案

你可以使用和方法:

you could use unstack() and fillna() methods:

>>> g = df.groupby([["%s-%s" % (d.year, d.month) for d in df.date], df.tag]).size()
>>> g
        tag
2011-2  A      2
2011-3  B      1
2011-4  C      1
2011-5  Z      1
2011-6  A      2
dtype: int64
>>> g.unstack().fillna(0)
tag     A  B  C  Z
2011-2  2  0  0  0
2011-3  0  1  0  0
2011-4  0  0  1  0
2011-5  0  0  0  1
2011-6  2  0  0  0

这篇关于使用Pandas GroupBy和size()/ count()来生成聚合的DataFrame的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!

08-11 14:04