为 pandas.DataFrame 复制 GROUP_CONCAT
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Replicating GROUP_CONCAT for pandas.DataFrame
提问by Mitch Flax
I have a pandas DataFrame df:
我有一个Pandas数据帧 df：
+------+---------+
| team | user |
+------+---------+
| A | elmer |
| A | daffy |
| A | bugs |
| B | dawg |
| A | foghorn |
| B | speedy |
| A | goofy |
| A | marvin |
| B | pepe |
| C | petunia |
| C | porky |
+------+---------
I want to find or write a function to return a DataFrame that I would return in MySQL using the following:
我想找到或编写一个函数来返回一个我将使用以下命令在 MySQL 中返回的数据帧：
SELECT
team,
GROUP_CONCAT(user)
FROM
df
GROUP BY
team
for the following result:
对于以下结果：
+------+---------------------------------------+
| team | group_concat(user) |
+------+---------------------------------------+
| A | elmer,daffy,bugs,foghorn,goofy,marvin |
| B | dawg,speedy,pepe |
| C | petunia,porky |
+------+---------------------------------------+
I can think of nasty ways to do this by iterating over rows and adding to a dictionary, but there's got to be a better way.
我可以想到通过迭代行并添加到字典中来做到这一点的讨厌的方法，但必须有更好的方法。
回答by Phillip Cloud
Do the following:
请执行下列操作：
df.groupby('team').apply(lambda x: ','.join(x.user))
to get a Seriesof strings or
得到一个Series字符串或
df.groupby('team').apply(lambda x: list(x.user))
to get a Seriesof lists of strings.
得到Series的list字符串秒。
Here's what the results look like:
结果如下：
In [33]: df.groupby('team').apply(lambda x: ', '.join(x.user))
Out[33]:
team
a elmer, daffy, bugs, foghorn, goofy, marvin
b dawg, speedy, pepe
c petunia, porky
dtype: object
In [34]: df.groupby('team').apply(lambda x: list(x.user))
Out[34]:
team
a [elmer, daffy, bugs, foghorn, goofy, marvin]
b [dawg, speedy, pepe]
c [petunia, porky]
dtype: object
Note that in general any further operations on these types of Serieswill be slow and are generally discouraged. If there's another way to aggregate without putting a listinside of a Seriesyou should consider using that approach instead.
请注意，通常对这些类型的任何进一步操作Series都会很慢并且通常不鼓励。如果有另一种聚合方式而不将 alist放入内部，Series您应该考虑使用该方法。
回答by Kamil Sindi
A more general solution if you want to use agg:
如果您想使用更通用的解决方案agg：
df.groupby('team').agg({'user' : lambda x: ', '.join(x)})
