pandas 熊猫计算唯一行
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Pandas Counting Unique Rows
提问by qwertylpc
I have a pandas data frame similar to:
我有一个类似于以下内容的Pandas数据框：
ColA ColB
1 1
1 1
1 1
1 2
1 2
2 1
3 2
I want an output that has the same function as Counter. I need to know how many time each row appears (with all of the columns being the same.
我想要一个与Counter具有相同功能的输出。我需要知道每行出现多少次（所有列都相同。
In this case the proper output would be:
在这种情况下，正确的输出将是：
ColA ColB Count
1 1 3
1 2 2
2 1 1
3 2 1
I have tried something of the sort:
我尝试过这样的事情：
df.groupby(['ColA','ColB']).ColA.count()
but this gives me some ugly output I am having trouble formatting
但这给了我一些丑陋的输出我在格式化时遇到问题
回答by jezrael
You can use sizewith reset_index:
你可以用size与reset_index：
print df.groupby(['ColA','ColB']).size().reset_index(name='Count')
ColA ColB Count
0 1 1 3
1 1 2 2
2 2 1 1
3 3 2 1
回答by eddygeek
I only needed to count the unique rowsand have used this alternative:
我只需要计算唯一行并使用了这个替代方案：
len(df[['ColA','ColB']].drop_duplicates())
For this task, on my data, it was twice faster than len(df.groupby(['ColA','ColB']))like in the above, more general solution.
对于这个任务，在我的数据上，它比len(df.groupby(['ColA','ColB']))上面更通用的解决方案快两倍。
