Python：计算pandas系列中值的累积出现次数

Question

我有一个看起来像这样的DataFrame：

    fruit
0  orange
1  orange
2  orange
3    pear
4  orange
5   apple
6   apple
7    pear
8    pear
9  orange

我想添加一个列，计算每个值的累积次数，即

    fruit  cum_count
0  orange          1
1  orange          2
2  orange          3
3    pear          1
4  orange          4
5   apple          1
6   apple          2
7    pear          2
8    pear          3
9  orange          5

目前我这样做：

df['cum_count'] = [(df.fruit[0:i+1] == x).sum() for i, x in df.fruit.iteritems()]

...这对10行很好，但是当我尝试用几百万行做同样的事情时需要很长时间。 有没有更有效的方法来做到这一点？

Answer 1

你可以使用groupby和cumcount ：

df['cum_count'] = df.groupby('fruit').cumcount() + 1

In [16]: df
Out[16]:
    fruit  cum_count
0  orange          1
1  orange          2
2  orange          3
3    pear          1
4  orange          4
5   apple          1
6   apple          2
7    pear          2
8    pear          3
9  orange          5

定时

In [8]: %timeit [(df.fruit[0:i+1] == x).sum() for i, x in df.fruit.iteritems()]
100 loops, best of 3: 3.76 ms per loop

In [9]: %timeit df.groupby('fruit').cumcount() + 1
1000 loops, best of 3: 926 µs per loop

所以它的速度提高了4倍。

Answer 2

也许更好的是将groupby与cumcount一起使用指定列，因为它更有效：

df['cum_count'] = df.groupby('fruit' )['fruit'].cumcount() + 1
print df

    fruit  cum_count
0  orange          1
1  orange          2
2  orange          3
3    pear          1
4  orange          4
5   apple          1
6   apple          2
7    pear          2
8    pear          3
9  orange          5

比较len(df) = 10 ，我的解决方案是最快的：

In [3]: %timeit df.groupby('fruit')['fruit'].cumcount() + 1
The slowest run took 11.67 times longer than the fastest. This could mean that an intermediate result is being cached 
1000 loops, best of 3: 299 µs per loop

In [4]: %timeit df.groupby('fruit').cumcount() + 1
The slowest run took 12.78 times longer than the fastest. This could mean that an intermediate result is being cached 
1000 loops, best of 3: 921 µs per loop

In [5]: %timeit [(df.fruit[0:i+1] == x).sum() for i, x in df.fruit.iteritems()]
The slowest run took 4.47 times longer than the fastest. This could mean that an intermediate result is being cached 
100 loops, best of 3: 2.72 ms per loop

比较len(df) = 10k ：

In [7]: %timeit df.groupby('fruit')['fruit'].cumcount() + 1
The slowest run took 4.65 times longer than the fastest. This could mean that an intermediate result is being cached 
1000 loops, best of 3: 845 µs per loop

In [8]: %timeit df.groupby('fruit').cumcount() + 1
The slowest run took 5.59 times longer than the fastest. This could mean that an intermediate result is being cached 
100 loops, best of 3: 1.59 ms per loop

In [9]: %timeit [(df.fruit[0:i+1] == x).sum() for i, x in df.fruit.iteritems()]
1 loops, best of 3: 5.12 s per loop