我有一个如下所示的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,计算每个值的累计出现次数,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?
答案 0 :(得分:4)
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
<强>时序强>
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4次更快。
答案 1 :(得分:1)
使用带有groupby
的cumcount
和指定列可能更好,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