pandas 中值的唯一从零开始的 id [英] Unique zero-based id for values in pandas

问题描述

我在带有标识符列的 DataFrame 中有一些数据.

I have some data in a DataFrame with an identifier column.

data = DataFrame({'id' : [50,50,30,10,50,50,30]})

对于每个唯一的 id，我想提出一个新的唯一标识符.我希望 id 是从 0 开始的连续整数.这是我目前所拥有的:

For each unique id, I want to come up with a new unique identifier. I'd like the ids to be sequential integers starting at 0. Here's what I have so far:

unique = data[['id']].drop_duplicates()   
unique['group'] = np.arange(len(unique))
unique.set_index('id')
data = data.merge(unique, 'inner', on = 'id')

这可行，但似乎有点脏.有没有更好的办法?

This works but seems a little dirty. Is there a better way?

推荐答案

这就是 pandas.factorize 确实:

That is what pandas.factorize does:

data = pd.DataFrame({'id' : [50,50,30,10,50,50,30]})
print pd.factorize(data.id)[0]

输出:

[0 0 1 2 0 0 1]

numpy.unique 也可以这样做:

numpy.unique can also do this:

import numpy as np
print np.unique([50,50,30,10,50,50,30], return_inverse=True)[1]

输出:

array([2, 2, 1, 0, 2, 2, 1])

numpy.unique 输出的索引是按值排序的，所以将最小值 10 赋给索引 0.如果要使用 factorize 得到这个结果，请设置sort 给 True 的参数:

the index outputed by numpy.unique is sorted by value, so the smallest value 10 is assigend to index 0. If you want this result by using factorize, set sort argument to True:

pandas.factorize(data.id, sort=True)[0]

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