根据值对列表编号的更优雅方式 [英] More Elegant way to number a list according to values

问题描述

我想根据值将列表映射成数字.

I would like to map a list into numbers according to the values.

例如:

['aa', 'b', 'b', 'c', 'aa', 'b', 'a'] -> [0, 1, 1, 2, 0, 1, 3]

我正在尝试通过使用numpy和映射字典来实现这一目标.

I'm trying to achieve this by using numpy and a mapping dict.

def number(lst):
    x = np.array(lst)
    unique_names = list(np.unique(x))
    mapping = dict(zip(unique_names, range(len(unique_names)))) # Translating dict
    map_func = np.vectorize(lambda name: d[name])
    return map_func(x)

是否有更优雅/更快捷的方式来做到这一点?

Is there a more elegant / faster way to do this?

更新:奖金问题-保持订单顺序即可.

Update: Bonus question -- do it with the order maintained.

推荐答案

您可以使用return_inverse关键字:

x = np.array(['aa', 'b', 'b', 'c', 'aa', 'b', 'a'])
uniq, map_ = np.unique(x, return_inverse=True)
map_
# array([1, 2, 2, 3, 1, 2, 0])

保留订单的版本:

x = np.array(['aa', 'b', 'b', 'c', 'aa', 'b', 'a'])
uniq, idx, map_ = np.unique(x, return_index=True, return_inverse=True)
mxi = idx.max()+1
mask = np.zeros((mxi,), bool)
mask[idx] = True
oidx = np.where(mask)[0]
iidx = np.empty_like(oidx)
iidx[map_[oidx]] = np.arange(oidx.size)
iidx[map_]
# array([0, 1, 1, 2, 0, 1, 3])

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