numpy用整数逐步替换元素组 [英] numpy replace groups of elements with integers incrementally

问题描述

import numpy as np
data = np.array(['b','b','b','a','a','a','a','c','c','d','d','d'])

我需要像这样逐步将每组字符串替换为整数

I need to replace each group of strings with an integer incrementally like this

data = np.array([0,0,0,1,1,1,1,2,2,3,3,3])

我正在寻找一个麻木的解决方案

I'm looking for a numpy solution

使用此数据集 http://www.uploadmb.com/dw.php?id = 1364341573

import numpy as np
f = open('test.txt','r')
lines = np.array([ line.strip() for line in f.readlines() ])
lines100 = lines[0:100]
_, ind, inv = np.unique(lines100, return_index=True, return_inverse=True)
print ind
print inv
nums = np.argsort(ind)[inv]
print nums

[ 0 83 62 40 19]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4
 4 4 4 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 2 2 2 2 2 2 2 2 2 2 2 2
 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3 3
 3 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4]

lines200 = lines[0:200]
_, ind, inv = np.unique(lines200, return_index=True, return_inverse=True)
print ind
print inv
nums = np.argsort(ind)[inv]
print nums
[167   0  83 124 104 144 185  62  40  19]
[1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9
 9 9 9 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 7 7 7 7 7 7 7 7 7 7 7 7
 7 7 7 7 7 7 7 7 7 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 4 4 4 4 4 4 4
 4 4 4 4 4 4 4 4 4 4 4 4 4 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 5 5 5 5
 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6]
[9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6
 6 6 6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 5 5 5 5 5 5 5 5 5 5 5 5
 5 5 5 5 5 5 5 5 5 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 2 2 2 2 2 2 2
 2 2 2 2 2 2 2 2 2 2 2 2 2 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 4 4 4 4
 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3]

推荐答案

并非总是如此:

>>> a,b,c = np.unique(data, return_index=True, return_inverse=True)
>>> c # almost!!!
array([1, 1, 1, 0, 0, 0, 0, 2, 2, 3, 3, 3])
>>> np.argsort(b)[c]
array([0, 0, 0, 1, 1, 1, 1, 2, 2, 3, 3, 3], dtype=int64)

---

但这确实可行:

---

But this does work:

def replace_groups(data):
    a,b,c, = np.unique(data, True, True)
    _, ret = np.unique(b[c], False, True)
    return ret

，并且比字典替换方法要快，对于较大的数据集，大约是33％:

and is faster than the dictionary replacement approach, about 33% for larger datasets:

def replace_groups_dict(data):
    _, ind = np.unique(data, return_index=True)
    unqs = data[np.sort(ind)]
    data_id = dict(zip(unqs, np.arange(data.size)))
    num = np.array([data_id[datum] for datum in data])
    return num

In [7]: %timeit replace_groups_dict(lines100)
10000 loops, best of 3: 68.8 us per loop

In [8]: %timeit replace_groups_dict(lines200)
10000 loops, best of 3: 106 us per loop

In [9]: %timeit replace_groups_dict(lines)
10 loops, best of 3: 32.1 ms per loop

In [10]: %timeit replace_groups(lines100)
10000 loops, best of 3: 67.1 us per loop

In [11]: %timeit replace_groups(lines200)
10000 loops, best of 3: 78.4 us per loop

In [12]: %timeit replace_groups(lines)
10 loops, best of 3: 23.1 ms per loop

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