既然你提到你对你的值的数据结构不挑剔，t要得到像你在问题中发布的字典一样的东西，你可以对x中的唯一值进行字典理解，其中np.where是值：

>>> {i:np.where(x == i)[0] for i in set(x)}

{1: array([0, 5, 6]),
 2: array([1, 2]),
 4: array([3]),
 7: array([4, 7]),
 16: array([8])}

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将其与通过列表的更普通的循环进行比较，对于较大的数组，这将显着更快：

def list_method(x):
    res = {i:[] for i in set(x)}
    for i, value in enumerate(x):
         res[value].append(i)
    return res

def np_method(x):
    return {i:np.where(x == i)[0] for i in set(x)}

x = np.random.randint(1, 50, 1000000)

In [5]: %timeit list_method(x)
259 ms ± 4.03 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

In [6]: %timeit np_method(x)
120 ms ± 4.15 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

型

python会是这样的：

result = {}
for idx,val in enumerate(x):
    arr = result.get(val,[])
    arr.append(idx)
    result[val] = arr

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x = [1,2,2,4,7,1,1,7,16]
numlist = []
numdict = {}
c = 0
for n in x:
    if n not in numlist:
        numlist.append(n)
        numdict[n] = [c]
    else:
        numdict[n].append(c)
    c += 1
print(numlist, numdict)

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输出为：[1, 2, 4, 7, 16] {1: [0, 5, 6], 2: [1, 2], 4: [3], 7: [4, 7], 16: [8]}写入文件用途：

with open('file.txt', 'w') as f:
    f.write(str(numdict))

型

最快的方法是使用defaultdict（纯Python，O（n））

from collections import defaultdict

def list_to_unique_dict(data):
    unique_dict = defaultdict(list)
    for index, value in enumerate(data):
        unique_dict[value].append(index)
    return dict(unique_dict)

x = [1,2,2,4,7,1,1,7,16]
list_to_unique_dict(x)

Out[62]: {1: [0, 5, 6], 2: [1, 2], 4: [3], 7: [4, 7], 16: [8]}

x = np.random.randint(1, 50, 1000000)
%timeit list_to_unique_dict(x)
232 ms ± 13.5 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

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