使用 Python 从三维数组构建字典

Question

我目前无法根据数组的第一列将我的 3d 数组数据分类到不同的字典中。我想检查数组的第一列，并用其中的字符串名称制作一个字典。我还想用同一行上的值制作一个二维数组。我的数据看起来像这样：

allValues=[["str1","str1","str1","str1","str1","str1","str1","str2","str2","str2","str2","str2"],[1,2,3,4,5,6,7,8,9,10,11,12],[1,2,3,4,5,6,7,8,9,10,11,12]] 

而我的目标是：

allValuesDict = {'str1': [[1,2,3,4,5,6,7], [1,2,3,4,5,6,7]], 'str2': [[8,9,10,11,12], [8,9,10,11,12]]}

Answer 1

Pandas 是一个很好的工具。它产生相当可读的代码。

>>> import pandas as pd

>>> df = pd.DataFrame(allValues).T
>>> df
       0   1   2
0   str1   1   1
1   str1   2   2
2   str1   3   3
3   str1   4   4
4   str1   5   5
5   str1   6   6
6   str1   7   7
7   str2   8   8
8   str2   9   9
9   str2  10  10
10  str2  11  11
11  str2  12  12

>>> grouped = df.groupby(0)[[1,2]].apply(lambda l: list(zip(*pd.Series.tolist(l))))
>>> grouped
0
str1    [(1, 2, 3, 4, 5, 6, 7), (1, 2, 3, 4, 5, 6, 7)]
str2          [(8, 9, 10, 11, 12), (8, 9, 10, 11, 12)]
dtype: object

>>> grouped.to_dict()
{'str1': [(1, 2, 3, 4, 5, 6, 7), (1, 2, 3, 4, 5, 6, 7)], 'str2': [(8, 9, 10, 11, 12), (8, 9, 10, 11, 12)]}

Answer 2

zip 可以将您的列表翻译成[['str1',1,1],['str2',2,2],...]，这可以使迭代更容易：

allValues = [['str1','str1','str1','str1','str1','str1','str1','str2','str2','str2','str2','str2'],
             [1,2,3,4,5,6,7,8,9,10,11,12],
             [1,2,3,4,5,6,7,8,9,10,11,12]]

D = {}
for k,v1,v2 in zip(*allValues):
    if k in D:
        D[k][0].append(v1)
        D[k][1].append(v2)
    else:
        D[k] = [[v1],[v2]]

print D

输出：

{'str2': [[8, 9, 10, 11, 12], [8, 9, 10, 11, 12]], 'str1': [[1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7]]}

使用defaultdict 可以进一步简化：

from collections import defaultdict

allValues = [['str1','str1','str1','str1','str1','str1','str1','str2','str2','str2','str2','str2'],
             [1,2,3,4,5,6,7,8,9,10,11,12],
             [1,2,3,4,5,6,7,8,9,10,11,12]]

D = defaultdict(lambda:[[],[]])
for k,v1,v2 in zip(*allValues):
    D[k][0].append(v1)
    D[k][1].append(v2)

print D
print dict(D)  # If you don't want the final type to be defaultdict.

输出：

defaultdict(<function <lambda> at 0x00000000070A5128>, {'str2': [[8, 9, 10, 11, 12], [8, 9, 10, 11, 12]], 'str1': [[1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7]]})
{'str2': [[8, 9, 10, 11, 12], [8, 9, 10, 11, 12]], 'str1': [[1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7]]}

Answer 3

您可以使用collections.defaultdict 获得 O(n) 解决方案。

请注意，这是非常具体您当前拥有的数据结构：

from collections import defaultdict

d = defaultdict(lambda: [[], []])

for i, j, k in zip(*allValues):
    d[i][0].append(j)
    d[i][1].append(k)

如果你需要转换成普通的dict：

res = dict(d)

print(res)

{'str1': [[1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7]],
 'str2': [[8, 9, 10, 11, 12], [8, 9, 10, 11, 12]]}

Answer 4

您可以使用itertools.groupby 和zip：

import itertools
allValues=[["str1","str1","str1","str1","str1","str1","str1","str2","str2","str2","str2","str2"],[1,2,3,4,5,6,7,8,9,10,11,12],[1,2,3,4,5,6,7,8,9,10,11,12]] 
main, *rest = allValues
grouped_data = list(itertools.chain.from_iterable([[[a, [h for _, h in b]] for a, b in itertools.groupby(zip(main, i), key=lambda x:x[0])] for i in rest]))
final_grouping = {a:[i for _, i in b] for a, b in itertools.groupby(sorted(grouped_data, key=lambda x:x[0]), key=lambda x:x[0])}

输出：

{'str1': [[1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7]], 'str2': [[8, 9, 10, 11, 12], [8, 9, 10, 11, 12]]}