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Accepted
Violet
Asked:2020-06-20 17:18:42 +0000 UTC2020-06-20 17:18:42 +0000 UTC

匹配三个列表

  • 772

有3个列表：

a = [12, 12, 12, 12, 1, 1, 1]
b = [20, 21, 22, 23, 200, 201, 202]
c = ['04-04', '05-05', '06-06', '07-07', '008-008', '009-009', '100-100']

需要一个唯一值 froma来导出一个映射 fromb等于c值的重复次数a，即：

uid=12

  • 20 04-04
  • 21 05-05
  • 22 06-06
  • 23 07-07

uid=1

  • 200 008-008
  • 201 009-009
  • 202 100-100

输出唯一值a：

    for i in range(0, len(list(set(a)))):
        print('uid=' + str(sorted(set(a), key=a.index)[i]))
    >>> uid=12
    >>> uid=1

中的重复元素数a：

for i in range(0, len(list(set(a)))):
    print(a.count(sorted(set(a), key=a.index)[i]))
>>> 4
>>> 1

匹配b和c：

x = []
for j, o in zip(b, c):
    x.append(str(j) + ':' + o)
print(x)
>>> ['20:04-04', '21:05-05', '22:06-06', '23:07-07', '200:008-008', '201:009-009', '202:100-100']

进一步昏迷，如何比较我不知道

python

3 个回答

  1. Best Answer

    假设重复的元素在一行中并且不再出现：

    from collections import Counter

a = [12, 12, 12, 12, 1, 1, 1]
b = [20, 21, 22, 23, 200, 201, 202]
c = ['04-04', '05-05', '06-06', '07-07', '008-008', '009-009', '100-100']

result = {}
start = 0
stop = 0

for key, value in Counter(a).items():
  start = stop
  stop = start + value

  result[key] = list(zip(b[start:stop], c[start:stop]))

print(result)
    • 4 
  2. In [1]: a = [12, 12, 12, 12, 1, 1, 1]
   ...: b = [20, 21, 22, 23, 200, 201, 202]
   ...: c = ['04-04', '05-05', '06-06', '07-07', '008-008', '009-009', '100-100']

In [2]: result = {}

In [3]: for x, y, z in zip(a, b, c):
   ...:     result.setdefault(x, []).append((y, z))  # Группируем
   ...:

In [4]: for key, values in result.items():  # Выводим
   ...:     print(f"uid={key}")
   ...:     for y, z in values:
   ...:         print(f" - {y} {z}")
   ...:

    结论：

    uid=12
 - 20 04-04
 - 21 05-05
 - 22 06-06
 - 23 07-07
uid=1
 - 200 008-008
 - 201 009-009
 - 202 100-100
    • 4 
  3. import numpy as np

for elem in (np.unique(a)):
    print(elem)
    for i in [x[0] for x in np.argwhere(np.array(a)==elem)]:
        print(f"\t{b[i]}", end="")
        print(f"\t{c[i]}")

    我们得到：

    1
    200 008-008
    201 009-009
    202 100-100
12
    20  04-04
    21  05-05
    22  06-06
    23  07-07
    • 3

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