如何使用另一个元组中的元素从一个元组中查找索引

Question

我有以下信息元组，如何从名称元组中找到元组[？]的索引？

info  = [('Water SPF 15', '12B', '#f2c9b5', 31, '-100.1', '-100.1', '-100.1', '12N', '', '', '27B'), ('Water Spectrum SPF 15', '12B', '#f2c9b5', 31, '7..7', '7..7', '7..7', '12N', '', '', '27B'), ('Water Foundation SPF 15', '12N', '#e5be9d', 44, ' ', ' ', ' ', '', '12B', '', '15N'), ('FACE15', '12N', '#e5be9d', 44, '**', '**', '**', '12S', '14H', '', '16N'), ('FACE', '12B', '#f2c9b5', 31, 'asd', 'asd', 'asd', '14H', '', '8B', '18B'), ('hydrator', '10N', '#e5be9d', 44, '', '', '', '', '', '', '13N'), ('APE', '12B', '#e7cbb3', 39, '-100', '-100', '-100', '', '', '', '')]
name = [('Water Spectrum SPF 15', '12B', '#f2c9b5'),('FACE15', '12N', '#e5be9d')]

matched_index = []
n = 0
for i in info :
    if name[n][0] and name[n][1] and name[n][2] == i[n][0] and i[n][1] and i[n][2]:
        print("match")
        matched_index.append(n)
        n = n + 1
    else:
        n = n + 1
print("DONE")

有没有办法只匹配上面的前 3 个元素？当 n 达到 3 时我会出错...

Answer 1

类似下面的

infos = [('Water SPF 15', '12B', '#f2c9b5', 31, '-100.1', '-100.1', '-100.1', '12N', '', '', '27B'),
         ('Water Spectrum SPF 15', '12B', '#f2c9b5', 31, '7..7', '7..7', '7..7', '12N', '', '', '27B'),
         ('Water Foundation SPF 15', '12N', '#e5be9d', 44, ' ', ' ', ' ', '', '12B', '', '15N'),
         ('FACE15', '12N', '#e5be9d', 44, '**', '**', '**', '12S', '14H', '', '16N'),
         ('FACE', '12B', '#f2c9b5', 31, 'asd', 'asd', 'asd', '14H', '', '8B', '18B'),
         ('hydrator', '10N', '#e5be9d', 44, '', '', '', '', '', '', '13N'),
         ('APE', '12B', '#e7cbb3', 39, '-100', '-100', '-100', '', '', '', '')]
names = [('Water Spectrum SPF 15', '12B', '#f2c9b5'), ('FACE15', '12N', '#e5be9d')]

for name in names:
    for idx, info in enumerate(infos):
        if name[0] == info[0] and name[1] == info[1] and name[2] == info[2]:
            print(f'{name} {idx}')

输出

('Water Spectrum SPF 15', '12B', '#f2c9b5') 1
('FACE15', '12N', '#e5be9d') 3

Answer 2

您可以通过首先将info 中的元组转换为一个字典，该字典由每个元组的前 3 项以索引为值，从而避免嵌套循环。然后一个简单的字典查找提供了结果。总体而言，嵌套循环的运行时间复杂度为 O(n) vs O(n²)：

info  = [('Water SPF 15', '12B', '#f2c9b5', 31, '-100.1', '-100.1', '-100.1', '12N', '', '', '27B'), ('Water Spectrum SPF 15', '12B', '#f2c9b5', 31, '7..7', '7..7', '7..7', '12N', '', '', '27B'), ('Water Foundation SPF 15', '12N', '#e5be9d', 44, ' ', ' ', ' ', '', '12B', '', '15N'), ('FACE15', '12N', '#e5be9d', 44, '**', '**', '**', '12S', '14H', '', '16N'), ('FACE', '12B', '#f2c9b5', 31, 'asd', 'asd', 'asd', '14H', '', '8B', '18B'), ('hydrator', '10N', '#e5be9d', 44, '', '', '', '', '', '', '13N'), ('APE', '12B', '#e7cbb3', 39, '-100', '-100', '-100', '', '', '', '')]
name = [('Water Spectrum SPF 15', '12B', '#f2c9b5'),('FACE15', '12N', '#e5be9d')]

lookup = {t[:3]: i for i,t in enumerate(info)}
for n in name:
    if n in lookup:
        print(n, lookup[n])

输出：

('Water Spectrum SPF 15', '12B', '#f2c9b5') 1
('FACE15', '12N', '#e5be9d') 3

如果您只想将索引存储在列表中：

lookup = {t[:3]: i for i,t in enumerate(info)}
matched_index = [lookup[n] for n in name if n in lookup]
print(matched_index)

输出

[1, 3]

Answer 3

您还可以考虑使用列表推导，它比 Python 中的传统循环更受欢迎：

只获取索引：

lst = [idx for idx, info in enumerate(infos) for name in names if name == info[:3]]
print(lst)

输出：[1, 3]

作为字典列表：

lst = [{name:idx} for idx, info in enumerate(infos) for name in names if name == info[:3]]
print(lst)

输出：

[{('Water Spectrum SPF 15', '12B', '#f2c9b5'): 1}, {('FACE15', '12N', '#e5be9d'): 3}]