Question

我有一个包含以下结构的列表：

A1, B2, C3, 66
A1, B2, C3, 00
A2, B2, C3, 77
A3, B3, C4, 44
A4, B4, C5, 11
A4, B4, C5, 12
A4, B4, C5, 13

我需要枚举唯一的1-3列元素以获得如下输出：

A1, B2, C3, 66, 1
A1, B2, C3, 00, 2
A2, B2, C3, 77, 1
A3, B3, C4, 44, 1
A4, B4, C5, 11, 1
A4, B4, C5, 12, 2
A4, B4, C5, 13, 3

如您所知，我希望第四列中的序号以1-3列的唯一值排序。

在阅读说明后，我得出结论，我需要使用collections模块。这是正确的决定吗？

我试试这个：

new = ['A1,B2,C3,66','A1,B2,C3,00','A2, B2, C3, 77','A3, B3, C4, 44','A4, B4, C5, 11','A4, B4, C5, 12','A4, B4, C5, 13']
test=[]
i = 0
for a in new:
    i+=1
    test.append('{},{}'.format(i, a))
    print(test)
    if a[i]!=a[i-1]:
        continue

Answer 1

您可以使用itertools.groupby：

import itertools
import re
new = ['A1,B2,C3,66','A1,B2,C3,00','A2, B2, C3, 77','A3, B3, C4, 44','A4, B4, C5, 11','A4, B4, C5, 12','A4, B4, C5, 13']
n = map(lambda x:re.split(',\s*', x), new)
s = [list(b) for _, b in itertools.groupby(n, key=lambda x:x[:-1])]
last_data = map(lambda x:', '.join(x[:-1]+[str(x[-1])]), [i for b in [[b+[i] for i, b in enumerate(c, start=1)] for c in s] for i in b])

输出：

['A1, B2, C3, 66, 1', 
'A1, B2, C3, 00, 2', 
'A2, B2, C3, 77, 1', 
'A3, B3, C4, 44, 1', 
'A4, B4, C5, 11, 1', 
'A4, B4, C5, 12, 2', 
'A4, B4, C5, 13, 3']

Answer 2

如果你喜欢自己做List/Dictionnary/YourFavoriteICollectionImplementation，下面是一个选项（不是完美的解决方案）。

它使用一个名为group by的dict来存储一个键的出现次数。然后利用此dict在循环时获得预期值。

index_cache

输出：

import re
test = ['A1,B2,C3,66','A1,B2,C3,00','A2, B2, C3, 77','A3, B3, C4, 44','A4, B4, C5, 11','A4, B4, C5, 12','A4, B4, C5, 13']
index_cache = {} #store all occurrences count for each key

def prepare(data):
    formatted = re.split(',\s*', data.strip())
    index_cache['-'.join(formatted[:3])] = 0 # init the count to 0
    return formatted

test1 = list(map(prepare, test))

for item in test1:
    index_cache['-'.join(item[:3])] += 1
    item.append(index_cache['-'.join(item[:3])])
print (test1)

Answer 3

无需转到itertools即可轻松编码。 itertools可以更快 - 但不知道，需要测试 - 但这更容易阅读/理解/操纵。您的枚举规则非常明确：对于名为new的列表中的每个元素，如果前三个项与列表中前一个元素的前三个相同，则将索引增加1。如果它们不同，请中断序列并从1开始。

以下是代码：

new = ['A1, B2, C3, 66','A1, B2, C3, 00','A2, B2, C3, 77','A3, B3, C4, 44','A4, B4, C5, 11','A4, B4, C5, 12','A4, B4, C5, 13']
result = []
prev = None
c = 1
for x in new:
    current = ", ".join(x.split(", ")[:3])
    if current == prev:
        c += 1
    else:
        c = 1
    result.append(", ".join((x, str(c))))
    prev = current

列表中的编号重合

3 个答案: