如何使用 itertools groupby 打印？

Question

我有这个代码：

from itertools import groupby

text = ["a", "a", "b", "b", "c"]

group = groupby(text)

for k, g in group:
    print(k, end= " ")
    print(sum(1 for _ in g), end=" ")

例如我需要什么：

A B C
2 2 1

我的itertools只显示如下：

A 2 B 2 C 1

Answer 1

将多个打印语句与转置一起使用

from itertools import groupby

text = ["a", "a", "b", "b", "c"]

group = groupby(text)

# Transpose so we have row 0 with group names, and row 1 with grouping items
transposed = [(k, list(g)) for k, g in group]

for k in transposed:
    print(k[0], end = " ")
print()
for k in transposed:
    print(sum(1 for _ in k[1]), end=" ")

Answer 2

您可以这样做，它对结果groupby返回的结果进行后处理，以便轻松获取输出的每一行所需的值：

from itertools import groupby

text = ["a", "a", "b", "b", "c"]

groups = [(k, str(sum(1 for _ in g))) for k, g in groupby(text)]
a, b = zip(*groups)
print(' '.join(a))  # -> a b c
print(' '.join(b))  # -> 2 2 1

Answer 3

这是另一种选择：

from itertools import groupby

text = ["a", "a", "b", "b", "c"]

group = groupby(text)
dictionary = {}

for k, g in group:
    dictionary[k] = sum(1 for _ in g)

keys = " ".join(list(dictionary.keys()))
values = " ".join(str(v) for v in list(dictionary.values()))

print(keys)
print(values)

Answer 4

我知道你问过 itertools.groupby，但我会使用 Counter 来完成这样的任务：

>>> text = ["a", "a", "b", "b", "c"]
>>> from collections import Counter
>>> c = Counter(text)
>>> print(c)
Counter({'a': 2, 'b': 2, 'c': 1})
>>> headers = c.keys()
>>> values = [str(val) for val in c.values()]
>>> print(' '.join(headers))
a b c
>>> print(' '.join(values))
2 2 1

Answer 5

由于groupby()不适合部署tee()来制作副本，因此最好的解决方案似乎是创建一个元组理解。 请注意，我们对每个组 g 中包含的值不感兴趣，因此我们只存储从组中动态构建的元组的长度。

import itertools

text = ["a", "a", "b", "b", "c"]

group = tuple((k,len(tuple(v))) for k,v in itertools.groupby(text))

for t in group:
    print(t[0], end= " ")
print()
for t in group:
    print(t[1], end=" ")
print()
# a b c
# 2 2 1