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如何使用 itertools groupby 打印?

[英]How to print with itertools groupby?

I have this code: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=" ")

Example what I need:例如我需要什么:

A B C
2 2 1

My itertools only shows like this:我的itertools只显示如下:

A 2 B 2 C 1

Using multiple print statements with transpose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=" ")

You can do it like this which post-processes the results groupby returns to make it easy to get the values needed for each row of the output: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

Here is one more alternative: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)

I know you asked about itertools.groupby, BUT I would use a Counter for such task: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

Since groupby() does not lend itself to deploying tee() to make a copy, the best solution seems to be to create a tuple comprehension.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。 Note that we're not interested in the values contained in each group g, so we'll just store the length of the tuple constructed on the fly from the group.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

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