格式化连续数字

问题描述

我正在尝试使用 Python 格式化整数列表，但在实现我想要的结果时遇到了一些困难.

I'm trying to format a list of integers with Python and I'm having a few difficulties achieving what I'd like.

输入是整数的排序列表:

Input is a sorted list of Integers:

list = [1, 2, 3, 6, 8, 9]

我希望它的输出是这样的字符串:

I would like it the output to be a String looking like this:

outputString = "1-3, 6, 8-9"

到目前为止，我设法实现的是:

So far all I managed to achieve is this:

outputString = "1-2-3, 6, 8-9"

如果 Int 已经是连续的，我很难告诉我的代码忽略它.

I'm having trouble to tell my code to ignore a Int if it was already consecutive.

到目前为止，这是我的代码:

Here is my code so far:

def format(l):
    i = 0
    outputString = str(l[i])
    for x in range(len(l)-1):
        if l[i + 1] == l[i]+1 :
            outputString += '-' + str(l[i+1])
        else :
            outputString += ', ' + str(l[i+1])
        i = i + 1
    return outputString

感谢您的帮助和见解:)

Thanks for your help and insights :)

推荐答案

你可以像这样使用 itertools 模块中的 groupby 和 count:

You can use groupby and count from itertools module like this way:

感谢 @ason​​gtoruin 评论.要从输入中删除重复项，您可以使用:sorted(set(a)).

Thanks to @asongtoruin comment. For removing duplicates from the input you can use: sorted(set(a)).

from itertools import groupby, count

a = [1, 2, 3, 6, 8, 9]
clustered = [list(v) for _,v in groupby(sorted(a), lambda n, c = count(): n-next(c))]

for k in clustered:
    if len(k) > 1:
        print("{0}-{1}".format(k[0], k[-1]))
    else:
        print("{0}".format(k[0]))

输出:

1-3
6
8-9

或者你可以做这样的事情来获得漂亮的输出:

Or maybe you can do something like this in order to have a pretty output:

from itertools import groupby, count

a = [1, 2, 3, 6, 8, 9]
clustered = [list(v) for _,v in groupby(sorted(a), lambda n, c = count(): n-next(c))]
out = ", ".join(["{0}-{1}".format(k[0], k[-1]) if len(k) > 1 else "{0}".format(k[0]) for k in clustered ])

print(out)

输出:

1-3, 6, 8-9

更新:

我猜，使用 itertools 模块可能会让许多 Python 的新开发人员感到困惑.这就是为什么我决定重写相同的解决方案而不导入任何包并尝试显示 groupby 和 count 在幕后所做的事情:

I'm guessing, using itertools modules may confuse many Python's new developers. This is why i decided to rewrite the same solution without importing any package and trying to show what groupby and count are doing behind the scenes:

def count(n=0, step=1):
    """Return an infinite generator of numbers"""
    while True:
        n += step
        yield n


def concat(lst):
    """Group lst elements based on the result of elm - next(_count)"""
    _count, out = count(), {}
    for elm in sorted(lst):
        c = elm - next(_count)
        if c in out:
            out[c].append(elm)
        else:
            out[c] = [elm]
    return out


def pretty_format(dct):
    for _, value in dct.items():
        if len(value) > 1:
            yield '{}-{}'.format(value[0], value[-1])
        else:
            yield '{}'.format(value[0])


lst = [1, 2, 3, 6, 8, 9]
dct = concat(lst)
formatted = list(pretty_format(dct))
print(formatted)

输出:

['1-3', '6', '8-9']

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