Python直方图一线 [英] python histogram one-liner

问题描述

有很多方法可以编写用于计算直方图的Python程序。

There are many ways to write a Python program that computes a histogram.

通过直方图，我的意思是一个函数可以计算<$ c中对象的出现$ c> iterable 并在字典中输出计数。例如：

By histogram, I mean a function that counts the occurrence of objects in an iterable and outputs the counts in a dictionary. For example:

>>> L = 'abracadabra'
>>> histogram(L)
{'a': 5, 'b': 2, 'c': 1, 'd': 1, 'r': 2}

编写此功能的一种方法是：

One way to write this function is:

def histogram(L):
    d = {}
    for x in L:
        if x in d:
            d[x] += 1
        else:
            d[x] = 1
    return d

有没有更简洁的编写此函数的方法？

Are there more concise ways of writing this function?

如果我们在Python中具有字典理解功能，则可以编写：

If we had dictionary comprehensions in Python, we could write:

>>> { x: L.count(x) for x in set(L) }

但自Python 2.6起没有它们，我们必须写：

but since Python 2.6 doesn't have them, we have to write:

>>> dict([(x, L.count(x)) for x in set(L)])

尽管此方法可能可读，但效率不高：L被多次遍历。此外，这不适用于单寿命发电机；函数对于迭代器生成器应该同样有效，例如：

Although this approach may be readable, it is not efficient: L is walked-through multiple times. Furthermore, this won't work for single-life generators; the function should work equally well for iterator generators such as:

def gen(L):
    for x in L:
        yield x

我们可能会尝试使用 reduce 函数（RIP）：

We might try to use the reduce function (R.I.P.):

>>> reduce(lambda d,x: dict(d, x=d.get(x,0)+1), L, {}) # wrong!

糟糕，此操作无效：键名称为'x'，而不是 x 。 ：（

Oops, this does not work: the key name is 'x', not x. :(

我以：

>>> reduce(lambda d,x: dict(d.items() + [(x, d.get(x, 0)+1)]), L, {})

（在Python 3中，我们必须编写 list（d.items（））的 d.items（），但这是假设的，因为那里没有 reduce 。）

(In Python 3, we would have to write list(d.items()) instead of d.items(), but it's hypothethical, since there is no reduce there.)

请用更好的，更具可读性的单线打败我！;）

Please beat me with a better, more readable one-liner! ;)

推荐答案

Python 3.x确实具有 reduce ，您只需从functools import reduce 中进行即可。它还具有 dict理解，其语法与您的示例中的语法完全相同。

Python 3.x does have reduce, you just have to do a from functools import reduce. It also has "dict comprehensions", which have exactly the syntax in your example.

Python 2.7和3.x也具有 Counter 类，它确实满足您的要求：

Python 2.7 and 3.x also have a Counter class which does exactly what you want:

from collections import Counter
cnt = Counter("abracadabra")

在Python 2.6或更早版本中，我个人使用 defaultdict 并分两行进行：

In Python 2.6 or earlier, I'd personally use a defaultdict and do it in 2 lines:

d = defaultdict(int)
for x in xs: d[x] += 1

这是干净，高效，Python式的，对大多数人来说更容易比涉及到 reduce 的任何事物都要理解。

That's clean, efficient, Pythonic, and much easier for most people to understand than anything involving reduce.

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