用嵌套的for循环替换重复的if语句?

问题描述

在我编写的以下代码中，n = 4，因此有五个if语句，因此，如果我想将n增加为10，那么将会有很多if.因此，我的问题是:如何用更优雅的东西替换所有的if语句?

In the following code that I wrote, n = 4, and so there are five if statements, so if I would like to increase n to be, say 10, then there will be a lot of if's. Therefore my question: how can I replace all the if statements with something more elegant?

n, p = 4, .5  # number of trials, probability of each trial
s = np.random.binomial(n, p, 100)
# result of flipping a coin 10 times, tested 1000 times.

d = {"0" : 0, "1" : 0, "2" : 0, "3" : 0, "4" : 0 }

for i in s:
    if i == 0:
        d["0"] += 1
    if i == 1:
        d["1"] += 1
    if i == 2:
        d["2"] += 1
    if i == 3:
        d["3"] += 1
    if i == 4:
        d["4"] += 1

我尝试使用嵌套的for循环

I tried using nested for loops,

 for i in s:
     for j in range(0,5):
         if i == j:
             d["j"] += 1

但是我得到这个错误:

d["j"] += 1

KeyError: 'j'

推荐答案

您可以使用 collections.Counter 理解为:

You could use collections.Counter with a comprehension:

from collections import Counter

Counter(str(i) for i in s)

Counter在这里起作用，因为您要加一.但是，如果您希望它更通用，也可以使用 collections.defaultdict :

Counter works here because you're incrementing by one. However if you want it more general you could also use collections.defaultdict:

from collections import defaultdict

dd = defaultdict(int)   # use int as factory - this will generate 0s for missing entries
for i in s:
    dd[str(i)] += 1  # but you could also use += 2 or whatever here.

，或者如果您希望将其作为纯字典，则将其包装在dict调用中，例如:

or if you want it as plain dictionary, wrap it inside a dict call, for example:

dict(Counter(str(i) for i in s))

在键不存在时都避免使用KeyError，并且避免了双循环.

Both avoid KeyErrors when the key isn't present yet and you avoid the double loop.

作为旁注:如果您想要简单的命令，也可以使用 dict.get :

As a side note: If you want plain dicts you could also use dict.get:

d = {}  # empty dict
for i in d:
    d[str(i)] = d.get(str(i), 0) + 1

但是Counter和defaultdict的行为几乎像普通字典一样，因此几乎不需要最后一本字典，因为它(可能)比较慢，而且我认为可读性较低.

However Counter and defaultdict behave almost like plain dictionaries so there's almost no need for this last one, because it is (probably) slower and in my opinion less readable.

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