本文介绍了如何在不调用 RuntimeError 的情况下使用循环删除 Counter 对象中的条目?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

from collections import *
ignore = ['the','a','if','in','it','of','or']
ArtofWarCounter = Counter(ArtofWarLIST)
for word in ArtofWarCounter:
    if word in ignore:
        del ArtofWarCounter[word]

ArtofWarCounter 是一个 Counter 对象,其中包含《孙子兵法》中的所有单词.我正在尝试从 ArtofWarCounter 中删除 ignore 中的单词.

ArtofWarCounter is a Counter object containing all the words from the Art of War. I'm trying to have words in ignore deleted from the ArtofWarCounter.

追溯:

  File "<pyshell#10>", line 1, in <module>
    for word in ArtofWarCounter:
RuntimeError: dictionary changed size during iteration

推荐答案

为了最少的代码更改,使用 list,这样你正在迭代的对象与 Counter 解耦代码>

For minimal code changes, use list, so that the object you are iterating over is decoupled from the Counter

ignore = ['the','a','if','in','it','of','or']
ArtofWarCounter = Counter(ArtofWarLIST)
for word in list(ArtofWarCounter):
    if word in ignore:
        del ArtofWarCounter[word]

在 Python2 中,您可以使用 ArtofWarCounter.keys() 代替 list(ArtofWarCounter),但是当编写面向未来的代码如此简单时,为什么不做吗?

In Python2, you can use ArtofWarCounter.keys() instead of list(ArtofWarCounter), but when it is so simple to write code that is futureproofed, why not do it?

最好不要计算您希望忽略的项目

It is a better idea to just not count the items you wish to ignore

ignore = {'the','a','if','in','it','of','or'}
ArtofWarCounter = Counter(x for x in ArtofWarLIST if x not in ignore)

请注意,我将 ignore 设置为 set,这使得测试 x not in ignore 更加高效

note that I made ignore into a set which makes the test x not in ignore much more efficient

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05-20 04:30