python中的lambda可以迭代dict吗? [英] lambda in python can iterate dict?

问题描述

我最近有一次采访.面试官问我如何迭代python中的dict.我说过所有方式都使用 for 语句.但是他告诉我 lambda 怎么样?

I have an interview recently. The interviewer asked me the ways to iterate dict in python. I said all the ways use for statement. But he told me that how about lambda?

我感到非常困惑，我认为lambda是一个匿名函数，但是如何迭代dict?像这样的一些代码:

I feel confused very much and I consider lambda as an anonymity function, but how it iterates a dict? some code like this:

new_dict = sorted(old_dict.items(), lambda x: x[1]) # sorted by value in dict

但是在此代码中，lambda用作提供比较键的函数.您如何看待这个问题?

But in this code, the lambda is used as a function to provide the compared key. What do you think this question?

推荐答案

您不必使用lambda进行迭代.有以下几种在Python中迭代可迭代对象的方法:

You don't iterate with lambda. There are following ways to iterate an iterable object in Python:

1. for陈述(您的答案)
2. 理解，包括列表[x for x in y]，字典{key: value for key, value in x}和设置{x for x in y}
3. 生成器表达式:(x for x in y)
4. 传递将对其进行迭代的函数(map，all，itertools模块)
5. 手动调用next函数，直到StopIteration发生.

1. for statement (your answer)
2. Comprehension, including list [x for x in y], dictionary {key: value for key, value in x} and set {x for x in y}
3. Generator expression: (x for x in y)
4. Pass to function that will iterate it (map, all, itertools module)
5. Manually call next function until StopIteration happens.

注意:3不会对其进行迭代，除非您稍后对该生成器进行迭代.如果是4，则取决于功能.

Note: 3 will not iterate it unless you iterate over that generator later. In case of 4 it depends on function.

要遍历dict或list之类的特定集合，可以使用诸如while col: remove element或带有索引切片技巧的更多技术.

For iterating specific collections like dict or list there can be more techniques like while col: remove element or with index slicing tricks.

现在lambda出现在图片中.您可以在其中一些函数中使用lambda，例如:map(lambda x: x*2, [1, 2, 3]).但是这里的lambda与迭代过程本身无关，您可以传递常规函数map(func, [1, 2, 3]).

Now lambda comes into the picture. You can use lambdas in some of those functions, for example: map(lambda x: x*2, [1, 2, 3]). But lambda here has nothing to do with iteration process itself, you can pass a regular function map(func, [1, 2, 3]).

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