帮助 python 列表理解

发布于 2024-08-28 06:07:28 字数 570 浏览 6 评论 0原文

我的问题的简化版本：

我有一个列表理解，我用它在二维列表上设置位标志，所以：

s = FLAG1 | FLAG2 | FLAG3
[[c.set_state(s) for c in row] for row in self.__map]

所有 set_state 所做的是：

self.state |= f

这工作正常，但我必须在 __map 中的每个单元格中都有这个函数“set_state”。 __map 中的每个单元格都有一个 .state，所以我想做的是：

[[c.state |= s for c in row] for row in self.map]

或

map(lambda c: c.state |= s, [c for c in row for row in self.__map])

两者都不起作用（语法错误）。也许我用map/lamda找错了树，但我想摆脱set_state。也许知道为什么赋值在列表理解中不起作用

原文

A simplified version of my problem:

I have a list comprehension that i use to set bitflags on a two dimensional list so:

s = FLAG1 | FLAG2 | FLAG3
[[c.set_state(s) for c in row] for row in self.__map]

All set_state does is:

self.state |= f

This works fine but I have to have this function "set_state" in every cell in __map. Every cell in __map has a .state so what I'm trying to do is something like:

[[c.state |= s for c in row] for row in self.map]

map(lambda c: c.state |= s, [c for c in row for row in self.__map])

Except that neither works (Syntax error). Perhaps I'm barking up the wrong tree with map/lamda but I would like to get rid on set_state. And perhaps know why assignment does not work in the list-comprehension

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

时光沙漏 2024-09-04 06:07:28

列表推导式用于创建列表。您似乎并不关心您正在创建的实际列表，因此您应该只使用 for 语句，如下所示：

for row in self.__map:
    for c in row:
        c.state |= s

List comprehensions are for creating lists. You don't seem to care about the actual lists you are making, so you should just use a for statement, like so:

for row in self.__map:
    for c in row:
        c.state |= s

回复收藏 0 原文

椒妓 2024-09-04 06:07:28

是的，您使用了错误的工具。列表理解返回一个全新的值，因此您也许可以这样做：

self.__map = [[c.state | s for c in row] for row in self.__map]

但我的直觉是您应该只使用两个 for 循环：

for row in self.__map:
    for c in row:
        c.state |= s

在列表理解中，结果必须是一个表达式。那是因为你的双重理解只是糖：

list1 = []
for row in self.__map:
    list2 = []
    for c in row:
        list2.append(c.state | s)
    list1.append(list2)
self.__map = list1

说

        list2.append(c.state |= s)

因为最里面的表达式必须返回要附加到 list2 的内容是没有意义的。

基本上，每次更新标志时，列表推导都会生成 self.__map 的完整新副本。如果这就是你想要的，那就随它去吧。但我怀疑你只是想改变现有的地图。在这种情况下，请使用双 for 循环。

Yes, you're using the wrong tool. A list comprehension returns a completely new value, so you might be able to do this:

self.__map = [[c.state | s for c in row] for row in self.__map]

But my instinct is that you should just be using two for loops:

for row in self.__map:
    for c in row:
        c.state |= s

In a list comprehension, the result has to be an expression. That's because your double comprehension is just sugar for this:

list1 = []
for row in self.__map:
    list2 = []
    for c in row:
        list2.append(c.state | s)
    list1.append(list2)
self.__map = list1

It makes no sense to say

        list2.append(c.state |= s)

Because the innermost expression has to return something to be appended to list2.

Basically, the list comprehensions make a complete new copy of self.__map each time you update the flags. If that's what you want, then go with it. But I suspect you just want to change the existing map. In that case, use the double for loops.

回复收藏 0 原文