如何将字符串空格分隔的键，唯一字的值对转换为字典 [英] How to transform string of space-separated key,value pairs of unique words into a dict

问题描述

我有一个字符串，其中的单词用空格分隔（所有单词都是唯一的，没有重复）。我将此字符串转换为列表：

I've got a string with words that are separated by spaces (all words are unique, no duplicates). I turn this string into list:

s = "#one cat #two dogs #three birds"
out = s.split()

并计算创建的值的数量：

And count how many values are created:

print len(out) # Says 192 

然后我尝试从列表中删除所有内容：

Then I try to delete everything from the list:

for x in out:
     out.remove(x)

再次计算：

print len(out) # Says 96 

有人可以解释请问为什么它说96而不是0？

Can someone explain please why it says 96 instead of 0?

更多信息

每一行以'＃'开头并且是实际上是一对空格分隔的单词：对中的第一个是键，第二个是值。

Each line starts with '#' and is in fact a space-separated pair of words: the first in the pair is the key and second is the value.

所以，我正在做的是：

for x in out:
     if '#' in x: 
          ind = out.index(x) # Get current index 
          nextValue = out[ind+1] # Get next value 
          myDictionary[x] = nextValue
          out.remove(nextValue)
          out.remove(x) 

问题是我无法将所有键值对移动到字典中，因为我只迭代了96个项目。

The problem is I cannot move all key,value-pairs into a dictionary since I only iterate through 96 items.

推荐答案

我觉得你真的想要这样的东西：

I think you actually want something like this:

s = '#one cat #two dogs #three birds'
out = s.split()
entries = dict([(x, y) for x, y in zip(out[::2], out[1::2])])

这段代码在做什么？让我们分解吧。首先，我们按原样将 s 按空格分成 out 。

What is this code doing? Let's break it down. First, we split s by whitespace into out as you had.

接下来我们遍历 out 中的对，称它们为 x，y 。这些对成为元组/对的列表。 dict（）接受一个大小为两个元组的列表，并将它们视为 key，val 。

Next we iterate over the pairs in out, calling them "x, y". Those pairs become a list of tuple/pairs. dict() accepts a list of size two tuples and treats them as key, val.

以下是我试用时的结果：

Here's what I get when I tried it:

$ cat tryme.py

s = '#one cat #two dogs #three birds'
out = s.split()
entries = dict([(x, y) for x, y in zip(out[::2], out[1::2])])

from pprint import pprint
pprint(entries)

$ python tryme.py
{'#one': 'cat', '#three': 'birds', '#two': 'dogs'}

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