如何将空格分隔的键值对字符串转换为字典

问题描述

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()

And count how many values are created:

print len(out) # Says 192 

Then I try to delete everything from the list:

for x in out:
     out.remove(x)

And then count again:

print len(out) # Says 96 

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

MORE INFO

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.

So, what I am doing is:

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) 

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

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])])

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

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.

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'}

这篇关于如何将空格分隔的键值对字符串转换为字典的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持跟版网！