擅长:python、mysql、java
<p>这只是数学,从实际位置减去<code>x</code>中的“h”位置,该位置可由<code>enumerate</code>在列表上迭代时确定:</p>
<pre><code>x = "vnu4)rhg&3j"
qList = ["%s='%s'" % (i-x.find('h'), x[i]) for (i, j) in enumerate(x)]
print(qList) # as list
print(','.join(qList)) # as string
</code></pre>
<p>输出:</p>
<pre><code>["-6='v'", "-5='n'", "-4='u'", "-3='4'", "-2=')'", "-1='r'", "0='h'", "1='g'", "2='&'", "3='3'", "4='j'"]
-6='v',-5='n',-4='u',-3='4',-2=')',-1='r',0='h',1='g',2='&',3='3',4='j'
</code></pre>