擅长:python、mysql、java
<p>您可以使用str.join并使用切片在其他元素上进行压缩:</p>
<pre><code>list(map(' '.join, zip(l[::2], l[1::2])))
</code></pre>
<hr/>
<pre><code>['scalar2_q1p2_114(-2,-2) = 0',
'scalar2_q1p2_114(-2,-1) = 0',
'scalar2_q1p2_114(-2,0) = 0',
'scalar2_q1p2_114(-2,1) = 0',
'scalar2_q1p2_114(-2,2) = 0',
'scalar2_q1p2_114(-1,-2) = 0',
'scalar2_q1p2_114(-1,-1) = 0',
'scalar2_q1p2_114(-1,0) = 0',
'scalar2_q1p2_114(-1,1) = 0',
'scalar2_q1p2_114(-1,2) = 0',
'scalar2_q1p2_114(0,-2) = 0',
'scalar2_q1p2_114(0,-1) = 0',
'scalar2_q1p2_114(0,0) = 0',
'scalar2_q1p2_114(0,1) = 0',
'scalar2_q1p2_114(0,2) = 0',
'scalar2_q1p2_114(1,-2) = 0',
'scalar2_q1p2_114(1,-1) = 0',
'scalar2_q1p2_114(1,0) = 0',
'scalar2_q1p2_114(2,-2) = 0',
'scalar2_q1p2_114(2,-1) = 0',
'scalar2_q1p2_114(2,0) = 0',
'scalar2_q1p2_114(1,2) = scalar2_q1p2_114(1,1)*((-d+3)/(SPD[p2,p2]))']
</code></pre>