擅长:python、mysql、java
<p>您必须一次遍历列表中的2个元素</p>
<p>试试这个</p>
<pre><code>[my_list[i]+my_list[i+1] for i in range(0,len(my_list)-1,2)] #my_list is the original list you've posted in the question
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<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]))']
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