如何简化这个糟糕的代码？

Question

我是编程新手，在简化代码方面遇到一些问题。 我之前的部分没有问题，只需要知道如何简化此代码即可。 我希望下面的代码在“ 11”的位置也绘制出12,18,20,27和28。 我将不胜感激！

simrecno1inds11 = nonzero(datasim11[:,1]==no1)[0]
simrecno2inds11 = nonzero(datasim11[:,1]==no2)[0]
simrecno3inds11 = nonzero(datasim11[:,1]==no3)[0]
simrecno4inds11 = nonzero(datasim11[:,1]==no4)[0]
simrecno5inds11 = nonzero(datasim11[:,1]==no5)[0]

simrecno7inds11 = nonzero(datasim11[:,1]==no7)[0]
simrecno8inds11 = nonzero(datasim11[:,1]==no8)[0]
simrecno9inds11 = nonzero(datasim11[:,1]==no9)[0]
simrecno10inds11 = nonzero(datasim11[:,1]==no10)[0]
simrecno11inds11 = nonzero(datasim11[:,1]==no11)[0]
simrecno12inds11 = nonzero(datasim11[:,1]==no12)[0]
simrecno13inds11 = nonzero(datasim11[:,1]==no13)[0]
simrecno14inds11 = nonzero(datasim11[:,1]==no14)[0]
simrecno15inds11 = nonzero(datasim11[:,1]==no15)[0]
simrecno16inds11 = nonzero(datasim11[:,1]==no16)[0]
simrecno17inds11 = nonzero(datasim11[:,1]==no17)[0]
simrecno18inds11 = nonzero(datasim11[:,1]==no18)[0]
simrecno19inds11 = nonzero(datasim11[:,1]==no19)[0]
simrecno20inds11 = nonzero(datasim11[:,1]==no20)[0]
simrecno21inds11 = nonzero(datasim11[:,1]==no21)[0]
simrecno22inds11 = nonzero(datasim11[:,1]==no22)[0]
simrecno23inds11 = nonzero(datasim11[:,1]==no23)[0]
simrecno24inds11 = nonzero(datasim11[:,1]==no24)[0]
simrecno25inds11 = nonzero(datasim11[:,1]==no25)[0]
simrecno26inds11 = nonzero(datasim11[:,1]==no26)[0]
simrecno27inds11 = nonzero(datasim11[:,1]==no27)[0]
simrecno28inds11 = nonzero(datasim11[:,1]==no28)[0]
simrecno29inds11 = nonzero(datasim11[:,1]==no29)[0]
simrecno30inds11 = nonzero(datasim11[:,1]==no30)[0]

recno1inds11 = nonzero(data11[:,1]==no1)[0]
recno2inds11 = nonzero(data11[:,1]==no2)[0]
recno3inds11 = nonzero(data11[:,1]==no3)[0]
recno4inds11 = nonzero(data11[:,1]==no4)[0]
recno5inds11 = nonzero(data11[:,1]==no5)[0]

recno7inds11 = nonzero(data11[:,1]==no7)[0]
recno8inds11 = nonzero(data11[:,1]==no8)[0]
recno9inds11 = nonzero(data11[:,1]==no9)[0]
recno10inds11 = nonzero(data11[:,1]==no10)[0]
recno11inds11 = nonzero(data11[:,1]==no11)[0]
recno12inds11 = nonzero(data11[:,1]==no12)[0]
recno13inds11 = nonzero(data11[:,1]==no13)[0]
recno14inds11 = nonzero(data11[:,1]==no14)[0]
recno15inds11 = nonzero(data11[:,1]==no15)[0] 
recno16inds11 = nonzero(data11[:,1]==no16)[0]
recno17inds11 = nonzero(data11[:,1]==no17)[0]
recno18inds11 = nonzero(data11[:,1]==no18)[0]
recno19inds11 = nonzero(data11[:,1]==no19)[0]
recno20inds11 = nonzero(data11[:,1]==no20)[0]
recno21inds11 = nonzero(data11[:,1]==no21)[0] 
recno22inds11 = nonzero(data11[:,1]==no22)[0]
recno23inds11 = nonzero(data11[:,1]==no23)[0]
recno24inds11 = nonzero(data11[:,1]==no24)[0]
recno25inds11 = nonzero(data11[:,1]==no25)[0]
recno26inds11 = nonzero(data11[:,1]==no26)[0]
recno27inds11 = nonzero(data11[:,1]==no27)[0]
recno28inds11 = nonzero(data11[:,1]==no28)[0]
recno29inds11 = nonzero(data11[:,1]==no29)[0]
recno30inds11 = nonzero(data11[:,1]==no30)[0]

Answer 1

无论no1 - no30是什么，您都需要将其放入序列对象（如list ，然后在该序列对象上循环以产生输出，您也将其放入序列对象中（在这种情况下，我认为dict将最好）。

nos = [no1, no2, ..., no30]
simrecno_inds11 = {}
recno_inds11 = {}
exclude_nums = [6]
for k, no in enumerate(nos):
    if k in exclude_nums:
        continue
    simrecno_inds11[k] = nonzero(datasim11[:,1]==no)[0]
    recno28inds11[k] = nonzero(data11[:,1]==no)[0]

现在，代替访问simrecno17inds11 ，您将访问simrecno_inds11[17] ，依此类推。

Answer 2

看来您的“ simrecnoXXindsYY”变量可以用二维数组simrec[n, i] （n-> no，i-> inds）。 以同样的方式，“ recnoXXindsYY”可以变成rec[n,i] 。 最后，“ noXX”可以是一维数组no[n] 。 因此，您的作业具有以下形式：

simrec[n, i] = nonzero(datasim[i][:,1]==no[n])[0]
rec[n, i] = nonzero(data[i][:,1]==no[n])[0]

因此，您需要为n （nos）和i （index）的所有可能值循环。

indexes = [11, 12, 18, 20, 27, 28]
nos = range(1,31)

for i in index:
    for n in nos:
        simrec[n, i] = nonzero(datasim[i][:,1]==no[n])[0]
        rec[n, i] = nonzero(data[i][:,1]==no[n])[0]