Question

Goodmorning，我对以下功能有疑问：

aminoacids = ['A', 'R', 'N', 'D', 'C', 'Q', 'E', 'G', 'H', 'I', 'L', 'K', 'M', 'F', 'P', 'S', 'T', 'W', 'Y', 'V']

pair_no_change = ['A', 'K']

original_pairs = [['A', 'K'], ['A', 'K'], ['A', 'K'], ['A', 'K'], ['A', 'K'],['D', 'E'], ['S', 'F'], ['B', 'C'], ['Y', 'V'], ['K', 'W']]

fq = 5

def frequency(original_pairs, pair_no_change, fq):
    updated_pairs = list(list())
    for pair in original_pairs:
        if pair != pair_no_change:
            pair[0] *= fq
            pair[1] *= fq
            updated_pairs.append([pair[0], pair[1]])
        else:
            continue
    return updated_pairs

updated_pairs = frequency(original_pairs, pair_no_change, fq)
#print(updated_pairs)
for pair in updated_pairs:
    print(pair)

这个函数给出了以下输出：

['DDDDD', 'EEEEE']
['SSSSS', 'FFFFF']
['BBBBB', 'CCCCC']
['YYYYY', 'VVVVV']
['KKKKK', 'WWWWW']

我需要一个输出：

['D','E']
['D','E']
['D','E']
['D','E']
['D','E']
['S','F']
['S','F']
['S','F']
['S','F']
['S','F']
etc.

我认为编写for循环的方式有问题，当我写对[0] * = fq时。谢谢你的时间和答案！

Answer 1

你正在通过乘以而不是那个来制作数组，只需追加。试试这个，

aminoacids = ['A', 'R', 'N', 'D', 'C', 'Q', 'E', 'G', 'H', 'I', 'L', 'K', 'M', 'F', 'P', 'S', 'T', 'W', 'Y', 'V']

pair_no_change = ['A', 'K']

original_pairs = [['A', 'K'], ['A', 'K'], ['A', 'K'], ['A', 'K'], ['A', 'K'],['D', 'E'], ['S', 'F'], ['B', 'C'], ['Y', 'V'], ['K', 'W']]

fq = 5

def frequency(original_pairs, pair_no_change, fq):
    updated_pairs = list(list())

    for pair in original_pairs:
        fq1 = fq
        if pair != pair_no_change:
            # pair[0] *= fq
            # pair[1] *= fq
            while fq1 > 0:
                updated_pairs.append([pair[0], pair[1]])
                fq1 -= 1
        else:
            continue
    return updated_pairs

updated_pairs = frequency(original_pairs, pair_no_change, fq)
#print(updated_pairs)
for pair in updated_pairs:
    print(pair)

Answer 2

你想要这样的东西：

>>> pair_no_change = ['A', 'K']
>>> original_pairs = [['A', 'K'], ['A', 'K'], ['A', 'K'], ['A', 'K'], ['A', 'K'],['D', 'E'], ['S', 'F'], ['B', 'C'], ['Y', 'V'], ['K', 'W']]
>>> fq = 5
>>> [[x]*fq for x in original_pairs if "".join(pair_no_change)!="".join(x)]
[[['D', 'E'], ['D', 'E'], ['D', 'E'], ['D', 'E'], ['D', 'E']], [['S', 'F'], ['S', 'F'], ['S', 'F'], ['S', 'F'], ['S', 'F']], [['B', 'C'], ['B', 'C'], ['B', 'C'], ['B', 'C'], ['B', 'C']], [['Y', 'V'], ['Y', 'V'], ['Y', 'V'], ['Y', 'V'], ['Y', 'V']], [['K', 'W'], ['K', 'W'], ['K', 'W'], ['K', 'W'], ['K', 'W']]]

Answer 3

只需在最后添加另一个for循环。

aminoacids = ['A', 'R', 'N', 'D', 'C', 'Q', 'E', 'G', 'H', 'I', 'L', 'K', 'M', 'F', 'P', 'S', 'T', 'W', 'Y', 'V']

pair_no_change = ['A', 'K']

original_pairs = [['A', 'K'], ['A', 'K'], ['A', 'K'], ['A', 'K'], ['A', 'K'],['D', 'E'], ['S', 'F'], ['B', 'C'], ['Y', 'V'], ['K', 'W']]

fq = 5

def frequency(original_pairs, pair_no_change, fq):
    updated_pairs = list(list())
    for pair in original_pairs:
        if pair != pair_no_change:
            pair[0] *= fq
            pair[1] *= fq
            updated_pairs.append([pair[0], pair[1]])
        else:
            continue
    return updated_pairs

updated_pairs = frequency(original_pairs, pair_no_change, fq)
#print(updated_pairs)
for pair in updated_pairs:
    for i ,j in zip(list(pair[0]), list(pair[1])):
        print [i,j]

<强>输出：

['D', 'E']
['D', 'E']
['D', 'E']
['D', 'E']
['D', 'E']
['S', 'F']
['S', 'F']
['S', 'F']
['S', 'F']
['S', 'F']
['B', 'C']
['B', 'C']
['B', 'C']
['B', 'C']
['B', 'C']
['Y', 'V']
['Y', 'V']
['Y', 'V']
['Y', 'V']
['Y', 'V']
['K', 'W']
['K', 'W']
['K', 'W']
['K', 'W']
['K', 'W']

Answer 4

再添加一个for循环

def frequency(original_pairs, pair_no_change, fq):
    updated_pairs = list(list())
    for pair in original_pairs:
        if pair != pair_no_change:

            for i in range(fq):

                updated_pairs.append([pair[0], pair[1]])

        else:
            continue
    return updated_pairs

Answer 5

您正在使用pair[0] *= fq。如果pair[0] = 'Q'和fq = 3则结果将是字符串'QQQ'，您只是将结果推送到updated_pairs。因此，使用for循环来推入updated_pairs数组。

def frequency(original_pairs, pair_no_change, fq):
    updated_pairs = list(list())
    for pair in original_pairs:
       if pair != pair_no_change:
           for i in range(fq):
               updated_pairs.append([pair[0], pair[1]])
       else:
           continue
return updated_pairs

python3，如何重复一个字符列表

5 个答案: