元组打印元素,如果条件否则打印范围内的另一个值



我有 2 个列表,其中 1 个是嵌套的。我想用条件将其导出到 txt。每个不同的字母都以...即:"字母,A",后跟其元组的第三个元素存在,否则打印一个","。存在的条件是元组的第二个元素与var在范围内(从 0 到 5(:

var=5
letter=['A','B','C','D','E','F','G','H']
nested_list=[
('A', 1, 0),
('A', 2, 0),
('B', 1, 9),
('B', 3, 9),
('C', 2, 0),
('C', 4, 0),
('C', 5, 0),
('D', 2, 9),
('E', 3, 0),
('F', 3, 9)]

到目前为止我的代码:

bd="Letter,"
for i in range(0,len(nested_list)-1):
    if nested_list[i][0]!=nested_list[i+1][0]:
        bd+="nLetter,%s,"%(nested_list[i][0])
        for j in range(0,var):
            if nested_list[i][1]==j:
                bd+="%s,"%nested_list[i][2]
            else:
                bd+=","
    elif nested_list[i][0]==nested_list[i+1][0]:
        bd+="n"
        for j in range(0,var):
            if nested_list[i+1][1]==j:
                bd+="%s,"%nested_list[i+1][2]
            else:
                bd+=","
print bd

电流输出:

Letter,A,,,0,,,
Letter,A,,,0,,,B,,,,9,,
Letter,B,,,,9,,C,,,,,0,C,,,,,,
Letter,C,,,,,,
Letter,D,,,9,,,
Letter,E,,,,0,,

预期输出

Letter,A,0,0,,,
Letter,B,9,,9,,
Letter,C,,0,,0,0
Letter,D,,9,,,
Letter,E,,,0,,
Letter,F,,,0,,

有什么建议吗?

如果你的列表总是正确排序,我认为这是itertools.groupby的一个很好的用例:

In [14]: from itertools import groupby
In [15]: from operator import itemgetter
In [16]: for k, group in groupby(nested_list, itemgetter(0)):
    ...:     plist = ['']*5
    ...:     for _, idx, val in group:
    ...:         plist[idx-1] = str(val)
    ...:     print("Letter,{},{}".format(k, ','.join(plist)))
    ...:
Letter,A,0,0,,,
Letter,B,9,,9,,
Letter,C,,0,,0,0
Letter,D,,9,,,
Letter,E,,,0,,
Letter,F,,,9,,

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