Python append方法工作不正常

Question

def sem1Sort1(semester1, selectionSEM1):

    list = []

    for period in semester1:
        if period == 1:
            for index in semester1[period]:
                if index in selectionSEM1:
                    list.append(index)

    return list

def sem1Sort2(semester1, selectionSEM1):

    list = []

    for period in semester1:
        if semester1 == 2:
            for index in semester1[period]:
                if index in selectionSEM1:
                    list.append(index)

    return list

def main():

    selectionSEM1 = ["a", "b", "c", "d", "e", "f", "g", "h"]
    selectionSEM2 = []

    semester1 = {

    1: ["e", "f", "g", "h"], 2: ["a", "b", "c", "d"] ,

    }

    SEM1period1 = sem1Sort1(semester1, selectionSEM1)
    SEM1period2 = sem1Sort2(semester1, selectionSEM1)

    print SEM1period1
    print SEM1period2

main()

当我运行这段代码时，它会输出SEM1period1 fine，如[“e”，“f”，“g”，“h”]，但是第二个方法sem1Sort2似乎不会将任何内容保存到SEM1period2中-因为print语句会输出[]

更新：

def sem1Sort1(semester1, selectionSEM1):

    list = []

    for period in semester1:
        if period == 1:
            for index in semester1[period]:
                if index in selectionSEM1:
                    list.append(index)

    return list

def sem1Sort2(semester1, selectionSEM1):

    list = []

    for period in semester1:
        if period == 2:
            for index in semester1[period]:
                if index in selectionSEM1:
                    list.append(index)

    return list

def sem1Sort3(semester1, selectionSEM1):

    list = []

    for period in semester1:
        if period == 3:
            for index in semester1[period]:
                if index in selectionSEM1:
                    list.append(index)

def main():

    selectionSEM1 = ["a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l"]
    selectionSEM2 = []


    semester1 = {
    1: ["e", "f", "g", "h"], 2: ["a", "b", "c", "d"] ,
    3: ["i", "j", "k", "l"]
    }

    SEM1period1 = sem1Sort1(semester1, selectionSEM1)
    SEM1period2 = sem1Sort2(semester1, selectionSEM1)
    SEM1period3 = sem1Sort3(semester1, selectionSEM1)

    print SEM1period1
    print SEM1period2
    print SEM1period3

main()

为什么print SEM1period3不返回none？你知道吗

Answer 1

为什么这么复杂？你知道吗

你不需要循环所有的dict条目并找出匹配的条目-这使得dict的整个用法毫无意义。你知道吗

相反，您可以告诉dict为您提供与给定键相关联的值。您可以在main()函数中这样做，将函数简化为

def semSort(semester, selection):
    list = []
    for index in semester:
        if index in selection:
            list.append(index)
    return list

可以在main()函数中调用

def main():
    selectionSEM1 = ["a", "b", "c", "d", "e", "f", "g", "h"]
    selectionSEM2 = []
    semester1 = {
        1: ["e", "f", "g", "h"], 2: ["a", "b", "c", "d"] ,
    }

    SEM1period1 = semSort(semester1[1], selectionSEM1)
    SEM1period2 = semSort(semester1[2], selectionSEM1)

    print SEM1period1
    print SEM1period2

main()

你会得到你想要的。你知道吗

您甚至可以对其进行优化：

def semSort(semester, selection):

    result = []
    sel_set = set(selection)    
    for index in semester:
        if index in sel_set:
            result.append(index)

    return result

集合使查找更快。你知道吗

def semSort(semester, selection):
    sel_set = set(selection)
    return [index for index in semester if index in sel_set]

更紧凑。你知道吗

Answer 2

def sem1Sort1(semester1, selectionSEM1):
 list = []
 for period in semester1:
   if period == 1:
     for index in semester1[period]:
       if index in selectionSEM1:
          list.append(index)

 return list

def sem1Sort2(semester1, selectionSEM1):
 list = []
 for period in semester1:
  if period == 2:
   for index in semester1[period]:
    if index in selectionSEM1:
       list.append(index)
 return list

def main():
 selectionSEM1 = ["a", "b", "c", "d", "e", "f", "g", "h"]
 selectionSEM2 = []
 semester1 = {
 1: ["e", "f", "g", "h"], 2: ["a", "b", "c", "d"] ,
 }
 SEM1period1 = sem1Sort1(semester1, selectionSEM1)
 SEM1period2 = sem1Sort2(semester1, selectionSEM1)
 print SEM1period1
 print SEM1period2
main()

Answer 3

这里比较semester1和integer，但是semester1是sem1Sort2函数中的dict对象

    for period in semester1:
        if semester1 == 2:

实际上，您必须将整数与dict的键进行比较，如下所示

    for period in semester1:
        if period == 2:

你剩下的时间都是这样

    def sem1Sort1(semester1, selectionSEM1):

    list = []

    for period in semester1:
        if period == 1:
            for index in semester1[period]:
                if index in selectionSEM1:
                    list.append(index)

    return list

def sem1Sort2(semester1, selectionSEM1):

    list = []

    for period in semester1:
        if period == 2:
            for index in semester1[period]:
                if index in selectionSEM1:
                    list.append(index)

    return list

def main():

    selectionSEM1 = ["a", "b", "c", "d", "e", "f", "g", "h"]
    selectionSEM2 = []

    semester1 = {

    1: ["e", "f", "g", "h"], 2: ["a", "b", "c", "d"] ,

    }

    SEM1period1 = sem1Sort1(semester1, selectionSEM1)
    SEM1period2 = sem1Sort2(semester1, selectionSEM1)

    print SEM1period1
    print SEM1period2

main()

输出：

['e', 'f', 'g', 'h']
['a', 'b', 'c', 'd']