Question

我无法获得结果以在列表中生成整数，而不是它所属的索引。

#this function takes, as input, a list of numbers and an option, which is either 0 or 1.
#if the option is 0, it returns a list of all the numbers greater than 5 in the list
#if the option is 1, it returns a list of all the odd numbers in the list
def splitList(myList, option):
    #empty list for both possible options
    oddList = []
    greaterList = []
    #if option is 0, use this for loop
    if int(option) == 0:
        #create for loop the length of myList
        for i in range(0, len(myList)):
            #check if index is greater than 5
            if ((myList[i])>5):
                #if the number is greater than 5, add to greaterList
                greaterList.append(i)
        #return results
        return greaterList
    #if option is 1, use this for loop
    if int(option) == 1:
        #create for loop the length of myList
        for i in range(0, len(myList)):
            #check if index is odd by checking if it is divisible by 2
            if ((myList[i])%2!=0):
                #if index is not divisible by 2, add the oddList
                oddList.append(i)
        #return results
        return oddList

我收到的结果如下：

>>>splitList([1,2,6,4,5,8,43,5,7,2], 1)
   [0, 4, 6, 7, 8]

我想把结果变成[1,5,43,5,7]

Answer 1

您正在迭代索引的范围。相反，迭代列表。

for i in myList:
    #check if index is greater than 5
    if i >5:
        #if the number is greater than 5, add to greaterList
        greaterList.append(i)

因此，您的代码会被重写为（稍作修改）

def splitList(myList, option):
    final_list = []
    if int(option) == 0:
        for i in myList:
            if i > 5:
                final_list.append(i)
    elif int(option) == 1:
        for i in myList:
            if i%2 != 0:
                final_list.append(i)
    return final_list

您可以通过

来减少它

def splitList(myList, option):
    if int(option) == 0:
        return [elem for elem in myList if elem > 5]
    elif int(option) == 1:
        return [elem for elem in myList if elem % 2 != 0]

输出 -

>>> splitList([1,2,6,4,5,8,43,5,7,2], 1)
[1, 5, 43, 5, 7]

Answer 2

列表推导大大简化了您的代码。

def split_list(xs, option):
    if option:
        return [x for x in xs if x % 2]
    else:
        return [x for x in xs if x > 5]

Answer 3

if ((myList[i])>5):
    #if the number is greater than 5, add to greaterList
    greaterList.append(i)

不添加索引i，而是添加值（myList[i]）：

if ((myList[i])>5):
    #if the number is greater than 5, add to greaterList
    greaterList.append(myList[i])

oddList案例也是如此。

注意：@Sukrit Kalra的解决方案更可取，但我要离开这一点，以表明有多种解决方法。

Answer 4

在你正在使用的比较中仔细查看你的.append（）命令......

if ((mylList[i])%2!=0)

或

if ((myList[i])>5)

...但是当你把它放到列表中时，你只是使用

greaterList.append(i)

而不是

greaterList.append(myList[i])

这必须是某个地方的家庭作业或课堂吗？

如何创建一个列表，从而得到给定列表中的数字，而不是索引？

4 个答案: