如何創建一個從給定列表而不是索引中得出數字的列表？

Question

我在獲取結果以生成列表中的整數時遇到麻煩，而不是它屬於哪個索引。

#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

我收到的結果如下：

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

我正在嘗試將結果設為[1、5、43、5、7]

Answer 1

您正在遍歷索引范圍。 而是遍歷列表。

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，您的代碼被重寫為（有一些小的更改）

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

你可以通過做來減少它

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]

輸出-

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

Answer 2

列表推導極大地簡化了您的代碼。

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)

代替添加索引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])

對於oddList情況oddList如此。

---

注意：@Sukrit Kalra的解決方案是更可取的，但我在此保留它，以表明有多種解決方法。

Answer 4

仔細看看.append（）命令...在比較中，您正在使用：

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

要么

if ((myList[i])>5)

...但是當您將其放入列表中時，您只是在使用

greaterList.append(i)

代替

greaterList.append(myList[i])

這一定是家庭作業或某個地方的課程嗎？