Question

def purify(num):
    for number in num:
        if number%2 == 0:
            number = list(str(number))
            print(number)


purify([1,8,9,6,5,3,4,6,5,8,8,9,9,2,3,1,5])

o / p - [＆＃39; 8＆＃39;] [＆＃39; 6＆＃39;] [＆＃39; 4＆＃39;] [＆＃39; 6＆＃39;] [＆＃39; 8＆＃39;] [＆＃39; 8＆＃39;] [＆＃39; 2＆＃39;]

我正在尝试从列表中删除所有奇数，我已将所有数字转换为字符串，但在打印后我得到多个列表。

Answer 1

这是以同样的方式完成的：

num = [1,8,9,6,5,3,4,6,5,8,8,9,9,2,3,1,5]
newlist = [str(i) for i in num if i%2==0]
print(newlist)

它给出了这样的o / p：

['8', '6', '4', '6', '8', '8', '2']

Answer 2

def purify(num):
    even = []
    for number in num:
        if number%2 == 0:
            even.append(str(number))
    print("['{}']".format("']['".join(even)))

然后

>>> purify([1,8,9,6,5,3,4,6,5,8,8,9,9,2,3,1,5])
['8']['6']['4']['6']['8']['8']['2']

Answer 3

def purify(num):
    newlist=[]
    for number in num:
        if number%2 == 0:
            number = str(number)
            newlist.append(number)
    print(newlist)
    return newlist

试试这个。您需要一个列表来存储数字。输出：

['8', '6', '4', '6', '8', '8', '2']

Answer 4

如果您上次的结果是奇数，那么您可以尝试：

ls=[1,8,9,6,5,3,4,6,5,8,8,9,9,2,3,1,5]
print(list(filter(lambda x:not x%2,ls)))

输出：

[8, 6, 4, 6, 8, 8, 2]

如果你想要字符串：

[['8'], ['6'], ['4'], ['6'], ['8'], ['8'], ['2']]

如何在python中将输出更改为单个列表？

4 个答案: