Question

我写了一个看似简单的循环。列表中有6个项目，它应该循环六次。但是，它只循环3次。的为什么吗

list1 = 'one two three four five six'
newlist = list1.split(' ')
print (newlist)
list2 = ['seven', 'eight', 'nine', 'ten', 'eleven', 'twelve']

for number in list2:
    nextnumber = list2.pop()
    print ("Adding number ", nextnumber)
    newlist.append(nextnumber)

print (newlist)

Answer 1

如评论中所述，您在迭代时删除项目。您可以使用while循环更好地建模此模式：

while list2:
    newlist.append(list2.pop())

Answer 2

使用list2循环将newlist添加到for可以这样做：

for number in list2:
    print ("Adding number ", number)
    newlist.append(number)

但短暂，快速和pythonic的方式是

newlist.extend(list2)

Answer 3

请改用以下代码：

for number in list2:
    print ("Adding number ", number)
    newlist.append(number)
print (newlist)

观察：list2.pop（）正在减少你的列表元素，它也减少了循环次数

Answer 4

它只会循环三次，因为您正在主动修改list2。一个简单的修正就是复制!!

导入副本

list2_c = copy.copy(list2)

for num in list2_c:
.
.
.
.

Answer 5

快速回答，复制list2：

list1 = 'one two three four five six'
newlist = list1.split(' ')
print (newlist)
list2 = ['seven', 'eight', 'nine', 'ten', 'eleven', 'twelve']

for number in list(list2):
    nextnumber = list2.pop()
    print ("Adding number ", nextnumber)
    newlist.append(nextnumber)

print (newlist)

Answer 6

像先前的答案所说的那样，你正在使用list2，换句话说，当你为list2中的数字做时，你正在服用7,8,9但从不使用它们，而是你要删除12,11,10从列表中打印出来。你可以这样做：

    for number in list2:
    #Unless you want to append in reverse order(see below for that)
    #comment the next line
        #nextnumber = list2.pop() 
        print ("Adding number ", number) 
        newlist.append(number)#Append the number you just got from list2
    print (newlist)

  out[:]: ['one', 'two', 'three', 'four', 'five', 'six']
  Adding number  seven
  Adding number  eight
  Adding number  nine
  Adding number  ten
  Adding number  eleven
  Adding number  twelve
  ['one', 'two', 'three', 'four', 'five', 'six', 'seven', 'eight', 'nine', 
  'ten', 'eleven', 'twelve']

如果您想以相反的顺序追加，请执行@thachnb建议的内容。

仅循环循环3次。为什么呢？

6 个答案: