Question

因此，我尝试将一次用于list1的项目打印一次，然后item2的第一个元素指示要打印列表3中的多少项目，然后围绕下一个循环，从它停止的地方开始。然后重复，直到我用完list1中的所有项目为止。我想我所有列表的格式都正确，因为list1是一个字符串列表，list2是一个整数列表，list3是另一个字符串列表

prevval = 0
for i in list1:
    print(i)
    for j in list2:
        val = j
        print(list3[prevval:val])
        prevval = val

它打印出列表3的前x个项目，总共8次（这是列表2中的元素个数）

示例：

list1 = ["test","please"]
list2 = [2, 4, 6]
list3 = ["hard", "to", "come", "up","with","values"]

它将打印：

test
hard to
hard to
hard to
hard to come

预期为：

test
hard to
please
come up with values

等

Answer 1

对于list1中的每个单词，您要打印list3的一个切片，该切片由list2中的相应值确定。您可以通过压缩list1和list2并计算适当的切片来做到这一点：

list1 = ["test","please"]
list2 = [2, 4, 6]
list3 = ["hard", "to", "come", "up","with","values"]

prevval = 0
for w1, i in zip(list1, list2):
    print(w1)
    if prevval < len(list3): 
        print(list3[prevval:prevval + i])
        prevval += i

输出

test
['hard', 'to']
please
['come', 'up', 'with', 'values']

如果要将列表设置为字符串格式，可以使用join：

' '.join(print(list3[prevval:prevval + i]))

Answer 2

prevval = 0

val=0

for i in list1:

    print(i)

    val=list2[val]

    print(list3[prevval:val])

    prevval = val

Answer 3

这将摧毁list3：

list1 = ["test","please"]
list2 = [2, 4, 6]
list3 = ["hard", "to", "come", "up","with","values"]

l1_idx = 0
l2_idx = 0

while l3_idx < len(list3):
  print(list1[l1_idx])
  l1_idx += 1
  outarr = []
  for x in range(list2[l2_idx]):
    try:
      outarr.append(list3.pop(0))
    except IndexError:
      break
  print(" ".join(outarr))
  l2_idx += 1

Answer 4

这里最短的一个：

prevval=0
for x,y in zip(list1,list2):
   print(x)
   print(' '.join(list3[prevval:prevval+y]))
   prevval+=y

输出为：

test
hard to
please
come up with values

zip遍历它们，像往常一样添加prevval，然后删除内部循环，这就是所有更改，请注意，这使其变短了（到目前为止）。

Answer 5

#!/bin/python

list1 = ["test","please"]
list2 = [2, 4, 6]
list3 = ["hard", "to", "come", "up","with","values"]

for i in list1:
    print(i)
    s = []
    for j in range(list2.pop(0)):
        s.append(list3.pop(0))
    print(" ".join(s))

输出：

test
hard to
please
come up with values

Answer 6

list1 = ['test', 'please', 'this', 'works']
list2 = [2,4,6]
list3 = tmp = ['hard', 'to', 'come', 'up', 'with', 'values', 'because', 'it', 'is', 'simple', 'with', 'python']
for index, item in enumerate(list1):
    try:
       print(tmp[:list2[index]])
       tmp = tmp[list2[index]:]
    except:
       break

打印“ x”个项目“ y”次？

6 个答案: