Python在列表中移动多个元素

Question

如果从某个索引开始有空格，我需要移动列表中的元素，移动元素右侧的元素不能从它们的索引移动，一个演示：

# desired output
["not end","x","y","","","","don't move"]


# works here
l = ["not end","","","","x","y","don't move"]
start = 1
aext_len = 3
end = start + aext_len + 1

for empty, cell in enumerate(l[start:end - 1], 1):
    if cell:
        break

for z in range(aext_len + 2 - empty):
    l.insert(start + z, l.pop(start + empty + z))
print (l)
#['not end', 'x', 'y', '', '', '', "don't move"]


# not here
l = ["not end","","x","y","","","don't move"]
start = 1
aext_len = 3
end = start + aext_len + 1

for empty, cell in enumerate(l[start:end - 1], 1):
    if cell:
        break

for z in range(aext_len + 2 - empty):
    l.insert(start + z, l.pop(start + empty + z))
print (l)
#['not end', 'y', '', '', '', 'x', "don't move"]

Answer 1

解决方案实际上要简单得多。

l = ["not end", "", "", "", "x", "y", "don't move"]

定义“请勿移动”区域的末端：

MOVE = 6

“挤出”空字符串：

part1 = [x for x in l[:MOVE] if x]

将“挤出”的空字符串移到末尾：

part2 = (MOVE - len(part1)) * [""]

组合碎片：

part1 + part2 + l[MOVE:]
#['not end', 'x', 'y', '', '', '', "don't move"]

或者，您可以通过为空字符串的属性对列表的第一部分进行排序：

sorted(l[:MOVE], key=lambda x: x=="") + l[MOVE:]
#['not end', 'x', 'y', '', '', '', "don't move"]