Python 删除相邻重复元素后打印列表 new_list=[] i=0 def移除（nums）： 全球i 而i
问题：给定一个数字列表，返回一个列表，其中所有相邻==元素已减少为单个元素，因此Python 删除相邻重复元素后打印列表 new_list=[] i=0 def移除（nums）： 全球i 而i,python,recursion,Python,Recursion,问题：给定一个数字列表，返回一个列表，其中所有相邻==元素已减少为单个元素，因此[1,2,2,3]返回[1,2,3]。您可以创建新列表或修改传入的列表 问题：打印的最终列表由[1,2,3,5]组成，而不是[1,2,3,5,4,5,6,7,8,9,8]您想要的问题最好由 itertools.groupby所做的是，它通过生成元素的元组和作为列表的连续组将连续键分组在一起 为了清楚地理解itertools.groupby，您可以转储通过对连续数字列表进行分组而生成的元组结果列表 l Out[35]:
[1,2,2,3][1,2,3]问题：打印的最终列表由
[1,2,3,5][1,2,3,5,4,5,6,7,8,9,8]itertools.groupbyitertools.groupbyl
Out[35]: [1, 2, 3, 5, 4, 4, 5, 5, 6, 7, 7, 7, 8, 8, 9, 8, 8]
from itertools import groupby
[k for k, _ in groupby(l)]
Out[36]: [1, 2, 3, 5, 4, 5, 6, 7, 8, 9, 8]
你想要的是一个最好的解决方法
itertools.groupbyitertools.groupbyl
Out[35]: [1, 2, 3, 5, 4, 4, 5, 5, 6, 7, 7, 7, 8, 8, 9, 8, 8]
from itertools import groupby
[k for k, _ in groupby(l)]
Out[36]: [1, 2, 3, 5, 4, 5, 6, 7, 8, 9, 8]
我创建了一个函数，它获取一个列表并迭代其项，然后添加与上一次迭代中已经添加到列表中的项不同的项
[1, 2, 3, 5, 4, 5, 6, 7, 8, 9, 8]
我创建了一个函数，它获取一个列表并迭代其项，然后添加与上一次迭代中已经添加到列表中的项不同的项
[1, 2, 3, 5, 4, 5, 6, 7, 8, 9, 8]
这是一个完美的发电机。我不会改变原来的名单。相反，我返回的是一个新列表，相邻值之间没有相等的值
l = [1, 2, 3, 5, 4, 4, 5, 5, 6, 7, 7, 7, 8, 8, 9, 8, 8] # List that we want to change
def remove_adjacent(l): # Define a new function and accept an argument: the list to check.
new = [l[0]] # Make a new list (temporal) and assing like it's first item the first item of the main list. It's the same as new = [] and new.append(l[0]).
for item in l[1:]: # We iterate across the list, but we don't iterate on the first item because we've alreaday added it to the list, if you want you can delete the slice in [1:] since it will only make the iteration a really small fraction more slowly.
if new[-1] != item: # We check if the new item is the same as the last item added to the new list, if not, we add it.
new.append(item) # We add the item to the new list.
return new # We return the list.
print(remove_adjacent(l)) # We check it.
# [1, 2, 3, 5, 4, 5, 6, 7, 8, 9, 8]
设置
def removerator(l):
last = None
for x in l:
if x != last:
last = x
yield x
list(removerator(l))
[1, 2, 3, 5, 4, 5, 6, 7, 8, 9, 8]
这是一个完美的发电机。我不会改变原来的名单。相反，我返回的是一个新列表，相邻值之间没有相等的值
l = [1, 2, 3, 5, 4, 4, 5, 5, 6, 7, 7, 7, 8, 8, 9, 8, 8] # List that we want to change
def remove_adjacent(l): # Define a new function and accept an argument: the list to check.
new = [l[0]] # Make a new list (temporal) and assing like it's first item the first item of the main list. It's the same as new = [] and new.append(l[0]).
for item in l[1:]: # We iterate across the list, but we don't iterate on the first item because we've alreaday added it to the list, if you want you can delete the slice in [1:] since it will only make the iteration a really small fraction more slowly.
if new[-1] != item: # We check if the new item is the same as the last item added to the new list, if not, we add it.
new.append(item) # We add the item to the new list.
return new # We return the list.
print(remove_adjacent(l)) # We check it.
# [1, 2, 3, 5, 4, 5, 6, 7, 8, 9, 8]
设置
def removerator(l):
last = None
for x in l:
if x != last:
last = x
yield x
list(removerator(l))
[1, 2, 3, 5, 4, 5, 6, 7, 8, 9, 8]
在循环中，打印一些变量以查看发生了什么。获取列表中相邻项对的一种简单方法是使用zip（nums，nums[1:]：print（a，b）-不知道这是否有帮助。当然，我不是在为您分配任务，但看起来您试图决定是使用循环还是递归，但最终您同时使用了这两种方法。（注：去掉全局变量。使用返回值。）@KennyOstrom我从未要求任何人做我的作业。在循环中，打印一些变量以查看发生了什么。获取列表中相邻项对的一种简单方法是
for a，b，In-zip（nums，nums[1:]：print（a，b）