我有两份名单:
data = [[1,2,3,4], [5,6,7], [8,9,10,11,12,13,14]]
splitters = [3,7,10,13]我希望使用以下条件将数据中的嵌套列表按拆分器中的值拆分:
最终的结果应该是:
results = [[1,2,3],[3,4],[5,6,7],[8,9,10],[10,11,12,13],[13,14]我的第一次尝试是这样的:
temp = []
for route in data:
for node in route:
if node in splitter and ((route.index(node) !=0) and (route.index(node) != (len(route)-1))):
#route should be splitted and save it for now with the splitter
temp.append([route, node])
#here a big part is missing
#start a new subroute
#maybe something like a whileloop with len(route)
#check the same if-statement for the remaining subroute
else:
#no splitter in this route, so keep the original route
temp.append([route, 0])临时工看起来是这样的:
[[[1, 2, 3, 4], 0],
[[1, 2, 3, 4], 0],
[[1, 2, 3, 4], 3],
[[1, 2, 3, 4], 0],...]在此基础上,我可以删除多余的路由并分割路由,但我认为我的方法不必要地复杂,如果我想要实现一些满足其他条件的东西,就会变得越来越混乱。
到目前为止,我的研究还没有成功(使用itertools.groupby等)。这是有关联的:values/
将欣赏一些想法/方法如何解决这个问题或细分它在较小的部分。
面向未来读者的编辑:--我更喜欢maxiotic的解决方案,因为它甚至可以处理类似的数据。
data = [[1,2,3],[1,2,3,4,5,6,7]]
splitters = [1,2,3,4,7]嵌套列表的每个开始/结束都在拆分器中。Relondom解决方案中的问题是以下if语句,必须更改:
if inner[0] in splitters or inner[-1] in splitters: # check if first or last elemtn in splitters非常感谢!
发布于 2019-05-13 20:20:12
下面是我正在研究的解决方案:
results = []
for route in data:
found = 0
for idx, r in enumerate(route[1:-1], 1): # start idx at 1
if r in splitters:
temp = route[found:idx+1] # +1 to capture the splitter value
results.append(temp)
found = idx
remaining = route[found:]
results.append(remaining)发布于 2019-05-13 19:47:36
我不知道这是否是一种最佳的方法,但我决定编写这段代码,因为还没有人回答。
res = []
for inner in data:
if inner[0] in splitters or inner[-1] in splitters: # check if first or last elemtn in splitters
res.append(inner)
continue
else:
temp = []
for val in inner:
if val not in splitters:
temp.append(val)
else:
temp.append(val) # list ends with value from splitters
res.append(temp) # add new list to result
temp = [val] # new list starts with value from splitters
if temp not in res:
res.append(temp)https://stackoverflow.com/questions/56118707
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