[英]"group" items in list
I have [string, int,int],[string, int,int]...
kind of list that I want to group with a different list我有
[string, int,int],[string, int,int]...
类型的列表,我想用不同的列表分组
[string1, int1,int1] + [string2, int2,int2] = ["string+string2", int1+int1+int2+int2]
the History goes like I have already made import function that gets me compounds:历史就像我已经制作了让我复合的导入功能:
ex[Ch3, 15.3107,15.284] kinda like this... ex[Ch3, 15.3107,15.284] 有点像这样......
I have a function that gives me: dictionary{0:"CH3"}
and another that gives me: List ["CH3",30.594700000000003]
我有一个函数给我:
dictionary{0:"CH3"}
和另一个给我的函数: List ["CH3",30.594700000000003]
def group_selectec_compounds(numCompound,values)
values can be list of list that have everything I also have dic made that is something like this {0:["CH4"],...}
numCoumpound
should be various variables (I think) or tuple of keys?值可以是列表列表,其中包含我也有 dic 制作的所有内容,类似于
{0:["CH4"],...}
numCoumpound
应该是各种变量(我认为)或键元组? So I can do the math for the user.所以我可以为用户做数学计算。
In the end I want something like: ["CH3+CH4",61.573]
it can also be: ["CH3+CH4+H2SO4",138.773]
最后我想要的是:
["CH3+CH4",61.573]
也可以是: ["CH3+CH4+H2SO4",138.773]
I would solve this using '+'.join
, sum
and comprehensions:我会使用
'+'.join
、 sum
和 comprehensions 来解决这个问题:
>>> data = [['string1', 2, 3], ['string2', 4, 5], ['string3', 6, 7]]
>>> ['+'.join(s for s, _, _ in data), sum(x + y for _, x, y in data)]
['string1+string2+string3', 27]
First, create a dictionary that stores the location of the type:首先,创建一个存储类型位置的字典:
helper = {}
for elem in lst1:
elemType = str(type(elem))
if elemType not in helper:
helper[elemType] = lst1.index[elem]
Now you have a dictionary that indexes your original list by type, you just have to run the second list and append accordingly:现在你有一个按类型索引原始列表的字典,你只需要运行第二个列表并相应地附加:
for elem in lst2:
elemType = str(type(elem))
if elemType not in helper:
#in case list 2 contains a type that list 1 doesn't have
lst1.append(elem)
helper[elemType] = lst1.index[elem]
else:
lst1[helper[elemType]] += elem
Hope this makes sense!希望这是有道理的! I have not vetted this for correctness but the idea is there.
我没有审查这个的正确性,但这个想法就在那里。
Edit: This also does not solve the issue of list 1 having more than 1 string or more than 1 int, etc., but to solve that should be trivial depending on how you wish to resolve that issue.编辑:这也不能解决列表 1 具有超过 1 个字符串或超过 1 个 int 等的问题,但根据您希望如何解决该问题,解决该问题应该很简单。
2nd Edit: This answer is generic, so it doesn't matter how you order the strings and ints in the list, in fact, lst1 can be [string, int, double] and lst2 can be [int, double, string] and this would still work, so it is robust in case the order of your list changes第二次编辑:这个答案是通用的,所以你如何对列表中的字符串和整数进行排序并不重要,事实上,lst1 可以是 [string, int, double] 而 lst2 可以是 [int, double, string] 和这仍然有效,因此在您的列表顺序发生变化的情况下它很强大
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