[ad_1] zipping names with list_price does most of the work for you: result = [] for name, prices in zip(names, list_price): result.append({“name”: name, “size_price”: prices, “size”: size}) 1 [ad_2] solved Data structure in python 2.7 [duplicate]
[ad_1] There is no problem with your usage of Iterator‘s next() and remove(). Your ConcurrentModificationException is caused by adding elements to the List after creating the Iterator. You should add elements to the List before the Iterator is created. Change: Iterator<Integer> iterator = list.iterator(); Collections.addAll(list, 1, 2, 3, 4, 5); to: Collections.addAll(list, 1, 2, 3, … Read more
[ad_1] You dont need to worry about memory leaks, in your case you use std::vector which uses RAII to make sure no memory is leaked. With STL containers you can “leak” in such case on;y if you do not shrink your vector after using remove algorithm. Below is fix to your code, and not entire … Read more
[ad_1] push_back might cause a reallocation of the data contained in the vector. And that reallocation will make all iterators to the vector invalid. Dereferencing invalid iterators leads to undefined behavior. Indexes into the vector will continue to stay valid, unless you remove elements from the vector. 2 [ad_2] solved Vector iterators incompatible error for … Read more
[ad_1] The Best answer would be that List has some kind of Tracking over its list items and can update its items as you ask but a simple IEnumerable does not so it will not allow you to change them. [ad_2] solved Why can’t you modify a collection in a for each loop
[ad_1] To understand what yield does, you must understand what generators are. And before you can understand generators, you must understand iterables. Iterables When you create a list, you can read its items one by one. Reading its items one by one is called iteration: >>> mylist = [1, 2, 3] >>> for i in … Read more