问题描述
问题是这样的,取两个列表,比如说这两个:
Problem is this, take two lists, say for example these two:
a = [1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89] b = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]
并编写一个程序,该程序返回一个列表,该列表仅包含列表之间共有的元素(没有重复).确保您的程序适用于两个不同大小的列表.
And write a program that returns a list that contains only the elements that are common between the lists (without duplicates). Make sure your program works on two lists of different sizes.
这是我的代码:
a = [1, 1, 2, 2, 3, 5, 8, 13, 21, 34, 55, 89] b = [1, 2, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13] c = [] for i in a: if i in b and i not in c: c.append([i]) print(c)
尽管有i not in c"语句,但我的输出仍然给我重复.为什么是这样?我敢肯定它很明显,我只是看不到它!
My output is still giving me duplicates despite the 'i not in c' statement. why is this? I'm sure its blatantly obvious, I just cant see it!
推荐答案
- 您将包含 i 的列表附加到 c,因此 i not in c 将始终返回 <代码>正确代码>.您应该单独附加 i:c.append(i)
- You are appending a list containing i to c, so i not in c will always return True. You should append i on its own: c.append(i)
或者
只需使用集合(如果顺序不重要):
Simply use sets (if order is not important):
a = [1, 1, 2, 2, 3, 5, 8, 13, 21, 34, 55, 89] b = [1, 2, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13] c = set(a) & set(b) # & calculates the intersection. print(c) # {1, 2, 3, 5, 8, 13}
编辑作为@Ev.Kounis 在评论中建议,您将通过使用获得一些速度
c = set(a).intersection(b).
EDIT As @Ev. Kounis suggested in the comment, you will gain some speed by using
c = set(a).intersection(b).