問題描述

問題是這樣的，取兩個列表，比如說這兩個:

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]

并編寫一個程序，該程序返回一個列表，該列表僅包含列表之間共有的元素(沒有重復).確保您的程序適用于兩個不同大小的列表.

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.

這是我的代碼:

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)

盡管有i not in c"語句，但我的輸出仍然給我重復.為什么是這樣?我敢肯定它很明顯，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!

推薦答案

1. 您將包含 i 的列表附加到 c，因此 i not in c 將始終返回 <代碼>正確.您應該單獨附加 i:c.append(i)

1. 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

1. 只需使用集合(如果順序不重要):

1. 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}

編輯作為@Ev.Kounis 在評論中建議，您將通過使用獲得一些速度
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).

這篇關于兩個列表之間沒有重復的公共元素的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網(wǎng)！