取不同字符串中的相同元素之一

Question

我想把我拥有的三个不同的系列结合起来。我还需要为每个元素提供奇点。输出顺序不是important.Example 3个数组：

a=["apple","banana","orange"]
b=["apple","banana"]
c=["cherry","grappe"]

结果

result=["apple","banana","orange","cherry","grappe"]

Answer 1

您可以使用 set 并这样做：

a=["apple","banana","orange"]
b=["apple","banana"]
c=["cherry","grappe"]

result = list(set(a + b + c))

>>> result = ["apple","banana","orange","cherry","grappe"]

Answer 2

a = ["apple", "banana", "orange"]
b = ["apple", "banana"]
c = ["cherry", "grappe"]

lst = list(set(a + b + c))
print(lst)

['banana', 'apple', 'cherry', 'grappe', 'orange']

查看 set 的文档以了解它在做什么。

Answer 3

您可以使用 itertools.chain 然后转换为设置以删除重复项，然后转换回列表（也进行了排序）：

import itertools
a=["apple","banana","orange"]
b=["apple","banana"]
c=["cherry","grappe"]
d = sorted(list(set(list(itertools.chain(a,b,c)))))
print(d)

Answer 4

您可以使用 set 删除重复项：

list(set(a+b+c))

Answer 5

如果你想保持第一次出现的顺序，正如你的示例输出所建议的那样，一种方法是创建一个以你的数据作为键的字典，利用字典自 Python 以来排序的事实3.7:

from itertools import chain

a = ["apple","banana","orange"]
b = ["apple","banana"]
c = ["cherry","grappe"]

d = {key: None for key in chain(a, b, c)}
result = list(d.keys())

print(result)
# ['apple', 'banana', 'orange', 'cherry', 'grappe']

Answer 6

a = ["apple", "banana", "orange"]
b = ["apple", "banana"]
c = ["cherry", "grappe"]
result = a + b + c  # marge all lsit
result = list(dict().fromkeys(result))  # remove duplicate
print(result)

参考：https://www.w3schools.com/python/python_howto_remove_duplicates.asp

取不同字符串中的相同元素之一

take one of the same elements in different strings

python

arrays

element

repeat

difference