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Question

我目前的输出[1,2,3,4,5,6] [2,1,3,4,5,6] [3,2,1,4,5,6] ...... [6,5,4,3,2,1] 所有数字都相同，但我只想要这个输出一个[1,2,3,4,5,6]我该怎么做

import    os
import    sys

numbers=[1,2,3,4,5,6]
x=[]
for i1 in numbers:
    for i2 in numbers:
        for i3 in numbers:
            for i4 in numbers:
                for i5 in numbers:
                    for i6 in numbers:
                        if i1 not in (i2,i3,i4,i5,i6) and i2 not in (i1,i3,i4,i5,i6) and i3 not in (i1,i2,i4,i5,i6) and i4 not in (i1,i2,i3,i5,i6) and i5 not in (i1,i2,i3,i4,i6) and i6 not in (i1,i2,i3,i4,i5):
                            x.append([i1,i2,i3,i4,i5,i6])

i=0
while i < len(x):
    print x[i]
    i=i+1

Answer 1

这是一个相当不明确的问题，但要删除列表中的重复项：

nums = [1,2,3,3,4,5,6]
x = []
for i in nums:
    if i not in x:
        x.append(i)

[print(i) for i in x] # output -> 1 2 3 4 5 6

只能从几个列表中获取独特内容：

nums = [1,2,3,3,4,5,6]
nums2 = [2,1,3,4,4,5,7,8]

lists = nums + nums2

x = []
for i in lists:
    if i not in x:
        x.append(i)

[print(i) for i in x] # output -> 1 2 3 4 5 6 7 8

编辑并回答

只获取一组唯一的数字：

nums1 = [1,2,3,4,5,6]
nums2 = [3,2,1,6,5,4]
nums3 = [4,3,2,5,1,6]
nums4 = [2,3,4,5,6,7]

s1 = set(nums1)
s2 = set(nums2)
s3 = set(nums3)
s4 = set(nums4)

sets = [s1, s2, s3, s4]
uniq = []
for s in sets:
    if s not in uniq:
        uniq.append(s)

for u in uniq:
    [print(i, end=' ') for i in u]
    print() # output -> {1,2,3,4,5,6} and {2,3,4,5,6,7}

不需要使用集合，但会确保原始列表中没有重复项，并且在其他方面也会更容易。

Answer 2

set()它！

只需在要删除重复项的列表中使用set()即可。然后list()将它投射到列表中。

Answer 3

您可以在python中使用list comprehensions来实现此目的。

假设您有多个列表，并希望将这些列表中的所有唯一元素放入一个数组中，您可以这样编码。

number_list_1 = [1,2,3,4,5]
number_list_2 = [4,3,2,1,6]
number_list_3 = [2,4,1,3,7]
no_dupes = []

[no_dupes.append(item) for item in number_list_1 if item not in no_dupes]
[no_dupes.append(item) for item in number_list_2 if item not in no_dupes]
[no_dupes.append(item) for item in number_list_3 if item not in no_dupes]

print str(no_dupes)

这将返回：[1, 2, 3, 4, 5, 6, 7]

我想不列出重复数量的数组

3 个答案:

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