Question

我有以下代码。它正在工作，除非我有一个重复3次或更多次的重复，程序正在考虑该元素作为新数字并显示两次。如何解决这个问题？

import collections
mylist = [1,2,3,4,5,5,6,7,7,7,8,9]
i = 0
count = 0


for i in range(len(mylist)):
    print1 = mylist.count(mylist[i])
    if print1 > 1:

        print("Duplicate Element: " , mylist[i])
        print("It is repeated " , print1, "times")
        mylist.remove(mylist[i])
        count += 1
        i += 1

我得到的输出是

Duplicate Element:  5
It is repeated  2 times
Duplicate Element:  7
It is repeated  3 times
Duplicate Element:  7
It is repeated  2 times
Traceback (most recent call last):
  File "C:/Users/sanantha/Documents/Python/Lab6/task3.py", line 10, in <module>
    print1 = mylist.count(mylist[i])
IndexError: list index out of range

如何解决这个问题？

Answer 1

以下是使用itertools.groupby：

进行操作的另一种方法

from itertools import groupby

print(*["Duplicate element: {}. It is repeated {} times."
         .format(k, len(list(group))) for k, group in groupby(mylist)
         if len(list(group)) > 1],
         sep='\n')

输出：

Duplicate element: 5. It is repeated 2 times.
Duplicate element: 7. It is repeated 3 times.

Answer 2

使用collections.Counter，它是为此而构建的：

from collections import Counter

mylist = [1,2,3,4,5,5,6,7,7,7,8,9]
for value, repeats in Counter(mylist).items():
    if repeats > 1:
        print("Value {} repeats {} times".format(value, repeats)

Answer 3

您的代码只需要修复。这种方式的最佳方法是使用sets

mylist = [1,2,3,4,5,5,6,7,7,7,8,9]

for i in set(mylist):
    count = mylist.count(i)
    if count > 1:
        print('Duplicate Element: %i' % i)
        print('It is repeated  %i times' % count)

Answer 4

索引超出范围意味着您正在尝试访问数据结构中超出范围的元素。

E.g。你有一个数组[1,2,3,4]（索引：0,1,2,3） - 你在索引1上执行删除。这将给你一个[1,3,4]数组（索引：0,1,2）。现在尝试访问索引3将导致索引超出范围异常，因为它不再存在。

显示重复项和重复项数

4 个答案: