Question

我想要做的是从列表中删除重复项，每次删除重复项时都插入一个空项目。

我有删除重复项的代码。它还会忽略空列表项

import csv

#Create new output file

new_file = open('addr_list_corrected.csv','w')
new_file.close()

with open('addr_list.csv', 'r') as addr_list:
    csv_reader = csv.reader(addr_list, delimiter=',')
    for row in csv_reader:

        print row
        print "##########################"
        seen=set()
        seen_add=seen.add
        #empty cell/element evaluates to false

        new_row = [ cell for cell in row if not (cell and cell in seen or seen_add(cell))]
        print new_row

        with open('addr_list_corrected.csv', 'a') as addr_list_corrected:
            csv_writer=csv.writer(addr_list_corrected, delimiter=',')
            csv_writer.writerow(new_row)

但是我需要用空字符串替换每个被删除的项目。

Answer 1

我会用迭代器做到这一点。像这样：

def dedup(seq):
    seen = set()
    for v in seq:
        yield '' if v in seen else v
        seen.add(v)

Answer 2

编辑：反转逻辑以使含义更清晰：

另一种选择是做这样的事情：

seen = dict()
seen_setdefault = seen.setdefault
new_row = ["" if cell in seen else seen_setdefault(cell, cell) for cell in row]

举个例子：

>>> row = ["to", "be", "or", "not", "to", "be"]
>>> seen = dict()
>>> seen_setdefault = seen.setdefault
>>> new_row = ["" if cell in seen else seen_setdefault(cell, cell) for cell in row]
>>> new_row
['to', 'be', 'or', 'not', '', '']

编辑2：出于好奇，我进行了快速测试，看看哪种方法最快：

>>> from random import randint
>>> from statistics import mean
>>> from timeit import repeat
>>>
>>> def standard(seq):
...     """Trivial modification to standard method for removing duplicates."""
...     seen = set()
...     seen_add = seen.add
...     return ["" if x in seen or seen_add(x) else x for x in seq]
...
>>> def dedup(seq):
...     seen = set()
...     for v in seq:
...         yield '' if v in seen else v
...         seen.add(v)
...
>>> def pedro(seq):
...     """Pedro's iterator based approach to removing duplicates."""
...     my_dedup = dedup
...     return [x for x in my_dedup(seq)]
...
>>> def srgerg(seq):
...     """Srgerg's dict based approach to removing duplicates."""
...     seen = dict()
...     seen_setdefault = seen.setdefault
...     return ["" if cell in seen else seen_setdefault(cell, cell) for cell in seq]
...
>>> data = [randint(0, 10000) for x in range(100000)]
>>>
>>> mean(repeat("standard(data)", "from __main__ import data, standard", number=100))
1.2130275770426708
>>> mean(repeat("pedro(data)", "from __main__ import data, pedro", number=100))
3.1519048346103555
>>> mean(repeat("srgerg(data)", "from __main__ import data, srgerg", number=100))
1.2611971098676882

从结果中可以看出，对this other stack-overflow question中描述的标准方法进行相对简单的修改是最快的。

Answer 3

您可以使用set来跟踪所查看的内容。使用上面使用的示例列表：

x = ['to', 'be', 'or', 'not', 'to', 'be']
seen = set()
for index, item in enumerate(x):
    if item in seen:
        x[index] = ''
    else:
        seen.add(item)
print x

Answer 4

如果元素不存在于新列表中，您可以创建一个新列表并附加该元素。如果该元素已存在于新列表中，则附加无。

oldList = [3, 1, 'a', 2, 4, 2, 'a', 5, 1, 3]
newList = []

for i in oldList:
    if i in newList:
        newList.append(None)
    else:
        newList.append(i)
print newList

输出：

[3, 1, 'a', 2, 4, None, None, 5, None, None]

如何在保留顺序和长度的同时从Python列表中删除重复项？

4 个答案: