如何在 Python 中删除重复打印？答案

【问题标题】：How to remove duplicate prints in Python?如何在 Python 中删除重复打印？
【发布时间】：2021-03-14 05:42:20
【问题描述】：

所以我想编写一个程序来打印出一个数字在列表中重复了多少次。我对 Python 很陌生，我使用 .count() 命令来获取元素重复的次数。我希望输出遵循“数字 n 重复 n 次”的格式。

这是我目前所拥有的：

x = [1,1,2,4,5,7,2,3,3,8,2,3,6,7]
index = 0 
while index < len(x):
print('The number {} is repeated {} times.'.format(x[index],x.count(x[index])))
index = index + 1

这是输出：

The number 1 is repeated 2 times
The number 1 is repeated 2 times
The number 2 is repeated 3 times
The number 4 is repeated 1 times
The number 5 is repeated 1 times
The number 7 is repeated 2 times
The number 2 is repeated 3 times
The number 3 is repeated 3 times
The number 3 is repeated 3 times
The number 8 is repeated 1 times
The number 2 is repeated 3 times
The number 3 is repeated 3 times
The number 6 is repeated 1 times
The number 7 is repeated 2 times

我希望输出显示一个数字仅重复一次的次数，并且按升序排列。我怎样才能使输出像这样：

The number 1 is repeated 2 times.
The number 2 is repeated 3 times. 
The number 3 is repeated 3 times. 
...

谢谢！ :)

【问题讨论】：

标签： python count duplicates

【解决方案1】：

试试这个：

x = [1,1,2,4,5,7,2,3,3,8,2,3,6,7]
index = 0 
while index < len(x):
    if index==x.index(x[index]):
        print('The number {} is repeated {} times.'.format(x[index],x.count(x[index])))
    index = index + 1

【讨论】：

【解决方案2】：

这会计算所有出现的数字并按升序打印。

from collections import Counter

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

for key, val in sorted(Counter(x).items()):
    print(f'The number {key} is repeated {val} times.')

代码输出：

The number 1 is repeated 2 times.
The number 2 is repeated 3 times.
The number 3 is repeated 3 times.
The number 4 is repeated 1 times.
The number 5 is repeated 1 times.
The number 6 is repeated 1 times.
The number 7 is repeated 2 times.
The number 8 is repeated 1 times.

【讨论】：

【解决方案3】：

使用Counter 进行计数和“去重”，然后迭代其结果：

from collections import Counter

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

for n, c in Counter(x).items():
    print('The number {} is repeated {} times.'.format(n, c))

对于按最常见到最常见排序的打印：

for n, c in reversed(Counter(x).most_common()):
    ...

对于按值排序的打印：

for n, c in sorted(Counter(x).items()):
    ...

【讨论】：

【解决方案4】：

你可以试试这个（尽管这些答案似乎都一样有效）：

x = [1,1,2,4,5,7,2,3,3,8,2,3,6,7]
x_sorted = sorted(set(x)) # remove duplicates with set, convert to list, and sort
for num in x_sorted:
    print('The number {} is repeated {} times.'.format(num,x.count(num)))

这可能不是最漂亮的解决方案（我不是 sorted(set(x)) 行的粉丝），但它会按升序打印结果。

这个输出：

The number 1 is repeated 2 times.
The number 2 is repeated 3 times.
The number 3 is repeated 3 times.
The number 4 is repeated 1 times.
The number 5 is repeated 1 times.
The number 6 is repeated 1 times.
The number 7 is repeated 2 times.
The number 8 is repeated 1 times.

【讨论】：

【解决方案5】：

如果您想模仿Counter 的功能，您可以填写自己的dict。我特意不使用try和except，或者setdefault，或者defaultdict，来说明基本逻辑很简单，只是一个循环和一个if-else块：

def count_elements(lst):
    ret = {}
    for elem in lst:
        if elem in ret:
            ret[elem] += 1
        else:
            ret[elem] = 1
    return ret

def print_counts(counts):
    for n, c in sorted(counts.items()):
        print(f'The number {n} is repeated {c} times')

x = [1,1,2,4,5,7,2,3,3,8,2,3,6,7]
c = count_elements(x)
print_counts(c)

【讨论】：