在python中有条件的列表中删除数字

如何解决在python中有条件的列表中删除数字

我试图删除位于列表中连续数字之间的数字。示例：

表格

[1,3,5,6,7,8,9,11,12,13,18]

到

[1,18]

你会丢弃 [6,8] 和 [12]

解决方法

你可以试试itertools.groupby：

from itertools import groupby

lst = [1,3,5,6,7,8,9,11,12,13,18]

out = []
for _,g in groupby(enumerate(lst),lambda k: k[0] - k[1]):
    g = list(g)
    out.append(g[0][1])
    if len(g) > 1:
        out.append(g[-1][1])

print(out)

打印：

[1,18]

,

我喜欢使用 zip 并行迭代列表：

def drop_consecutive(a):
    if len(a)<3:
        yield from a
        raise StopIteration

    yield a[0]
    for (left,center,right) in zip(a[:-2],a[1:-1],a[2:]):
        if center==left+1 and right==center+1:
            pass # skip because they're consecutive
        else:
           yield center
    yield a[-1]


assert([1,18] == list(drop_consecutive( [1,18]))

,

这是一种不使用任何库的方法：

def get_nums_with_dropped_internal_consecutive(nums):
    result = []
    if not nums:
        return result
    left = right = nums[0]
    for x in nums[1:]:
        if x - 1 != right:
            result.append(left) if left == right else result.extend([left,right])
            left = right = x
        else:
            right = x
    result.append(left) if left == right else result.extend([left,right])
    return result

nums = [1,18]
nums_with_dropped_internal_consecutive = get_nums_with_dropped_internal_consecutive(nums)
print(nums_with_dropped_internal_consecutive)

输出：

[1,18]

,

看看这个：

numbers = [2,15]
numbers_new = [numbers[0]]

for i in range(1,len(numbers)-1):
  if (numbers[i] == numbers[i-1]+1 and numbers[i] == numbers[i+1]-1):
    print(numbers[i],True)
  else:
    print(numbers[i],False)
    numbers_new.append(numbers[i])

numbers_new.append(numbers[-1])

numbers_new

这将返回数字并说明它是否位于连续数字之间，以及一个仅包含这些数字而非连续数字之间的新列表：

3 False
5 False
6 True
7 True
8 False
11 False
12 True
13 False
[2,15]