Question

我试图在python中创建一个重叠函数，它接受任何iterable，一个int n ，一个int m （默认值为1）作为参数，并生成 n 值的列表：

第一个列表包含第一个 n 值;
每个后续列表都会删除前一个列表中的第一个 m
从迭代器添加下一个 m 值，直到返回列表中的 n 值少于 n 。

示例：

for i in overlap('abcdefghijk',4,2):
    print(i,end=' ')

['a','b','c','d']和['c','d','e','f']的输出为['e','f','g','h']，['g','h',i','j']，n=4和m=2。在这种情况下，如果函数是这样的：

def overlap(iterable, n, m=1)

如何做到这一点？我面临的主要问题是当小于 n 值时如何停止。

当前代码：anmol_uppal's solution的分数。

def hide(iterable):
    for v in iterable:
        yield v

def overlap(word, n, m):
    start = 0
    while(start+n < len(word)):
        yield list(word[start:start+n])
        start += m

if __name__ == '__main__':   
    for i in overlap('abcdefghijk',3,2): 
        print(i, end=' ')
    print()
    for i in overlap(hide('abcdefghijk'),3,2): 
        print(i, end=' ')
    print()

但问题是我们不能将len()用于生成器。

我尝试过这段代码，但我不知道为什么它会给第二个测试用例带来意想不到的结果

def hide(iterable):
    for v in iterable:
        yield v

def window(seq, size, step=1):
    iters = [iter(seq) for i in range(size)]
    [next(iters[i]) for j in range(size) for i in range(-1, -j-1, -1)]
    while(True):
        yield [next(i) for i in iters]
        [next(i) for i in iters for j in range(step-1)]

if __name__ == '__main__':
    for i in window('abcdefghijk', 3, 2): 
        print(i, end=' ')
    print()
    for i in window(hide('abcdefghijk'), 3, 2): 
        print(i, end=' ')
    print()

Answer 1

def overlap(word, n,m):
    ans = []
    start = 0
    while(start+n<len(word)):
        ans.append(list(word[start:start+n]))
        start+=m

    return ans


>>> print overlap("abcdefghijk", 4, 2)
>>> [['a', 'b', 'c', 'd'], ['c', 'd', 'e', 'f'], ['e', 'f', 'g', 'h'], ['g', 'h', 'i', 'j']]
>>> print overlap("abcdefghi", 4, 2)
>>> [['a', 'b', 'c', 'd'], ['c', 'd', 'e', 'f'], ['e', 'f', 'g', 'h']]

Also if you would like to yield the results instead of returning them,then you can use the following simpler version:

def overlap(word, n,m):
    start = 0
    while(start+n<len(word)):
        yield list(word[start:start+n])
        start+=m

for i in overlap("abcdefghijk", 4, 2):
    print (i,end = " ")

To make it work with the hide() function you need to convert it to string first, so:

for i in overlap("".join(hide("abcdefghijk")),3, 2): 
    print i

给定n和m的重叠迭代没有使用len（）

1 个答案: