Generate a list of strings with a sliding window using itertools, yield, and iter() in Python 2.7.1?

问题

I'm trying to generate a sliding window function in Python. I figured out how to do it but not all inside the function. itertools, yield, and iter() are entirely new to me.

i want to input

a='abcdefg'
b=window(a,3)
print b
['abc','bcd','cde','def','efg']

the way i got it work was

def window(fseq, window_size=5):
    import itertools
    tentative=[]
    final=[]
    iteration=iter(fseq)
    value=tuple(itertools.islice(iteration,window_size))
    if len(value) == window_size:
        yield value
    for element in iteration:
        value = value[1:] + (element,)
        yield value

a='abcdefg'
result=window(a)
list1=[]
for k in result:
    list1.append(k)
list2=[]   
for j in list1:
    tentative=''.join(j)
    list2.append(tentative)
print list2

basically what im confused about is how to use the final value of the function inside the function?

here is my code for the function

def window(fseq, window_size=5):
    import itertools
    tentative=[]
    final=[]
    iteration=iter(fseq)
    value=tuple(itertools.islice(iteration,window_size))
    if len(value) == window_size:
        yield value
    for element in iteration:
        value = value[1:] + (element,)
        yield value
    for k in value:
        tentative.append(k)
    for j in tentative:
        tentative_string=''.join(j)
        final.append(tentative_string)
    return final



seq='abcdefg'
uence=window(seq)
print uence

i want it to return the joined list but when i press run it, it says "There's an error in your program * 'return' with argument inside generator"

I'm really confused . . .

回答1:

You mean you want to do this ? :

a='abcdefg'
b = [a[i:i+3] for i in xrange(len(a)-2)]
print b
['abc', 'bcd', 'cde', 'def', 'efg']

回答2:

Your generator could be much shorter:

def window(fseq, window_size=5):
    for i in xrange(len(fseq) - window_size + 1):
        yield fseq[i:i+window_size]


for seq in window('abcdefghij', 3):
    print seq


abc
bcd
cde
def
efg
fgh
ghi
hij

回答3:

Use zip function in one line code:

  [ "".join(j) for j in zip(*[fseq[i:] for i in range(window_size)])]

回答4:

def window(fseq,fn):
    alpha=[fseq[i:i+fn] for i in range(len(fseq)-(fn-1))]
    return alpha

回答5:

I don't know what your input or expected output are, but you cannot mix yield and return in one function. change return to yield and your function will not throw that error again.

def window(fseq, window_size=5):
    ....
        final.append(tentative_string)
    yield final

回答6:

>>>def window(data, win_size):
...    tmp = [iter(data[i:]) for i in range(win_size)]
...    return zip(*tmp)
>>> a = [1, 2, 3, 4, 5, 6]
>>> window(a, 3)
>>>[(1,2,3), (2,3,4), (3,4,5), (4,5,6)]

来源：https://stackoverflow.com/questions/8408117/generate-a-list-of-strings-with-a-sliding-window-using-itertools-yield-and-ite

标签

python

string

window

loops

sliding