增量列表

Question

我的零和一的名單，看起來像這樣：

lst = [0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 1, 1] 

我怎樣才能改變這個LST成這樣：

transformed_lst = lst = [0, 1, 1, 1, 1, 0, 0, 0, 2, 2, 0, 0, 0, 3, 0, 4, 4] 

基本上，在每次出現1，將其轉換爲n + 1整數。我確信有一個優雅的方式可以通過itertools/groupby/functools來完成。這是一種嘗試，但並不完全正確的：使用itertools.count()，itertools.chain()和itertools.groupby()

from itertools import cycle 

ints = cycle(range(len(lst))) 
transformed_lst = [next(ints) if i != 0 in lst else 0 for i in lst] 

>>> [0, 0, 1, 2, 3, 0, 0, 0, 4, 5, 0, 0, 0, 6, 0, 7, 8] 

Answer 1

：

In [14]: from itertools import * 

In [15]: c=count(1) 

In [16]: lis=[0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 1, 1] 

In [17]: list(chain(*[list(g) if k!=1 else [next(c)]*len(list(g)) for k,g in groupby(lis)])) 

Out[17]: [0, 1, 1, 1, 1, 0, 0, 0, 2, 2, 0, 0, 0, 3, 0, 4, 4] 

在這裏你還可以代替len(list(g))

使用sum(1 for _ in g)所要求，可讀使用發電機功能的版本：

In [27]: def func(l): 
    c=count(1) 
    for k,g in groupby(l): 
     if k==1: 
      for x in [next(c)]*sum(1 for _ in g): 
       yield x 
     else: 
      for x in g: 
       yield x 
    ....:     

In [28]: list(func(lis)) 
Out[28]: [0, 1, 1, 1, 1, 0, 0, 0, 2, 2, 0, 0, 0, 3, 0, 4, 4] 

Answer 2

你基本上有兩種狀態 - 「讀書0 s」和「讀書1的」 - 當你之間（即從那些到零）開關，然後將delta應用於後續1更迭：

reading_zeroes = True 
delta = 0 
for x in input: 
    if x: 
     reading_zeroes = False 
     x += delta 
    elif not reading_zeroes: 
     delta += 1 
     reading_zeroes = True 
    yield x 

Answer 3

>>> lst = [0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 1, 1] 
>>> transformed = [] 
>>> idx = 1 
>>> for key, grp in groupby(lst): 
... if key: 
...  transformed += [idx] * len(list(grp)) 
...  idx += 1 
... else: 
...  transformed += list(grp) 
... 
>>> transformed 
[0, 1, 1, 1, 1, 0, 0, 0, 2, 2, 0, 0, 0, 3, 0, 4, 4] 
>>> 

Answer 4

我覺得這是可讀性和簡潔之間很好地融合（不需要跟蹤狀態）：

from itertools import groupby 

def transform(numbers): 
    counter = 0 
    for value, iterator in groupby(numbers): 
     if value: 
      counter += 1 

     for i in iterator: 
      if value: 
       yield counter 
      else: 
       yield 0 

pre_transformed = [0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 1, 1] 
print list(transform(pre_transformed)) 

回報：

[0, 1, 1, 1, 1, 0, 0, 0, 2, 2, 0, 0, 0, 3, 0, 4, 4] 

Answer 5

>>> lst = [0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 1, 1] 
>>> from itertools import groupby 
>>> [i//2 if k else 0 for i, (k, g) in enumerate(groupby(lst), 2) for j in g] 
[0, 1, 1, 1, 1, 0, 0, 0, 2, 2, 0, 0, 0, 3, 0, 4, 4] 

Answer 6

請注意，你只有一個「下一組」的時候，當前元素爲1和上一個元素爲0

previous = 0 
grp = 0 
for elem in lst: 
    if elem and not previous: 
     grp += 1 
    previous = elem 
    yield (grp if elem else 0)