在递归中只打印一个列表

Question

在下面的代码中，我返回给定字符串中连续数字的整数值。

def consecutive_length(S): 
    if S == '': 
     return 0 
    if len(S) == 1: 
     return 1 
    if S[0] == S[1]: 
     return 1 + consecutive_length(S[1:]) 
    return 1 

def compress(S): 
    if S == '': 
     return 0 
    cons_length = consecutive_length(S) 
    return [cons_length] + [compress(S[cons_length:])] 

当我运行这个打印语句，返回以下：

>>> print (compress('1111000000001111000111111111111111')) 
[4, [8, [4, [3, [15, 0]]]]] 

在那里我真的希望下面的返回到：

>>> print (compress('1111000000001111000111111111111111')) 
[4, 8, 4, 3, 15] 

Answer 1

当你返回一个列表，[what_is_returned]将是一个嵌套列表，但是当你返回一个整数，它仅仅是一个列表。取而代之的是，（在compress()），始终返回一个列表，当你用什么返回删除括号：

def consecutive_length(S): 
    if S == '': 
     return 0 
    if len(S) == 1: 
     return 1 
    if S[0] == S[1]: 
     return 1 + consecutive_length(S[1:]) 
    return 1 

def compress(S): 
    if S == '': 
     return [] 
    cons_length = consecutive_length(S) 
    return [cons_length] + compress(S[cons_length:]) 

Answer 2

你还有另外一种方法来使用itertools.groupby()：

from itertools import groupby 

s = '1111000000001111000111111111111111' 
answer = [len([digit for digit in group[1]]) for group in groupby(s)] 
print(answer) 

输出

[4, 8, 4, 3, 15] 

Answer 3

在这里你去：

def consecutive_length(S): 
    if S == '': 
     return 0 
    if len(S) == 1: 
     return 1 
    if S[0] == S[1]: 
     return 1 + consecutive_length(S[1:]) 
    return 1 

def compress(S): 
    if S == '': 
     return [] 
    cons_length = consecutive_length(S) 
    return [cons_length] + compress(S[cons_length:])