Python的 - 逗号分隔字符串转换为减少字符串列表

Question

给出一个Python字符串是这样的：

location_in = 'London, Greater London, England, United Kingdom' 

我想将其转换成像这样的列表：

location_out = ['London, Greater London, England, United Kingdom', 
       'Greater London, England, United Kingdom', 
       'England, United Kingdom', 
       'United Kingdom'] 

换句话说，给出一个用逗号分隔的字符串（location_in），我想将它复制到一个列表（location_out），并逐渐删除第一个单词/短语每次分解它。

我是一个Python新手。任何想法写在这个好方法？谢谢。

Answer 1

location_in = 'London, Greater London, England, United Kingdom' 
locations = location_in.split(', ') 
location_out = [', '.join(locations[n:]) for n in range(len(locations))] 

Answer 2

的方式来做到这一点充足，但这里有一个：

def splot(data): 
    while True: 
    yield data 
    pre,sep,data=data.partition(', ') 
    if not sep: # no more parts 
     return 

location_in = 'London, Greater London, England, United Kingdom' 
location_out = list(splot(location_in)) 

更反常的解决方案：

def stringsplot(data): 
    start=-2    # because the separator is 2 characters 
    while start!=-1:  # while find did find 
    start+=2    # skip the separator 
    yield data[start:] 
    start=data.find(', ',start) 

Answer 3

这里的工作之一：

location_in = 'London, Greater London, England, United Kingdom' 
loci = location_is.spilt(', ') # ['London', 'Greater London',..] 
location_out = [] 
while loci: 
    location_out.append(", ".join(loci)) 
    loci = loci[1:] # cut off the first element 
# done 
print location_out 

Answer 4

>>> location_in = 'London, Greater London, England, United Kingdom' 
>>> location_out = [] 
>>> loc_l = location_in.split(", ") 
>>> while loc_l: 
...  location_out.append(", ".join(loc_l)) 
...  del loc_l[0] 
... 
>>> location_out 
['London, Greater London, England, United Kingdom', 
'Greater London, England, United Kingdom', 
'England, United Kingdom', 
'United Kingdom'] 
>>>