Ruby代碼的慣用Python版本，帶有一個用於測試語句的while循環？

Question

以下代碼Ruby代碼將迭代源字符串並生成由'。'分隔的累積字詞列表。字符，除了最後一個'。'之後的字符。

例如，給 'Company.Dept.Group.Team' 的源串，結果將是... [ 「Company.Dept.Group」， 「Company.Dept」， 「公司」]

鑑於Python中的while循環（我相信）只會測試一個表達式而不是如下所示的語句，那麼最好如何在慣用Python中編寫它？

#ruby 
source = 'Company.Dept.Group.Team' 
results = [] 

temp = source.clone 
while (i = temp.rindex('.')) # test statement not supported in Python? 
    temp = temp[0...i] 
    results << temp 
end 

p results # >> ["Company.Dept.Group", "Company.Dept", "Company"] 

Answer 1

Python的成語是這樣的：

while True: 
    i = temp.rindex('.') 
    if not i: 
     break 
    ... 

Answer 2

我會做

>>> import re 
>>> source = 'Company.Dept.Group.Team' 
>>> results = [source[:m.start()] for m in re.finditer(r"\.", source)] 
>>> results 
['Company', 'Company.Dept', 'Company.Dept.Group'] 

（使用reversed(results)如果你想顛倒順序）。

你的代碼的Python的或多或少的直譯是

source = 'Company.Dept.Group.Team' 
results = [] 
temp = source 

while True: 
    try: 
     i = temp.rindex('.') 
     temp = temp[:i]  
     results.append(temp) 
    except ValueError: 
     break 
print(results) 

Answer 3

>>> source = 'Company.Dept.Group.Team' 
>>> last = [] 
>>> [last.append(s) or '.'.join(last) for s in source.split('.')[:-1]] 
['Company', 'Company.Dept', 'Company.Dept.Group'] 

Answer 4

如果你習慣了Python中，你看到列表理解和迭代器/發電機無處不在！

Python的可能是

source = 'Company.Dept.Group.Team' 

# generate substrings 
temp = source.split(".") 
results = [".".join(temp[:i+1]) for i,s in enumerate(temp)] 

# pop the team (alternatively slice the team out above) 
results.pop() 

# reverse results 
result.reverse() 

print result # should yield ["Company.Dept.Group", "Company.Dept", "Company"] 

，但很可能有更地道的解決方案......

Answer 5

於一般做到這一點，我可能會做的事：

source = 'Company.Dept.Group.Team' 
split_source = source.split('.') 
results = ['.'.join(split_source[0:x]) for x in xrange(len(split_source) - 1, 0, -1)] 
print results 

一個文字翻譯會更像：

source = 'Company.Dept.Group.Team' 

temp = source 
results = [] 
while True: 
    i = temp.rfind('.') 
    if i < 0: 
     break 
    temp = temp[0:i] 
    results.append(temp) 

print results 

或者，如果你喜歡：

source = 'Company.Dept.Group.Team' 

temp = source 
results = [] 
try: 
    while True: 
     temp = temp[0:temp.rindex('.')] 
     results.append(temp) 
except ValueError: 
    pass 
print results 

或者：

source = 'Company.Dept.Group.Team' 

temp = source 
results = [] 
i = temp.rfind('.') 
while i > 0: 
    temp = temp[0:i] 
    results.append(temp) 
    i = temp.rfind('.') 

print results 

正如你指出，事實上，作爲一種表達，使這些情況有點不雅，你不能把分配。我認爲前一種情況 - 即「雖然是真的」 - 比最後一種更常見。

欲瞭解更多的背景，這個職位看起來還不錯：http://effbot.org/pyfaq/why-can-t-i-use-an-assignment-in-an-expression.htm