python - 如何在具有长度约束的 Python 中使用“in”

Question

ss = ''
for word in ['this','his','is','s']: # Attach the words
    if word not in ss:                          # if they are not already present
       ss = ss + word + ' '                     # to ss(substring) AFTER ss.

它给出的输出为：

'this '

但我想得到：

'this his is s '

如何使用'in'关键字来做到这一点？

Answer 1

ss = []
for word in ['this', 'his', 'is', 's']:
    if word not in ss:
        ss.append(word)
ss = ' '.join(ss)

Answer 2

使用集合，您的代码的问题是所有['his','is','s']实际上都是 的子字符串'this'，因此条件始终为 False。（in查找子字符串。）

>>> 'his' in 'this'
True
>>> 'is' in 'this'
True
>>> 's' in 'this'
True

解决方案1：

>>> seen = set()     #keep a track of seen word here.
>>> words = ['this','his','is','s']
>>> output = []
>>> for word in words:
...     if word not in seen:
...         output.append(word)
...         seen.add(word)
...         
>>> print " ".join(output) #This is better than normal string concatenation
this his is s

使用列表推导的上述代码的较小版本：

>>> seen = set() 
>>> " ".join([x for x in words if x not in seen and not seen.add(x)])
'this his is s'

解决方案2：

另一种方法（仅用于学习目的）是使用带有单词边界的正则表达式：

>>> import re
>>> ss = ''
for word in words:
    #now this regex looks for exact word match, not just substring         
    if not re.search(r'\b{}\b'.format(re.escape(word)), ss): 
        ss += word + ' '
...         
>>> ss
'this his is s '

Answer 3

正如所有其他答案中所解释的那样，问题是后面的“单词”都是第一个的子字符串，但是我想说问题是您正在比较粉笔和奶酪，或者在您的情况下，单词与字符串进行比较 - 如果您将单词与单词进行比较，问题就消失了：

>>> ss = ''
>>> for word in ['this','his','is','s']: # Attach the words
...     if word not in ss.split():       # if they are not already present in the list of words so far
...        ss = ss + word + ' '          # to ss(substring) AFTER ss.
...
>>> ss
'this his is s '

这正是你所要求的。

Answer 4

一个非常简单的技巧，只需对您的代码进行 2 次（好的，也许是 3 次）小改动：

ss = ' '
for word in ['this','his','is','s']:
    if ' '+word+' ' not in ss:
        ss = ss + word + ' '
result= ss[1:]

这甚至适用于限制情况 "word" == ''，它可以在结果中标识为两个连续的空格（或只有一个开头的空格）。

解释

为了视觉清晰，我将使用':'而不是空格。第一项的处理'this'是微不足道的，结果ss== ':this:'。':his:'将找不到下一项，并附加到=='his:'的结果。等等。ss':this:his:'