我有什么:
s='areyo uanap ppple'
我想要的是什么:
s='12345 12324 11123'
我应该使用词典并翻译i
中的每个s.split(' ')
吗?还是有更简单的方法?
答案 0 :(得分:2)
s='areyo uanap ppple'
incr=1
out=''
dict={}
for x in s:
if ' ' in x:
incr=1
dict={}
out+=' '
continue;
if x in dict.keys():
out+=str(dict[x])
continue;
out+=str(incr)
dict[x]=incr
incr=incr+1
print out //12345 12324 11123
答案 1 :(得分:1)
如果我正确理解OP,这可能是一个解决方案:
s='areyo uanap ppple'
def trans(word):
d = {}
for char in word:
if char not in d.keys():
d[char] = len(d.keys()) + 1
yield str(d[char])
o = ' '.join([ ''.join(trans(word)) for word in s.split(' ')])
print repr(o)
导致:
'12345 12324 11123'
基于unutbu答案的unique
,这也是可能的:
' '.join([''.join([ { a:str(i+1) for i,a in enumerate(unique(word)) }[char] for char in word]) for word in s.split(' ') ])
这是另一个,我认为我有点被带走了:))
' '.join([ w.translate(maketrans(*[ ''.join(x) for x in zip(*[ (a,str(i+1)) for i,a in enumerate(unique(w)) ]) ])) for w in s.split(' ') ])
答案 2 :(得分:1)
您可以使用unicode.translate:
import string
def unique(seq):
# http://www.peterbe.com/plog/uniqifiers-benchmark (Dave Kirby)
# Order preserving
seen = set()
return [x for x in seq if x not in seen and not seen.add(x)]
def word2num(word):
uniqs = unique(word)
assert len(uniqs) < 10
d = dict(zip(map(ord,uniqs),
map(unicode,string.digits[1:])))
return word.translate(d)
s = u'areyo uanap ppple'
for word in s.split():
print(word2num(word))
产量
12345
12324
11123
请注意,如果一个单词中有超过9个唯一字母,则不清楚您想要发生什么。如果assert
传递了这样的话,我就会使用word2num
来投诉。
答案 3 :(得分:1)
使用itertools recipes中的unique_everseen()
:
In [5]: def func(s):
for x in s.split():
dic={}
for i,y in enumerate(unique_everseen(x)):
dic[y]=dic.get(y,i+1)
yield "".join(str(dic[k]) for k in x)
dic={}
...:
In [6]: " ".join(x for x in func('areyo uanap ppple'))
Out[6]: '12345 12324 11123'
In [7]: " ".join(x for x in func('abcde fghij ffabc'))
Out[7]: '12345 12345 11234'