我有一个数组
["ejjjjmmtthh", "zxxuueeg", "aanlljrrrxx", "dqqqaaabbb", "oocccffuucccjjjkkkjyyyeehh"]
，并且需要在每个长度为k(在本例中为3)的字符串元素中提取连续字符，而不使用regex或groupby。
这是我目前为止写的:
s = ["ejjjjmmtthh", "zxxuueeg", "aanlljrrrxx", "dqqqaaabbb", "oocccffuucccjjjkkkjyyyeehhh"]
k = 3
output = []
for i in s:
result = ""
for j in range(1,len(i)-1):
if i[j]==i[j-1] or i[j]==i[j+1]:
result+=i[j]
if i[-1] == result[-1]:
result+=i[-1]
if i[0]==result[0]:
result=i[0]+result
output.append(result)
print(output)
#current output = ['jjjjmmtthh', 'xxuuee', 'aallrrrxx', 'qqqaaabbb', 'oocccffuucccjjjkkkyyyeehhh']
#expected outcome(for k =3) = ['rrr','qqq','aaa','bbb','ccc','ccc','jjj','kkk','yyy','hhh']
我的问题:
- 如何适应
k条件? - 是否有更优的方法来做到这一点?
这个解决方案更容易读，而且不太长。它适用于k>
0。s = ["ejjjjmmtthh", "zxxuueeg", "aanlljrrrxx", "dqqqaaabbb", "oocccffuucccjjjkkkjyyyeehhh"]
k = 3
output = []
for element in s:
state = "" #State variable (reset on every list item)
for char in element: #For each character
if state != "" and char == state[-1]: # Check if the last character is the same (only if state isn't empty)
state += char #Add it to the state
else:
if len(state) == k: #Otherwise, check if we have k characters
output.append(state) #Append te result if we do
state = char #Reset the state
#If there are no more characters (end of element), check too
if len(state) == k:
output.append(state)
print(output)
k = 3输出
['rrr', 'qqq', 'aaa', 'bbb', 'ccc', 'ccc', 'jjj', 'kkk', 'yyy', 'hhh']
k = 1输出
['e', 'z', 'g', 'n', 'j', 'd', 'j']
这里我按连续相同的字母手动分组。然后我只在它们与k长度相同的情况下才将它们计算到结果中。这是有效的，但我确信有一个更优化的方法:
s = ["ejjjjmmtthh", "zxxuueeg", "aanlljrrrxx", "dqqqaaabbb", "oocccffuucccjjjkkkjyyyeehhh"]
k = 3
def _next_group(st):
if not st:
return None
first = st[0]
res = [first]
for s in st[1:]:
if s == first:
res.append(s)
else:
break
return res
result = []
for st in s:
while True:
group = _next_group(st)
if not group:
break
if len(group) == k:
result.append("".join(group))
if len(group) == len(st):
break
st = st[len(group):]
print(result)
输出:['rrr', 'qqq', 'aaa', 'bbb', 'ccc', 'ccc', 'jjj', 'kkk', 'yyy', 'hhh']
for-loop方法
备注:我建议使用分治策略解决方案:关注一个字符串(而不是一个字符串列表),让一个函数,然后概括与循环/理解…
def repeated_chars(string, k=3):
out = []
c, tmp = 0, '' # counter, tmp char
for char in s:
if tmp == '':
tmp = char
c += 1
continue
if tmp == char:
c += 1
else:
if c == k:
out.append((tmp, c))
tmp = char
c = 1
# last term
if c == k:
out.append((tmp, c))
return [char * i for char, i in out]
data = ['jjjjmmtthh', 'xxuuee', 'aallrrrxx', 'qqqaaabbb', 'oocccffuucccjjjkkkyyyeehhh']
# apply the function to all strings
out = []
for s in data:
out.extend(repeated_chars(s, k=3))
print(out)
#['rrr', 'qqq', 'aaa', 'bbb', 'ccc', 'ccc', 'jjj', 'kkk', 'yyy', 'hhh']
编辑:是的，groubpy不应该按要求使用，但是做这项工作需要以某种方式分组(例如，请参阅接受的答案)，因此将责任分为多个功能似乎是一个好主意，这是良好的实践，通过重新实现groupby。
在这种情况下，groupby似乎是显而易见的，如果你不能使用itertools中的一个，就写一个。
核心函数也应该处理字符串，而不是字符串列表——只是for循环，以防你有字符串列表。
一旦你有了你的groubpy，它是直接的:
def extract_groups(s: str, k: int):
return [group for group in groupby(s) if len(group) == k]
让我们试一试:
input_strings = [
"ejjjjmmtthh",
"zxxuueeg",
"aanlljrrrxx",
"dqqqaaabbb",
"oocccffuucccjjjkkkjyyyeehhh",
]
expected_outputs = [
[],
[],
["rrr"],
["qqq", "aaa", "bbb"],
["ccc", "ccc", "jjj", "kkk", "yyy", "hhh"],
]
outputs = [extract_groups(s, k=3) for s in input_strings]
print(outputs == expected_outputs) # True
事实上，outputs是一个组列表:
In [ ]: outputs
Out[ ]: [[], [], ['rrr'], ['qqq', 'aaa', 'bbb'], ['ccc', 'ccc', 'jjj', 'kkk', 'yyy', 'hhh']]
如果你真的想让它变平，就把它变平:
In [ ]: from itertools import chain
... : list(chain.from_iterable(outputs))
Out[ ]: ['rrr', 'qqq', 'aaa', 'bbb', 'ccc', 'ccc', 'jjj', 'kkk', 'yyy', 'hhh']
In [ ]: [group for s in input_strings for group in extract_groups(s, k)]
Out[ ]: ['rrr', 'qqq', 'aaa', 'bbb', 'ccc', 'ccc', 'jjj', 'kkk', 'yyy', 'hhh']
groupby函数参考:
def groupby(s: str):
if not s:
return []
result = []
tgt = s[0]
counter = 1
for c in s[1:]:
if c == tgt:
counter += 1
else:
result.append(tgt * counter)
tgt = c
counter = 1
result.append(tgt * counter)
return result