如何在 Python 中获取一个字符串的所有连续子字符串？
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How To Get All The Contiguous Substrings Of A String In Python?
提问by lqhcpsgbl
Here is my code, but I want a better solution, how do you think about the problem?
这是我的代码，但我想要一个更好的解决方案，你如何看待这个问题？
def get_all_substrings(string):
length = len(string)
alist = []
for i in xrange(length):
for j in xrange(i,length):
alist.append(string[i:j + 1])
return alist
print get_all_substring('abcde')
采纳答案by thefourtheye
The only improvement I could think of is, to use list comprehension like this
我能想到的唯一改进是，像这样使用列表理解
def get_all_substrings(input_string):
length = len(input_string)
return [input_string[i:j+1] for i in xrange(length) for j in xrange(i,length)]
print get_all_substrings('abcde')
The timing comparison between, yours and mine
你和我的时间比较
def get_all_substrings(string):
length = len(string)
alist = []
for i in xrange(length):
for j in xrange(i,length):
alist.append(string[i:j + 1])
return alist
def get_all_substrings_1(input_string):
length = len(input_string)
return [input_string[i:j + 1] for i in xrange(length) for j in xrange(i,length)]
from timeit import timeit
print timeit("get_all_substrings('abcde')", "from __main__ import get_all_substrings")
# 3.33308315277
print timeit("get_all_substrings_1('abcde')", "from __main__ import get_all_substrings_1")
# 2.67816185951
回答by John La Rooy
You could write it as a generator to save storing all the strings in memory at once if you don't need to
如果您不需要，您可以将其编写为生成器以将所有字符串一次保存在内存中
def get_all_substrings(string):
length = len(string)
for i in xrange(length):
for j in xrange(i + 1, length + 1):
yield(string[i:j])
for i in get_all_substrings("abcde"):
print i
you can still make a list if you really need one
如果你真的需要，你仍然可以列一个清单
alist = list(get_all_substrings("abcde"))
The function can be reduced to return a generator expression
该函数可以简化为返回一个生成器表达式
def get_all_substrings(s):
length = len(s)
return (s[i: j] for i in xrange(length) for j in xrange(i + 1, length + 1))
Or of course you can change two characters to return a list if you don't care about memory
或者当然你可以改变两个字符来返回一个列表，如果你不关心内存
def get_all_substrings(s):
length = len(s)
return [s[i: j] for i in xrange(length) for j in xrange(i + 1, length + 1)]
回答by sanooj
Another solution:
另一种解决方案：
def get_all_substrings(string):
length = len(string)+1
return [string[x:y] for x in range(length) for y in range(length) if string[x:y]]
print get_all_substring('abcde')
回答by PaulMcG
I've never been fond of range(len(seq)), how about using enumerate and just using the index value:
我从来没有喜欢过range(len(seq))，如何使用 enumerate 并仅使用索引值：
def indexes(seq, start=0):
return (i for i,_ in enumerate(seq, start=start))
def gen_all_substrings(s):
return (s[i:j] for i in indexes(s) for j in indexes(s[i:], i+1))
def get_all_substrings(string):
return list(gen_all_substrings(string))
print(get_all_substrings('abcde'))
回答by PaulMcG
Use itertools.permutationsto generate all pairs of possible start and end indexes,
and filter out only those where the start index is less than then end index. Then
use these pairs to return slices of the original string.
使用itertools.permutations产生的所有对可能的开始和结束的指标，并过滤出那些开始指数小于然后结束索引。然后使用这些对返回原始字符串的切片。
from itertools import permutations
def gen_all_substrings(s):
lt = lambda pair: pair[0] < pair[1]
index_pairs = filter(lt, permutations(range(len(s)+1), 2))
return (s[i:j] for i,j in index_pairs)
def get_all_substrings(s):
return list(gen_all_substrings(s))
print(get_all_substrings('abcde'))
回答by PaulMcG
Python 3
蟒蛇 3
s='abc'
list(s[i:j+1] for i in range (len(s)) for j in range(i,len(s)))
['a', 'ab', 'abc', 'b', 'bc', 'c']
回答by DyRuss
can be done concisely with itertools.combinations
可以简洁地完成 itertools.combinations
from itertools import combinations
def get_all_substrings_2(string):
length = len(string) + 1
return [string[x:y] for x, y in combinations(range(length), r=2)]
回答by Nkromin989
Another solution using 2-D matrix approach
另一种使用二维矩阵方法的解决方案
p = "abc"
a = list(p)
b = list(p)
c = list(p)
count = 0
for i in range(0,len(a)):
dump = a[i]
for j in range(0, len(b)):
if i < j:
c.append(dump+b[j])
dump = dump + b[j]
