在Python中查找字符串中多次出现的字符串
如何在Python中查找字符串中多次出现的字符串?考虑这一点:在Python中查找字符串中多次出现的字符串,python,string,Python,String,如何在Python中查找字符串中多次出现的字符串?考虑这一点: >>> text = "Allowed Hello Hollow" >>> text.find("ll") 1 >>> 因此,ll的第一次出现如预期的那样为1。我如何找到下一次发生的情况 同样的问题也适用于列表。考虑: >>> x = ['ll', 'ok', 'll'] 如何找到所有的ll及其索引 使用正则表达式,您可以使用查找所有(非重叠)事件: &g
>>> text = "Allowed Hello Hollow"
>>> text.find("ll")
1
>>>
因此,ll
的第一次出现如预期的那样为1。我如何找到下一次发生的情况
同样的问题也适用于列表。考虑:
>>> x = ['ll', 'ok', 'll']
如何找到所有的
ll
及其索引 使用正则表达式,您可以使用查找所有(非重叠)事件:
>>> import re
>>> text = 'Allowed Hello Hollow'
>>> for m in re.finditer('ll', text):
print('ll found', m.start(), m.end())
ll found 1 3
ll found 10 12
ll found 16 18
或者,如果您不想增加正则表达式的开销,也可以重复使用以获取下一个索引:
>>text='Allowed Hello Hollow'
>>>索引=0
>>>当索引
这也适用于列表和其他序列。我认为您要查找的是
string.count
"Allowed Hello Hollow".count('ll')
>>> 3
希望这有帮助注意:对于您的列表示例,这仅捕获不重叠的事件:
>>> for n,c in enumerate(text):
... try:
... if c+text[n+1] == "ll": print n
... except: pass
...
1
10
16
In [1]: x = ['ll','ok','ll']
In [2]: for idx, value in enumerate(x):
...: if value == 'll':
...: print idx, value
0 ll
2 ll
如果希望列表中包含“ll”的所有项目,也可以这样做
In [3]: x = ['Allowed','Hello','World','Hollow']
In [4]: for idx, value in enumerate(x):
...: if 'll' in value:
...: print idx, value
...:
...:
0 Allowed
1 Hello
3 Hollow
对于列表示例,请使用理解:
>>> l = ['ll', 'xx', 'll']
>>> print [n for (n, e) in enumerate(l) if e == 'll']
[0, 2]
与字符串类似:
>>> text = "Allowed Hello Hollow"
>>> print [n for n in xrange(len(text)) if text.find('ll', n) == n]
[1, 10, 16]
这将列出“ll”的相邻运行,这可能是您想要的,也可能不是您想要的:
>>> text = 'Alllowed Hello Holllow'
>>> print [n for n in xrange(len(text)) if text.find('ll', n) == n]
[1, 2, 11, 17, 18]
FWIW,这里有一些非可再生能源的替代品,我认为比它们更整洁 第一个使用
str.index
并检查ValueError
:
def findall(sub, string):
"""
>>> text = "Allowed Hello Hollow"
>>> tuple(findall('ll', text))
(1, 10, 16)
"""
index = 0 - len(sub)
try:
while True:
index = string.index(sub, index + len(sub))
yield index
except ValueError:
pass
第二次测试使用str.find
并使用iter
检查-1
的哨兵:
def findall_iter(sub, string):
"""
>>> text = "Allowed Hello Hollow"
>>> tuple(findall_iter('ll', text))
(1, 10, 16)
"""
def next_index(length):
index = 0 - length
while True:
index = string.find(sub, index + length)
yield index
return iter(next_index(len(sub)).next, -1)
要将这些函数中的任何一个应用于列表、元组或其他字符串的iterable,可以使用更高级别的函数—将函数作为其参数之一—如下所示:
def findall_each(findall, sub, strings):
"""
>>> texts = ("fail", "dolly the llama", "Hello", "Hollow", "not ok")
>>> list(findall_each(findall, 'll', texts))
[(), (2, 10), (2,), (2,), ()]
>>> texts = ("parallellized", "illegally", "dillydallying", "hillbillies")
>>> list(findall_each(findall_iter, 'll', texts))
[(4, 7), (1, 6), (2, 7), (2, 6)]
"""
return (tuple(findall(sub, string)) for string in strings)
对于一般编程来说是全新的,并且通过在线教程进行工作。我也被要求这样做,但只使用到目前为止我所学的方法(基本上是字符串和循环)。不确定这是否会在这里增加任何价值,我知道这不是你将要做的,但我得到了它:
needle = input()
haystack = input()
counter = 0
n=-1
for i in range (n+1,len(haystack)+1):
for j in range(n+1,len(haystack)+1):
n=-1
if needle != haystack[i:j]:
n = n+1
continue
if needle == haystack[i:j]:
counter = counter + 1
print (counter)
这个版本的字符串长度应该是线性的,只要序列不太重复就可以了(在这种情况下,可以用while循环替换递归) bstpierre的列表理解对于短序列来说是一个很好的解决方案,但是看起来有二次复杂性,并且从未完成我使用的长文本
findall_lc = lambda txt, substr: [n for n in xrange(len(txt))
if txt.find(substr, n) == n]
对于非平凡长度的随机字符串,两个函数给出相同的结果:
import random, string; random.seed(0)
s = ''.join([random.choice(string.ascii_lowercase) for _ in range(100000)])
>>> find_all(s, 'th') == findall_lc(s, 'th')
True
>>> findall_lc(s, 'th')[:4]
[564, 818, 1872, 2470]
但是二次型的速度要慢300倍
%timeit find_all(s, 'th')
1000 loops, best of 3: 282 µs per loop
%timeit findall_lc(s, 'th')
10 loops, best of 3: 92.3 ms per loop
该程序统计所有子字符串的数量,即使它们在不使用正则表达式的情况下重叠。但这是一个简单的实现,为了在最坏的情况下获得更好的结果,建议遍历后缀树、KMP和其他字符串匹配数据结构和算法。这是我查找多个匹配项的函数。U与这里的其他解决方案一样,它支持用于切片的可选开始和结束参数,就像
str.index
:
def all_substring_indexes(string, substring, start=0, end=None):
result = []
new_start = start
while True:
try:
index = string.index(substring, new_start, end)
except ValueError:
return result
else:
result.append(index)
new_start = index + len(substring)
一个简单的迭代代码,返回子字符串所在的索引列表
def allindices(string, sub):
l=[]
i = string.find(sub)
while i >= 0:
l.append(i)
i = string.find(sub, i + 1)
return l
您可以拆分以获得相对位置,然后对列表中的连续数字求和,同时添加(字符串长度*出现顺序)以获得所需的字符串索引
>>> key = 'll'
>>> text = "Allowed Hello Hollow"
>>> x = [len(i) for i in text.split(key)[:-1]]
>>> [sum(x[:i+1]) + i*len(key) for i in range(len(x))]
[1, 10, 16]
>>>
也许不太像python,但有点不言自明。它返回单词在原始字符串中的位置
def retrieve_occurences(sequence, word, result, base_counter):
indx = sequence.find(word)
if indx == -1:
return result
result.append(indx + base_counter)
base_counter += indx + len(word)
return retrieve_occurences(sequence[indx + len(word):], word, result, base_counter)
我认为没有必要测试文本的长度;只要继续查找,直到没有任何内容可供查找。如下所示:
>>> text = 'Allowed Hello Hollow'
>>> place = 0
>>> while text.find('ll', place) != -1:
print('ll found at', text.find('ll', place))
place = text.find('ll', place) + 2
ll found at 1
ll found at 10
ll found at 16
您还可以使用条件列表理解来执行此操作,如下所示:
string1= "Allowed Hello Hollow"
string2= "ll"
print [num for num in xrange(len(string1)-len(string2)+1) if string1[num:num+len(string2)]==string2]
# [1, 10, 16]
我刚才随机得到这个想法。使用带字符串拼接和字符串搜索的while循环可以工作,即使对于重叠的字符串也是如此
findin = "algorithm alma mater alison alternation alpines"
search = "al"
inx = 0
num_str = 0
while True:
inx = findin.find(search)
if inx == -1: #breaks before adding 1 to number of string
break
inx = inx + 1
findin = findin[inx:] #to splice the 'unsearched' part of the string
num_str = num_str + 1 #counts no. of string
if num_str != 0:
print("There are ",num_str," ",search," in your string.")
else:
print("There are no ",search," in your string.")
我是Python编程(实际上是任何语言的编程)的业余爱好者,不确定它还会有什么其他问题,但我想它工作得很好吧
如果需要的话,我想lower()也可以在其中的某个地方使用。下面的函数会查找另一个字符串中出现的所有字符串,同时通知找到每个字符串的位置 您可以使用下表中的测试用例调用该函数。您可以尝试将单词、空格和数字混合在一起 该函数适用于重叠字符
| theString | aString |
| -------------------------- | ------- |
| "661444444423666455678966" | "55" |
| "661444444423666455678966" | "44" |
| "6123666455678966" | "666" |
| "66123666455678966" | "66" |
Calling examples:
1. print("Number of occurrences: ", find_all("123666455556785555966", "5555"))
output:
Found in position: 7
Found in position: 14
Number of occurrences: 2
2. print("Number of occorrences: ", find_all("Allowed Hello Hollow", "ll "))
output:
Found in position: 1
Found in position: 10
Found in position: 16
Number of occurrences: 3
3. print("Number of occorrences: ", find_all("Aaa bbbcd$#@@abWebbrbbbbrr 123", "bbb"))
output:
Found in position: 4
Found in position: 21
Number of occurrences: 2
def find_all(theString, aString):
count = 0
i = len(aString)
x = 0
while x < len(theString) - (i-1):
if theString[x:x+i] == aString:
print("Found in position: ", x)
x=x+i
count=count+1
else:
x=x+1
return count
|字符串| aString|
| -------------------------- | ------- |
| "661444444423666455678966" | "55" |
| "661444444423666455678966" | "44" |
| "6123666455678966" | "666" |
| "66123666455678966" | "66" |
调用示例:
1.打印(“出现次数:”,查找全部(“123666455556785555966”,“5555”))
输出:
找到位置:7
找到位置:14
发生次数:2
2.打印(“occorrence的数量:”,查找所有(“允许的Hello Hollow”,“ll”))
输出:
找到位置:1
找到位置:10
位置:16
发生次数:3
3.打印(“OCcorrence数:”,查找所有(“Aaa bbbcd$”@@ABWebBBBRR 123,“bbb”))
输出:
找到位置:4
位置:21
发生次数:2
def find_all(字符串、字符串):
计数=0
i=长度(收敛)
x=0
当x
此代码可能不是最短/最有效的,但它简单易懂
def findall(f, s):
l = []
i = -1
while True:
i = s.find(f, i+1)
if i == -1:
return l
l.append(s.find(f, i))
findall('test', 'test test test test')
# [0, 5, 10, 15]
对于第一个版本,请检查字符串:
def findall(文本,子项):
“”“返回文本中出现子字符串的所有索引”“”
返回[
指数
用于索引
findin = "algorithm alma mater alison alternation alpines"
search = "al"
inx = 0
num_str = 0
while True:
inx = findin.find(search)
if inx == -1: #breaks before adding 1 to number of string
break
inx = inx + 1
findin = findin[inx:] #to splice the 'unsearched' part of the string
num_str = num_str + 1 #counts no. of string
if num_str != 0:
print("There are ",num_str," ",search," in your string.")
else:
print("There are no ",search," in your string.")
| theString | aString |
| -------------------------- | ------- |
| "661444444423666455678966" | "55" |
| "661444444423666455678966" | "44" |
| "6123666455678966" | "666" |
| "66123666455678966" | "66" |
Calling examples:
1. print("Number of occurrences: ", find_all("123666455556785555966", "5555"))
output:
Found in position: 7
Found in position: 14
Number of occurrences: 2
2. print("Number of occorrences: ", find_all("Allowed Hello Hollow", "ll "))
output:
Found in position: 1
Found in position: 10
Found in position: 16
Number of occurrences: 3
3. print("Number of occorrences: ", find_all("Aaa bbbcd$#@@abWebbrbbbbrr 123", "bbb"))
output:
Found in position: 4
Found in position: 21
Number of occurrences: 2
def find_all(theString, aString):
count = 0
i = len(aString)
x = 0
while x < len(theString) - (i-1):
if theString[x:x+i] == aString:
print("Found in position: ", x)
x=x+i
count=count+1
else:
x=x+1
return count
def findall(f, s):
l = []
i = -1
while True:
i = s.find(f, i+1)
if i == -1:
return l
l.append(s.find(f, i))
findall('test', 'test test test test')
# [0, 5, 10, 15]