问题:如何获得字符在Python中的位置?
How can I get the position of a character inside a string in python?
回答 0
这有两个String的方法,find()
和index()
。两者之间的区别在于找不到搜索字符串时会发生什么。 find()
回报-1
和index()
加薪ValueError
。
使用 find()
>>> myString = 'Position of a character'
>>> myString.find('s')
2
>>> myString.find('x')
-1
使用 index()
>>> myString = 'Position of a character'
>>> myString.index('s')
2
>>> myString.index('x')
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
ValueError: substring not found
string.find(s, sub[, start[, end]])
返回s中找到子字符串sub的最低索引,以使sub完全包含在中s[start:end]
。-1
失败返回。开始和结束以及负值的解释默认值与切片相同。
和:
string.index(s, sub[, start[, end]])
喜欢,find()
但是ValueError
在找不到子字符串时提高。
There are two string methods for this, find()
and index()
. The difference between the two is what happens when the search string isn’t found. find()
returns -1
and index()
raises ValueError
.
Using find()
>>> myString = 'Position of a character'
>>> myString.find('s')
2
>>> myString.find('x')
-1
Using index()
>>> myString = 'Position of a character'
>>> myString.index('s')
2
>>> myString.index('x')
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
ValueError: substring not found
string.find(s, sub[, start[, end]])
Return the lowest index in s where the substring sub is found such that sub is wholly contained in s[start:end]
. Return -1
on failure. Defaults for start and end and interpretation of negative values is the same as for slices.
And:
string.index(s, sub[, start[, end]])
Like find()
but raise ValueError
when the substring is not found.
回答 1
仅出于完整性考虑,如果需要查找字符串中字符的所有位置,可以执行以下操作:
s = 'shak#spea#e'
c = '#'
print [pos for pos, char in enumerate(s) if char == c]
它将返回 [4, 9]
Just for a sake of completeness, if you need to find all positions of a character in a string, you can do the following:
s = 'shak#spea#e'
c = '#'
print [pos for pos, char in enumerate(s) if char == c]
which will return [4, 9]
回答 2
>>> s="mystring"
>>> s.index("r")
4
>>> s.find("r")
4
“长ed”方式
>>> for i,c in enumerate(s):
... if "r"==c: print i
...
4
得到子串,
>>> s="mystring"
>>> s[4:10]
'ring'
>>> s="mystring"
>>> s.index("r")
4
>>> s.find("r")
4
“Long winded” way
>>> for i,c in enumerate(s):
... if "r"==c: print i
...
4
to get substring,
>>> s="mystring"
>>> s[4:10]
'ring'
回答 3
只是为了完成,如果要在文件名中找到扩展名以进行检查,则需要找到最后一个“。”,在这种情况下,请使用rfind:
path = 'toto.titi.tata..xls'
path.find('.')
4
path.rfind('.')
15
就我而言,我使用以下命令,无论完整的文件名是什么,它都可以工作:
filename_without_extension = complete_name[:complete_name.rfind('.')]
Just for completion, in the case I want to find the extension in a file name in order to check it, I need to find the last ‘.’, in this case use rfind:
path = 'toto.titi.tata..xls'
path.find('.')
4
path.rfind('.')
15
in my case, I use the following, which works whatever the complete file name is:
filename_without_extension = complete_name[:complete_name.rfind('.')]
回答 4
当字符串包含重复字符时会发生什么?从我的经验中,index()
我看到重复的结果会返回相同的索引。
例如:
s = 'abccde'
for c in s:
print('%s, %d' % (c, s.index(c)))
会返回:
a, 0
b, 1
c, 2
c, 2
d, 4
在这种情况下,您可以执行以下操作:
for i, character in enumerate(my_string):
# i is the position of the character in the string
What happens when the string contains a duplicate character?
from my experience with index()
I saw that for duplicate you get back the same index.
For example:
s = 'abccde'
for c in s:
print('%s, %d' % (c, s.index(c)))
would return:
a, 0
b, 1
c, 2
c, 2
d, 4
In that case you can do something like that:
for i, character in enumerate(my_string):
# i is the position of the character in the string
回答 5
string.find(character)
string.index(character)
也许您想看一下文档,找出两者之间的区别。
string.find(character)
string.index(character)
Perhaps you’d like to have a look at the documentation to find out what the difference between the two is.
回答 6
一个字符可能在字符串中多次出现。例如,在字符串中sentence
,位置e
is是1, 4, 7
(因为索引通常从零开始)。但是我发现这是两个函数,find()
并且index()
返回字符的第一个位置。因此,这样做可以解决此问题:
def charposition(string, char):
pos = [] #list to store positions for each 'char' in 'string'
for n in range(len(string)):
if string[n] == char:
pos.append(n)
return pos
s = "sentence"
print(charposition(s, 'e'))
#Output: [1, 4, 7]
A character might appear multiple times in a string. For example in a string sentence
, position of e
is 1, 4, 7
(because indexing usually starts from zero). but what I find is both of the functions find()
and index()
returns first position of a character. So, this can be solved doing this:
def charposition(string, char):
pos = [] #list to store positions for each 'char' in 'string'
for n in range(len(string)):
if string[n] == char:
pos.append(n)
return pos
s = "sentence"
print(charposition(s, 'e'))
#Output: [1, 4, 7]
回答 7
more_itertools.locate
是一种第三方工具,用于查找满足条件的所有项目的索引。
在这里,我们找到字母的所有索引位置"i"
。
import more_itertools as mit
s = "supercalifragilisticexpialidocious"
list(mit.locate(s, lambda x: x == "i"))
# [8, 13, 15, 18, 23, 26, 30]
more_itertools.locate
is a third-party tool that finds all indicies of items that satisfy a condition.
Here we find all index locations of the letter "i"
.
import more_itertools as mit
s = "supercalifragilisticexpialidocious"
list(mit.locate(s, lambda x: x == "i"))
# [8, 13, 15, 18, 23, 26, 30]
回答 8
使用numpy的解决方案可以快速访问所有索引:
string_array = np.array(list(my_string))
char_indexes = np.where(string_array == 'C')
A solution with numpy for quick access to all indexes:
string_array = np.array(list(my_string))
char_indexes = np.where(string_array == 'C')