The input to this problem is a string and has a specific form. For example if s is a string then inputs can be s='3(a)2(b)' or s='3(aa)2(bbb)' or s='4(aaaa)'. The output should be a string, that is the substring inside the brackets multiplied by numerical substring value the substring inside the brackets follows.
For example,
Input ='3(a)2(b)'
Output='aaabb'
Input='4(aaa)'
Output='aaaaaaaaaaaa'
and similarly for other inputs. The program should print an empty string for wrong or invalid inputs.
This is what I've tried so far
s='3(aa)2(b)'
p=''
q=''
for i in range(0,len(s)):
#print(s[i],end='')
if s[i]=='(':
k=int(s[i-1])
while(s[i+1]!=')'):
p+=(s[i+1])
i+=1
if s[i]==')':
q+=k*p
print(q)
Can anyone tell what's wrong with my code?
A oneliner would be:
''.join(int(y[0])*y[1] for y in (x.split('(') for x in Input.split(')')[:-1]))
It works like this. We take the input, and split on the close paren
In [1]: Input ='3(a)2(b)'
In [2]: a = Input.split(')')[:-1]
In [3]: a
Out[3]: ['3(a', '2(b']
This gives us the integer, character pairs we're looking for, but we need to get rid of the open paren, so for each x in a, we split on the open paren to get a two-element list where the first element is the int (as a string still) and the character. You'll see this in b
In [4]: b = [x.split('(') for x in a]
In [5]: b
Out[5]: [['3', 'a'], ['2', 'b']]
So for each element in b, we need to cast the first element as an integer with int() and multiply by the character.
In [6]: c = [int(y[0])*y[1] for y in b]
In [7]: c
Out[7]: ['aaa', 'bb']
Now we join on the empty string to combine them into one string with
In [8]: ''.join(c)
Out[8]: 'aaabb'
Try this:
a = re.findall(r'[\d]+', s)
b = re.findall(r'[a-zA-Z]+', s)
c = ''
for i, j in zip(a, b):
c+=(int(i)*str(j))
print(c)
Here is how you could do it:
Step 1: Simple case, getting the data out of a really simple template
Let's assume your template string is 3(a). That's the simplest case I could think of. We'll need to extract pieces of information from that string. The first one is the count of chars that will have to be rendered. The second is the char that has to be rendered.
You are in a case where regex are more than suited (hence, the use of re module from python's standard library).
I won't do a full course on regex. You'll have to do that by our own. However, I'll explain quickly the step I used. So, count (the variable that holds the number of times we should render the char to render) is a digit (or several). Hence our first capturing group will be something like (\d+). Then we have a char to extract that is enclosed by parenthesis, hence \((\w+)\) (I actually enable several chars to be rendered at once). So, if we put them together, we get (\d+)\((\w+)\). For testing you can check this out.
Applied to our case, a straight forward use of the re module is:
import re
# Our template
template = '3(a)'
# Run the regex
match = re.search(r'(\d+)\((\w+)\)', template)
if match:
# Get the count from the first capturing group
count = int(match.group(1))
# Get the string to render from the second capturing group
string = match.group(2)
# Print as many times the string as count was given
print count * string
Output:
aaa
Yeah!
Step 2: Full case, with several templates
Okay, we know how to do it for 1 template, how to do the same for several, for instance 3(a)4(b)? Well... How would we do it "by hand"? We'd read the full template from left to right and apply each template one by one. Then this is what we'll do with python!
Hopefully for us the re module has a function just for that: finditer. It does exactly what we described above.
So, we'll do something like:
import re
# Our template
template = '3(a)4(b)'
# Iterate through found templates
for match in re.finditer(r'(\d+)\((\w+)\)', template):
# Get the count from the first capturing group
count = int(match.group(1))
# Get the string to render from the second capturing group
string = match.group(2)
print count * string
Output:
aaa
bbbb
Okay... Just remains the combination of that stuff. We know we can put everything at each step in an array, and then join each items of this array at the end, no?
Let's do it!
import re
template = '3(a)4(b)'
parts = []
for match in re.finditer(r'(\d+)\((\w+)\)', template):
parts.append(int(match.group(1)) * match.group(2))
print ''.join(parts)
Output:
aaabbb
Yeah!
Step 3: Final step, optimization
Because we can always do better, we won't stop. for loops are cool. But what I love (it's personal) about python is that there is so much stuff you can actually just write with one line! Is it the case here? Well yes :).
First we can remove the for loop and the append using a list comprehension:
parts = [int(match.group(1)) * match.group(2) for match in re.finditer(r'(\d+)\((\w+)\)', template)]
rendered = ''.join(parts)
Finally, let's remove the two lines with parts populating and then join and let's do all that in a single line:
import re
template = '3(a)4(b)'
rendered = ''.join(
int(match.group(1)) * match.group(2) \
for match in re.finditer(r'(\d+)\((\w+)\)', template))
print rendered
Output:
aaabbb
Yeah! Still the same output :).
Hope it helped!
The value of 'p' should be refreshed after each iteration.
s='1(aaa)2(bb)'
p=''
q=''
i=0
while i<len(s):
if s[i]=='(':
k=int(s[i-1])
p=''
while(s[i+1]!=')'):
p+=(s[i+1])
i+=1
if s[i]==')':
q+=k*p
i+=1
print(q)
The code is not behaving the way I want it to behave. The problem here is the placement of 'p'. 'p' is the variable that adds the substring inside the ( )s. I'm repeating the process even after sufficient adding is done. Placing 'p' inside the 'if' block will do the job.
s='2(aa)2(bb)'
q=''
for i in range(0,len(s)):
if s[i]=='(':
k=int(s[i-1])
p=''
while(s[i+1]!=')'):
#print(i,'first time')
p+=s[i+1]
i+=1
q+=p*k
#print(i,'second time')
print(q)
what you want is not print substrings . the real purpose is most like to generate text based regular expression or comands.
you can parametrize a function to read it or use something like it:
The python library rstr has the function xeger() to do what you need by using random strings and only returning ones that match:
Example
Install with pip install rstr
In [1]: from __future__ import print_function
In [2]: import rstr
In [3]: for dummy in range(10):
...: print(rstr.xeger(r"(a|b)[cd]{2}\1"))
...:
acca
bddb
adda
bdcb
bccb
bcdb
adca
bccb
bccb
acda
Warning
For complex re patterns this might take a long time to generate any matches.
Related
I have a few lines of string e.g:
AR0003242303
TR0402304004
CR0402340404
I want to create a dictionary from these lines.
And I need to create change it in regex to:
KOLAORM0003242303
KOLTORM0402304004
KOLCORM0402340404
So i need to split first 2 characters, before PUT KOL, between PUT O, and Afer second char put M. How can i reach it. Through many attempts I lose patience with the regex and unfortunately I now I have no time to learn it better now. Need some result now :(
Could someone help me with this case?
Using re.sub --> re.sub(r"^([A-Z])([A-Z])", r"KOL\1O\2M", string)
Ex:
import re
s = ["AR0003242303", "TR0402304004", "CR0402340404"]
for i in s:
print( re.sub(r"^([A-Z])([A-Z])", r"KOL\1O\2M", i) )
Output:
KOLAORM0003242303
KOLTORM0402304004
KOLCORM0402340404
You don't need regex for this, you can do it with getting the list of characters from the string, recreate the list, and join the string back
def get_convert_s(s):
li = list(s)
li = ['KOL', li[0], '0', li[1], 'M', *li[2:]]
return ''.join(li)
print(get_convert_s('AR0003242303'))
#KOLA0RM0003242303
print(get_convert_s('TR0402304004'))
#KOLT0RM0402304004
print(get_convert_s('CR0402340404'))
#KOLC0RM0402340404
import re
regex = re.compile(r"([A-Z])([A-Z])([0-9]+)")
inputs = [
'AR0003242303',
'TR0402304004',
'CR0402340404'
]
results = []
for input in inputs:
matches = re.match(regex, input)
groups = matches.groups()
results.append('KOL{}O{}M{}'.format(*groups))
print(results)
Assuming the length of the strings in your list will always be the same Devesh answers is pretty much the best approach (no reason to overcomplicate it).
My solution is similar to Devesh, I just like writing functions as oneliners:
list = ["AR0003242303", "TR0402304004", "CR0402340404"]
def convert_s(s):
return "KOL"+s[0]+"0"+s[1]+"M"+s[2:]
for str in list:
print(convert_s(str));
Altough it returns the same output.
How most effectively do I cut out a part of a word if the character '=#=' appears and then finish cutting the word if the character '=#=' appears? For example:
From a large string
'321#5=85#45#41=#=I-LOVE-STACK-OVER-FLOW=#=3234#41#=q#$^1=#=xx$q=#=xpa$=4319'
The python code returns:
'I-LOVE-STACK-OVER-FLOW'
Any help will be appreciated.
Using split():
s = '321#5=85#45#41=#=I-LOVE-STACK-OVER-FLOW=#=3234#41#=q#$^1=#=xx$q=#=xpa$=4319'
st = '=#='
ed = '=#='
print((s.split(st))[1].split(ed)[0])
Using regex:
import re
s = '321#5=85#45#41=#=I-LOVE-STACK-OVER-FLOW=#=3234#41#=q#$^1=#=xx$q=#=xpa$=4319'
print(re.search('%s(.*)%s' % (st, ed), s).group(1))
OUTPUT:
I-LOVE-STACK-OVER-FLOW
In addition to #DirtyBit's answer, if you want to also handle cases of more than 2 '=#='s, you can split the string, and then add every other element:
s = '321#5=85#45#41=#=I-LOVE-STACK-OVER-FLOW=#=3234#41#=q#$^1=#=xx$q=#=xpa$=4319=#=|I-ALSO-LOVE-SO=#=3123123'
parts = s.split('=#=')
print(''.join([parts[i] for i in range(1,len(parts),2)]))
Output
I-LOVE-STACK-OVER-FLOW|I-ALSO-LOVE-SO
The explanation is in the code.
import re
ori_list = re.split("=#=",ori_str)
# you can imagine your goal is to find the string wrapped between signs of "=#="
# so after the split, the even number position must be the parts outsides of "=#="
# and the odd number position is what you want
for i in range(len(ori_list)):
if i%2 == 1:#odd position
print(ori_list[i])
A small issue I've encountered during coding.
I'm looking to print out the name of a .txt file.
For example, the file is named: verdata_florida.txt, or verdata_newyork.txt
How can I exclude .txt and verdata_, but keep the string between? It must work for any number of characters, but .txt and verdata_ must be excluded.
This is where I am so far, I've already defined filename to be input()
print("Average TAM at", str(filename[8:**????**]), "is higher than ")
3 ways of doing it:
using str.split twice:
>>> "verdata_florida.txt".split("_")[1].split(".")[0]
'florida'
using str.partition twice (you won't get an exception if the format doesn't match, and probably faster too):
>>> "verdata_florida.txt".partition("_")[2].partition(".")[0]
'florida'
using re, keeping only center part:
>>> import re
>>> re.sub(".*_(.*)\..*",r"\1","verdata_florida.txt")
'florida'
all those above must be tuned if _ and . appear multiple times (must we keep the longest or the shortest string)
EDIT: In your case, though, prefixes & suffixes seem fixed. In that case, just use str.replace twice:
>>> "verdata_florida.txt".replace("verdata_","").replace(".txt","")
'florida'
Assuming you want it to split on the first _ and the last . you can use slicing and the index and rindex functions to get this done. These functions will search for the first occurrence of the substring in the parenthesis and return the index number. If no substring is found, they will throw a ValueError. If the search is desired, but not the ValueError, you can also use find and rfind, which do the same thing but always return -1 if no match is found.
s = 'verdata_new_hampshire.txt'
s_trunc = s[s.index('_') + 1: s.rindex('.')] # or s[s.find('_') + 1: s.rfind('.')]
print(s_trunc) # new_hampshire
Of course, if you are always going to exclude verdata_ and .txt you could always hardcode the slice as well.
print(s[8:-4]) # new_hampshire
You can leverage str.split() on strings. For example:
s = 'verdata_newyork.txt'
s.split('verdata_')
# ['', 'florida.txt']
s.split('verdata_')[1]
# 'florida.txt'
s.split('verdata_')[1].split('.txt')
['florida', '']
s.split('verdata_')[1].split('.txt')[0]
# 'florida'
You can just split string by dot and underscore like this:
string filename = "verdata_prague.txt";
string name = filename.split("."); //verdata_prague
name = name[0].split("_")[1]; //prague
or by replace function:
string filename = "verdata_prague.txt";
string name = filename.replace(".txt",""); //verdata_prague
name = name[0].replace("verdata_","")[1]; //prague
I have a string with a lot of recurrencies of a single pattern like
a = 'eresQQQutnohnQQQjkhjhnmQQQlkj'
and I have another string like
b = 'rerTTTytu'
I want to substitute the entire second string having as a reference the 'QQQ' and the 'TTT', and I want to find in this case 3 different results:
'ererTTTytuohnQQQjkhjhnmQQQlkj'
'eresQQQutnrerTTTytujhnmQQQlkj'
'eresQQQutnohnQQQjkhjrerTTTytu'
I've tried using re.sub
re.sub('\w{3}QQQ\w{3}' ,b,a)
but I obtain only the first one, and I don't know how to get the other two solutions.
Edit: As you requested, the two characters surrounding 'QQQ' will be replaced as well now.
I don't know if this is the most elegant or simplest solution for the problem, but it works:
import re
# Find all occurences of ??QQQ?? in a - where ? is any character
matches = [x.start() for x in re.finditer('\S{2}QQQ\S{2}', a)]
# Replace each ??QQQ?? with b
results = [a[:idx] + re.sub('\S{2}QQQ\S{2}', b, a[idx:], 1) for idx in matches]
print(results)
Output
['errerTTTytunohnQQQjkhjhnmQQQlkj',
'eresQQQutnorerTTTytuhjhnmQQQlkj',
'eresQQQutnohnQQQjkhjhrerTTTytuj']
Since you didn't specify the output format, I just put it in a list.
I wondering if there is some way to find only second quotes from each pair in string, that has paired quotes.
So if I have string like '"aaaaa"' or just '""' I want to find only the last '"' from it. If I have '"aaaa""aaaaa"aaaa""' I want only the second, fourth and sixth '"'s. But if I have something like this '"aaaaaaaa' or like this 'aaa"aaa' I don't want to find anything, since there are no paired quotes. If i have '"aaa"aaa"' I want to find only second '"', since the third '"' has no pair.
I've tried to implement lookbehind, but it doesn't work with quantifiers, so my bad attempt was '(?<=\"a*)\"'.
You don't really need regex for this. You can do:
[i for i, c in enumerate(s) if c == '"'][1::2]
To get the index of every other '"'. Example usage:
>>> for s in ['"aaaaa"', '"aaaa""aaaaa"aaaa""', 'aaa"aaa', '"aaa"aaa"']:
print(s, [i for i, c in enumerate(s) if c == '"'][1::2])
"aaaaa" [6]
"aaaa""aaaaa"aaaa"" [5, 12, 18]
aaa"aaa []
"aaa"aaa" [4]
import re
reg = re.compile(r'(?:\").*?(\")')
then
for match in reg.findall('"this is", "my test"'):
print(match)
gives
"
"
If your necessity is to change the second quote you can also match the whole string and put the pattern before the second quote into a capture group. Then making the substitution by the first match group + the substitution string would archive the issue.
For example, this regex will match everything before the second quote and put it into a group
(\"[^"]*)\"
if you replace whole the match (which includes the second quote) by only the value of the capture group (which does not include the second quote), then you would just cut it off.
See the online example
import re
p = re.compile(ur'(\"[^"]*)\"')
test_str = u"\"test1\"test2\"test3\""
subst = r"\1"
result = re.sub(p, subst, test_str)
print result #result -> "test1test2"test3
Please read my answer about why you don't want to use regular expressions for such a problem, even though you can do that kind of non-regular job with it.
Ok then you probably want one of the solutions I give in the linked answer, where you'll want to use a recursive regex to match all the matching pairs.
Edit: the following has been written before the update to the question, which was asking only for second double quotes.
Though if you want to find only second double quotes in a string, you do not need regexps:
>>> s1='aoeu"aoeu'
>>> s2='aoeu"aoeu"aoeu'
>>> s3='aoeu"aoeu"aoeu"aoeu'
>>> def find_second_quote(s):
... pos_quote_1 = s2.find('"')
... if pos_quote_1 == -1:
... return -1
... pos_quote_2 = s[pos_quote_1+1:].find('"')
... if pos_quote_2 == -1:
... return -1
... return pos_quote_1+1+pos_quote_2
...
>>> find_second_quote(s1)
-1
>>> find_second_quote(s2)
4
>>> find_second_quote(s3)
4
>>>
here it either returns -1 if there's no second quote, or the position of the second quote if there is one.
a parser is probably better, but depending on what you want to get out of it, there are other ways. if you need the data between the quotes:
import re
re.findall(r'".*?"', '"aaaa""aaaaa"aaaa""')
['"aaaa"',
'"aaaaa"',
'""']
if you need the indices, you could do it as a generator or other equivalent like this:
def count_quotes(mystr):
count = 0
for i, x in enumerate(mystr):
if x == '"':
count += 1
if count % 2 == 0:
yield i
list(count_quotes('"aaaa""aaaaa"aaaa""'))
[5, 12, 18]