Lets say with a string = "AABBAAAAABBBBAAABBBBAA"
I want to return string split by the odd lengths of the string (i.e when A = 5 or A = 3),
What I want returned is 1) AABBAAAAA 2)BBBBAAA 3)BBBBAA,
How can I do that?
I tried using regex [A]+[B]+ for a slightly different case
One option might be to regex iterate using re.finditer with the following pattern:
.*?(?:AAA(?:AA)?|$)
This pattern will non greedily consume until reaching either 3 A's, 5 A's, or the end of the string. Then, we can print out each complete match as we iterate.
input = 'AABBAAAAABBBBAAABBBBAA'
pattern = '.*?(?:AAA(?:AA)?|$)'
for match in re.finditer(pattern, input):
print match.group()
This prints:
AABBAAAAA
BBBBAAA
BBBBAA
You can use itertools.groupby:
s = 'BBAAAAABBBBAAABBBBAA'
from itertools import groupby
out = ['']
for v, g in groupby(s):
l = [*g]
out[-1] += ''.join(l)
if v == 'A' and len(l) in (3, 5):
out.append('')
print(out)
Prints:
['BBAAAAA', 'BBBBAAA', 'BBBBAA']
Related
How can I shuffle two strings s||t (shuffle(s, t)) with the given requirement that the first char always stands in front of the second one in s and t as well no matter we shuffle. The result returns as a set of strings without duplicates.
I have the following test:
print(shuffle('ab', 'cd'))
Result:
['abcd', 'acbd', 'acdb', 'cabd', 'cadb', 'cdab']
Thanks a lot.
This method will shuffle two strings and return a list of shuffles between them where the order of the characters is the same as in the original strings. If there are duplicate characters there will be duplicate results as well.
def shuffle(s1, s2):
if len(s1) == 1:
return [s2[:i] + s1 + s2[i:] for i in range(len(s2) + 1)]
if len(s2) == 1:
return [s1[:i] + s2 + s1[i:] for i in range(len(s1) + 1)]
return [s1[0]+ s for s in shuffle(s1[1:], s2)] + [s2[0] + s for s in shuffle(s1, s2[1:])]
print shuffle("ab", "cd")
It works by getting the first character of each string and recursively shuffling the rest and adding this character to each element in the list. When there is one character remaining on each of the strings it returns a list where the character is added in each position of the other string. Hope it helps.
So you can apply a condition on final shuffled list to generate a new list from the shuffled one:
S=shuffle('ab','cd')
nl=[]
for w in S:
if(w.index('a')<w.index('b') and w.index('c')<w.index('d')):
nl.append(w)
So nl is your new list as per your requirement:)
If I understood the question correctly, this should work. Note, as you add letters to this, it becomes a long running problem. 4 letters have 6 possible combination for each entry in the list. 8 letters have 5,040 possible combinations for each entry in the list.
import random
import math
InputList = ['ab','cd']
PossibleUniqueCombinations = math.factorial(len("".join(InputList))-1)
print (PossibleUniqueCombinations)
TargetList = []
UniqueCombinationList = []
for lst in InputList:
UniqueCnt = 0
FirstChar = lst[0]
TheRest = list(lst[1:])
while UniqueCnt < PossibleUniqueCombinations:
if InputList.index(lst) == 0:
LeftList = []
else:
LeftList = InputList[0:InputList.index(lst)]
RightList = list(InputList[InputList.index(lst)+1:])
TargetList = LeftList + TheRest + RightList
TargetStr = ''.join(TargetList)
TargetStr = ''.join(random.sample(TargetStr, len(TargetStr)))
ShuffledStr = FirstChar + ''.join(TargetStr)
try:
FndIdx = UniqueCombinationList.index(ShuffledStr)
except ValueError:
UniqueCombinationList.append(ShuffledStr)
UniqueCnt += 1
for combo in UniqueCombinationList:
print(combo)
i need sum in string letters value ex.
a = 1
b = 2
c = 3
d = 4
alphabet = 'abcdefghijklmnopqrstuvwxyz'
v1
string = "abcd"
# #result = sum(string) so
if string[0] and string[1] and string[2] and string[3] in alphabet:
if string[0] is alphabet[0] and string[1] is alphabet[1] and string[2] is alphabet[2] and string[3] is alphabet[3]:
print(a+b+c+d)
v2
string = ("ab","aa","dc",)
if string[0][0] and string[0][1] and string[1][0] and string[1][1] and string[2][0] and string[2][1] in alphabet:
if string[0] is alphabet[0] and string[1] is alphabet[1] and string[2] is alphabet[2] and string[3] is alphabet[3]:
print(a+b+c+d)
what is the solution? can you help me
Use the sum() function and a generator expression; a dictionary built from string.ascii_lowercase can serve as a means to getting an integer value per letter:
from string import ascii_lowercase
letter_value = {c: i for i, c in enumerate(ascii_lowercase, 1)}
wordsum = sum(letter_value.get(c, 0) for c in word if c)
The enumerate(ascii_lowercase, 1) produces (index, letter) pairs when iterated over, starting at 1. That gives you (1, 'a'), (2, 'b'), etc. That can be converted to c: i letter pairs in a dictionary, mapping letter to integer number.
Next, using the dict.get() method lets you pick a default value; for any character in the input string, you get to look up the numeric value and map it to an integer, but if the character is not a lowercase letter, 0 is returned instead. The sum(...) part with the loop then simply adds those values up.
If you need to support sequences with words, just use sum() again. Put the above sum() call in a function, and apply that function to each word in a sequence:
from string import ascii_lowercase
letter_value = {c: i for i, c in enumerate(ascii_lowercase, 1)}
def sum_word(word):
return sum(letter_value.get(c, 0) for c in word if c)
def sum_words(words):
return sum(sum_word(word) for word in words)
The old-fashioned way is to take advantage of the fact that lowercase letters are contiguous, so that ord(b) - ord(a) == 1:
data = "abcd"
print("Sum:", sum(ord(c)-ord("a")+1 for c in data))
Of course you could "optimize" it to reduce the number of computations, though it seems silly in this case:
ord_a = ord("a")
print("Sum:", sum(ord(c)-ord_a for c in data)+len(data))
I want to convert a string so that the pair positions will be in upper case characters and the impair positions will be in lower case characters.
Here is what I've tried so far:
def foldingo(chaine):
chaineuh=chaine[0::2].upper()
chaine=chaineuh[1::2].lower()
return chaine
your code takes every other character in chaine, uppercases them, and assigns those characters to chaineuh.
Then it takes every other character in chaineuh, lowercases them, and assigns those characters to chaine again. In other words:
abcdefg -> ACEG -> cg
You'll notice it's not keeping the characters that you're not trying to target.
You could try building all the uppercases and lowercases separately, then iterate with zip to get them together.
def fold(s):
uppers = s[0::2].upper()
lowers = s[1::2].lower()
return zip(uppers, lowers)
But this doesn't quit work either, since zip gives you tuples, not strings, and will drop the last character in odd-lengthed strings
abcdefg -> ACEG, bdf -> ('A', 'b'), ('C', 'd'), ('E', 'f')
We could fix that by using a couple calls to str.join and using itertools.zip_longest with a fillvalue='', but it's kind of like using a wrench to hammer in a nail. It's not really the right tool for the job. For the record: it would look like:
''.join([''.join(pair) for pair in itertools.zip_longest(uppers, lowers, fillvalue='')])
yuck.
Let's instead just iterate over the string and uppercase every other letter. We can use an alternating boolean to track whether we're upper'ing or lower'ing this time around.
def fold(s):
time_to_upper = True
result = ""
for ch in s:
if time_to_upper:
result += ch.upper()
else:
result += ch.lower()
time_to_upper = not time_to_upper
return result
You could also use enumerate and a modulo to keep track:
def fold(s):
result = ""
for i, ch in enumerate(s):
ch = ch.lower() if i % 2 else ch.upper()
result += ch
return result
Or by using itertools.cycle, str.join, and list comprehensions, we can make this a lot shorter (possibly at the cost of readability!)
import itertools
def fold(s):
return ''.join([op(ch) for op, ch in zip(itertools.cycle([str.upper, str.lower]), s)]
That's the source code:
def revers_e(str_one,str_two):
for i in range(len(str_one)):
for j in range(len(str_two)):
if str_one[i] == str_two[j]:
str_one = (str_one - str_one[i]).split()
print(str_one)
else:
print('There is no relation')
if __name__ == '__main__':
str_one = input('Put your First String: ').split()
str_two = input('Put your Second String: ')
print(revers_e(str_one, str_two))
How can I remove a letter that occurs in both strings from the first string then print it?
How about a simple pythonic way of doing it
def revers_e(s1, s2):
print(*[i for i in s1 if i in s2]) # Print all characters to be deleted from s1
s1 = ''.join([i for i in s1 if i not in s2]) # Delete them from s1
This answer says, "Python strings are immutable (i.e. they can't be modified). There are a lot of reasons for this. Use lists until you have no choice, only then turn them into strings."
First of all you don't need to use a pretty suboptimal way using range and len to iterate over a string since strings are iterable you can just iterate over them with a simple loop.
And for finding intersection within 2 string you can use set.intersection which returns all the common characters in both string and then use str.translate to remove your common characters
intersect=set(str_one).intersection(str_two)
trans_table = dict.fromkeys(map(ord, intersect), None)
str_one.translate(trans_table)
def revers_e(str_one,str_two):
for i in range(len(str_one)):
for j in range(len(str_two)):
try:
if str_one[i] == str_two[j]:
first_part=str_one[0:i]
second_part=str_one[i+1:]
str_one =first_part+second_part
print(str_one)
else:
print('There is no relation')
except IndexError:
return
str_one = input('Put your First String: ')
str_two = input('Put your Second String: ')
revers_e(str_one, str_two)
I've modified your code, taking out a few bits and adding a few more.
str_one = input('Put your First String: ').split()
I removed the .split(), because all this would do is create a list of length 1, so in your loop, you'd be comparing the entire string of the first string to one letter of the second string.
str_one = (str_one - str_one[i]).split()
You can't remove a character from a string like this in Python, so I split the string into parts (you could also convert them into lists like I did in my other code which I deleted) whereby all the characters up to the last character before the matching character are included, followed by all the characters after the matching character, which are then appended into one string.
I used exception statements, because the first loop will use the original length, but this is subject to change, so could result in errors.
Lastly, I just called the function instead of printing it too, because all that does is return a None type.
These work in Python 2.7+ and Python 3
Given:
>>> s1='abcdefg'
>>> s2='efghijk'
You can use a set:
>>> set(s1).intersection(s2)
{'f', 'e', 'g'}
Then use that set in maketrans to make a translation table to None to delete those characters:
>>> s1.translate(str.maketrans({e:None for e in set(s1).intersection(s2)}))
'abcd'
Or use list comprehension:
>>> ''.join([e for e in s1 if e in s2])
'efg'
And a regex to produce a new string without the common characters:
>>> re.sub(''.join([e for e in s1 if e in s2]), '', s1)
'abcd'
I have a string x: x = "{abc}{def}{ghi}"
And I need to print the string between second { and second }, in this case def. How can I do this without knowing the length of the string? For example, the string x could also be {abcde}{fghij}{klmno}"
This is where pattern matching is useful:
local x = "{abc}{def}{ghi}"
local result = x:match(".-{.-}.-{(.-)}")
print(result)
.- matches zero or more characters, non-greedy. The whole pattern .-{.-}.-{(.-)} captures what's between the second { and the second }.
Try also x:match(".-}{(.-)}"), which is simpler.
I would go about it in a different manner:
local i, x, result = 1, "{abc}{def}{ghi}"
for w in x:gmatch '{(.-)}' do
if i == 2 then
result = w
break
else
i = i + 1
end
end
print( result )