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I would like a function to detect string reputation, specifically
repetition("abcabcabc")
abc
repetition("aaaaaaa")
a
repetition("ababab")
ab
repetition("abcd")
abcd
I am thinking of doing it in a recursive way but I am confused
Thanks for any help in advance!
I am trying something like
def repetition(r):
if len(r) == 2:
if r[0] == r[1]:
return r[0]
half = len(r) / 2
repetition(r[:half])
if r[:half] == r[half:]:
return r[:half]
There probably is a better way to do this, but my first thought would be this:
def repetition(string):
substring = ''
for character in string:
substring += character
if len(string) % len(substring) == 0:
if (len(string) / len(substring)) * substring == string:
return substring
Using regular expressions:
import re
def repetitions(s):
r = re.compile(r"(.+?)\1+")
for match in r.finditer(s):
if len(match.group()) != len(s):
return s
return match.group(1)
Test:
repetitions("oblabla")
#output: "oblabla"
repetitions("blabla")
#output: "bla"
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I tried these, but its only for one column.
using notepad++ and python. Any solution that can do for multiple columns of text?
# Input:
tim
oso
d n
a d
y a
y
# Output:
today
is
monday
You can do this fairly easily with a loop, something like:
s = "tim\noso\nd n\na d\ny a\n y"
bits = s.split('\n')
lines = ['' for i in range(0, len(bits[0]))]
for bit in bits:
for i in range(0, len(bit)):
if bit[i] != ' ':
lines[i] += bit[i]
'\n'.join(lines)
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def CodelandUsernameValidation(s):
if len(s)>4 and len(s)<25 and s[0].isalpha() and [i for i in s if i.isalnum() or i=="_"]!=[] and s[-1]!="_":
return True
else:
return False
# keep this function call here
print(CodelandUsernameValidation(input()))
It's a list comprehension if you expand, it'll look like this ->
result = []
for i in s: # will fetch element one by one from iterable
if i.isalnum() or i=="_": # checking condition
result.append(i) # if true, then append it to the list
This above code can be rewritten as -
result = [i for i in s if i.isalnum() or i=="_"]
That produces a list of those characters in s that are either alphanumeric or underlines. The code is actually incorrect, because it will pass if ANY of the characters are alphanumeric or underline, whereas the intent surely was that ALL of the characters must be alphanumeric or underline. Here's a better way to write it:
def CodelandUsernameValidation(s):
return 4 < len(s) < 25 and s[0].isalpha() and all(i.isalnum() or i=='_' for i in s) and s[-1] != '_'
print(CodelandUsernameValidation(input()))
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Is there a quick method to find the shortest repeated substring and how many times it occurs? If there is non you only need to return the actual string ( last case ).
>>> repeated('CTCTCTCTCTCTCTCTCTCTCTCT')
('CT', 12)
>>> repeated('GATCGATCGATCGATC')
('GATC', 4)
>>> repeated('GATCGATCGATCGATCG')
('GATCGATCGATCGATCG', 1)
Because some people think it's 'homework' I can show my efforts:
def repeated(sequentie):
string = ''
for i in sequentie:
if i not in string:
string += i
items = sequentie.count(string)
if items * len(string) == len(sequentie):
return (string, items)
else:
return (sequentie, 1)
Your method unfortunately won't work, since it assumes that the repeating substring will have unique characters. This may not be the case:
abaabaabaabaabaaba
You were somewhat on the right track, though. The shortest way that I can think of is to just try and check over and over if some prefix indeed makes up the entire string:
def find_shorted_substring(s):
for i in range(1, len(s) + 1):
substring = s[:i]
repeats = len(s) // len(substring)
if substring * repeats == s:
return (substring, repeats)
It's not very efficient, but it works. There are better ways of doing it.
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I had a file called 'words.txt' which contains things such as. #+=%£%= and i need to go through and count each symbol without allowing any duplicates and then print my answer. for example it should look like this when printed:
# : 1
+ : 1
= : 2
% : 2
£ : 1
I know I need to use a for loop in this and that I need to use a set so It doesnt't allow any duplicates but how would I do it? Thanl you
A set isn't so useful here as it doesn't have a place to store the counts
from collections import Counter
with open("words.txt") as fin:
c = Counter(fin.read())
for item in c.items():
print("{} : {}".format(*item))
Use python dictionary:
symbols = {}
for c in string:
if c not in symbols:
symbols[c] = 0
symbols[c] += 1
Look up dictionary.
Don't want to post more as it would only spoil the exercise.
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Here is my function's objective:
Define a procedure, find_last, that takes as input two strings, a
search string and a target string,and returns the last position in the
search string where the target string appears, or -1 if there are no
occurrences.
Here is a link to the question.
And here's my function so far:
def find_last(target, search):
find = target.find(search, 0)
if find != -1:
targets = target.find(search, find)
while targets != -1:
find = find + 1
targets = target.find(search, find)
return find - 1
else:
return -1
The code returns the answer I'm looking for with return find - 1, but I know there is a better way to go about doing this.
Any help would be greatly appreciated!
You basically implemented target.rfind(search).
Why dont you use rfind ?
>>> d = "ball foo ball"
>>> f = "ball"
>>> d.rfind(f)
12
So your method becomes a 1 liner :)
def find_last(target, search):
return target.rfind(search)
You can return target.rfind(search) and in the calling method check for -1 and handle appropriately.
Can't resist to quote this awesome piece from XKCD