Mixing
encode repeated letters in a string with number
3
A string 'abc' must become 'a1b1c1'.
String 'aaabcca' - 'a3b1c2a1'
I wrote python function, but it fails to add the last letter and 'abc' is only 'a1b1'.
string = "aaabbcc"
coded = ''
if len(string) == 0:
print('')
else:
count = 1 #start with the first char, not zero!
prev = string[0]
for i in range(1,len(string)):
current = string[i]
if current == prev:
count +=1
else:
coded += prev
coded += str(count)
count = 1
prev = current
print("coded string: " + coded)
print(coded)
python
asked Nov 20 '18 at 20:42
ERJANERJAN
9,290123468
add a comment |
3
A string 'abc' must become 'a1b1c1'.
String 'aaabcca' - 'a3b1c2a1'
I wrote python function, but it fails to add the last letter and 'abc' is only 'a1b1'.
string = "aaabbcc"
coded = ''
if len(string) == 0:
print('')
else:
count = 1 #start with the first char, not zero!
prev = string[0]
for i in range(1,len(string)):
current = string[i]
if current == prev:
count +=1
else:
coded += prev
coded += str(count)
count = 1
prev = current
print("coded string: " + coded)
print(coded)
python
asked Nov 20 '18 at 20:42
ERJANERJAN
9,290123468
1
hint: itertoools.groupby
– Chris_Rands
Nov 20 '18 at 20:44
Counter?
– Jaba
Nov 20 '18 at 20:44
input letters may be unsorted, right?wwaaadvc
– RomanPerekhrest
Nov 20 '18 at 20:45
@RomanPerekhrest, yes
– ERJAN
Nov 20 '18 at 20:46
add a comment |
3
3
3
A string 'abc' must become 'a1b1c1'.
String 'aaabcca' - 'a3b1c2a1'
I wrote python function, but it fails to add the last letter and 'abc' is only 'a1b1'.
string = "aaabbcc"
coded = ''
if len(string) == 0:
print('')
else:
count = 1 #start with the first char, not zero!
prev = string[0]
for i in range(1,len(string)):
current = string[i]
if current == prev:
count +=1
else:
coded += prev
coded += str(count)
count = 1
prev = current
print("coded string: " + coded)
print(coded)
python
asked Nov 20 '18 at 20:42
ERJANERJAN
9,290123468
A string 'abc' must become 'a1b1c1'.
String 'aaabcca' - 'a3b1c2a1'
I wrote python function, but it fails to add the last letter and 'abc' is only 'a1b1'.
string = "aaabbcc"
coded = ''
if len(string) == 0:
print('')
else:
count = 1 #start with the first char, not zero!
prev = string[0]
for i in range(1,len(string)):
current = string[i]
if current == prev:
count +=1
else:
coded += prev
coded += str(count)
count = 1
prev = current
print("coded string: " + coded)
print(coded)
python
python
asked Nov 20 '18 at 20:42
ERJANERJAN
9,290123468
asked Nov 20 '18 at 20:42
ERJANERJAN
9,290123468
asked Nov 20 '18 at 20:42
ERJANERJAN
9,290123468
asked Nov 20 '18 at 20:42
ERJANERJAN
9,290123468
asked Nov 20 '18 at 20:42
ERJANERJAN
9,290123468
9,290123468
1
hint: itertoools.groupby
– Chris_Rands
Nov 20 '18 at 20:44
Counter?
– Jaba
Nov 20 '18 at 20:44
input letters may be unsorted, right?wwaaadvc
– RomanPerekhrest
Nov 20 '18 at 20:45
@RomanPerekhrest, yes
– ERJAN
Nov 20 '18 at 20:46
add a comment |
1
hint: itertoools.groupby
– Chris_Rands
Nov 20 '18 at 20:44
Counter?
– Jaba
Nov 20 '18 at 20:44
input letters may be unsorted, right?wwaaadvc
– RomanPerekhrest
Nov 20 '18 at 20:45
@RomanPerekhrest, yes
– ERJAN
Nov 20 '18 at 20:46
1
1
hint: itertoools.groupby
– Chris_Rands
Nov 20 '18 at 20:44
hint: itertoools.groupby
– Chris_Rands
Nov 20 '18 at 20:44
Counter?– Jaba
Nov 20 '18 at 20:44
Counter?– Jaba
Nov 20 '18 at 20:44
input letters may be unsorted, right?
wwaaadvc– RomanPerekhrest
Nov 20 '18 at 20:45
input letters may be unsorted, right?
wwaaadvc– RomanPerekhrest
Nov 20 '18 at 20:45
@RomanPerekhrest, yes
– ERJAN
Nov 20 '18 at 20:46
@RomanPerekhrest, yes
– ERJAN
Nov 20 '18 at 20:46
add a comment |
4 Answers
4
active
oldest
votes
1
You forget to explicitly add the very last iteration.
string = "aaabb"
coded = ''
if len(string) == 0:
print('')
else:
count = 1 #start with the first char, not zero!
prev = string[0]
for i in range(1,len(string)):
current = string[i]
if current == prev:
count +=1
else:
coded += prev
coded += str(count)
count = 1
prev = current
coded += prev # these two
coded += str(count) # lines
print(coded)
I would prefer a less complicated loop, though:
string = "aaabbcc"
coded = ''
while string:
i = 0
while i < len(string) and string[0] == string[i]:
i += 1
coded += string[0]+str(i)
string = string[i:]
print(coded)
answered Nov 20 '18 at 20:52
usr2564301usr2564301
17.7k73370
how can you write "while len(string)" ?
– ERJAN
Nov 20 '18 at 20:57
1
@ERJAN: uhm. Using my keyboard? Do you see something wrong with it? I did test my code.
– usr2564301
Nov 20 '18 at 20:58
python does not allow this? as condition? len(string) = True? can it be a condtion?
– ERJAN
Nov 20 '18 at 21:00
1
never imagined such a while cond...
– ERJAN
Nov 20 '18 at 21:00
1
while"tests the expression and, if it is true, executes the first suite".len(string)is a number ... ah – you don't needlen(string)😊 That's my C shining through I'm afraid.
– usr2564301
Nov 20 '18 at 21:01
add a comment |
3
Use itertools.groupby.
>>> from itertools import groupby
>>> s = 'aaabcca'
>>> ''.join('{}{}'.format(c, sum(1 for _ in g)) for c, g in groupby(s))
'a3b1c2a1'
Details on what groupby produces:
>>> groups = groupby(s)
>>> [(char, list(group)) for char, group in groups]
[('a', ['a', 'a', 'a']), ('b', ['b']), ('c', ['c', 'c']), ('a', ['a'])]
answered Nov 20 '18 at 20:46
timgebtimgeb
50.7k116393
add a comment |
3
Some regex magic:
import re
s = 'aaawbbbccddddd'
counts = re.sub(r'(.)1*', lambda m: m.group(1) + str(len(m.group())), s)
print(counts)
The output:
a3w1b3c2d5
Details:
regex pattern:
(.)- capturing a character.(any char) into the 1st captured group
1*- matches zero or more consecutive1which is a reference to the 1st captured group value (matching a potentially sequence of the same character)
replacement:
m.group(1)- contains the 1st matched group value
str(len(m.group()))- get length of the entire character sequence matched
answered Nov 20 '18 at 21:01
RomanPerekhrestRomanPerekhrest
55.6k32253
1
"Daily vote limit reached" – damn. I'll be back in a day. Wicked.
– usr2564301
Nov 20 '18 at 21:03
@usr2564301, no problem )
– RomanPerekhrest
Nov 20 '18 at 21:05
@RomanPerekhrest, can u elaborate on this expression in sub()?
– ERJAN
Nov 20 '18 at 21:17
@ERJAN, yes, see my update
– RomanPerekhrest
Nov 20 '18 at 21:30
add a comment |
1
If you wonder why your code didn't work or you don't want to use any external libraries here's working version of your code
string = "aaabbcc"
coded = ''
if len(string) == 0:
print('')
else:
count = 0
prev = string[0]
for i in range(1,len(string)):
current = string[i]
count +=1
if current != prev:
coded += prev
coded += str(count)
count = 0
prev = current
coded += current
coded += str(count+1)
print(coded) # -> a3b2c2
answered Nov 20 '18 at 20:53
Filip MłynarskiFilip Młynarski
1,7591413
Fixed, thanks for letting me know.
– Filip Młynarski
Nov 20 '18 at 21:09
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
You forget to explicitly add the very last iteration.
string = "aaabb"
coded = ''
if len(string) == 0:
print('')
else:
count = 1 #start with the first char, not zero!
prev = string[0]
for i in range(1,len(string)):
current = string[i]
if current == prev:
count +=1
else:
coded += prev
coded += str(count)
count = 1
prev = current
coded += prev # these two
coded += str(count) # lines
print(coded)
I would prefer a less complicated loop, though:
string = "aaabbcc"
coded = ''
while string:
i = 0
while i < len(string) and string[0] == string[i]:
i += 1
coded += string[0]+str(i)
string = string[i:]
print(coded)
answered Nov 20 '18 at 20:52
usr2564301usr2564301
17.7k73370
how can you write "while len(string)" ?
– ERJAN
Nov 20 '18 at 20:57
1
@ERJAN: uhm. Using my keyboard? Do you see something wrong with it? I did test my code.
– usr2564301
Nov 20 '18 at 20:58
python does not allow this? as condition? len(string) = True? can it be a condtion?
– ERJAN
Nov 20 '18 at 21:00
1
never imagined such a while cond...
– ERJAN
Nov 20 '18 at 21:00
1
while"tests the expression and, if it is true, executes the first suite".len(string)is a number ... ah – you don't needlen(string)😊 That's my C shining through I'm afraid.
– usr2564301
Nov 20 '18 at 21:01
add a comment |
1
You forget to explicitly add the very last iteration.
string = "aaabb"
coded = ''
if len(string) == 0:
print('')
else:
count = 1 #start with the first char, not zero!
prev = string[0]
for i in range(1,len(string)):
current = string[i]
if current == prev:
count +=1
else:
coded += prev
coded += str(count)
count = 1
prev = current
coded += prev # these two
coded += str(count) # lines
print(coded)
I would prefer a less complicated loop, though:
string = "aaabbcc"
coded = ''
while string:
i = 0
while i < len(string) and string[0] == string[i]:
i += 1
coded += string[0]+str(i)
string = string[i:]
print(coded)
answered Nov 20 '18 at 20:52
usr2564301usr2564301
17.7k73370
how can you write "while len(string)" ?
– ERJAN
Nov 20 '18 at 20:57
1
@ERJAN: uhm. Using my keyboard? Do you see something wrong with it? I did test my code.
– usr2564301
Nov 20 '18 at 20:58
python does not allow this? as condition? len(string) = True? can it be a condtion?
– ERJAN
Nov 20 '18 at 21:00
1
never imagined such a while cond...
– ERJAN
Nov 20 '18 at 21:00
1
while"tests the expression and, if it is true, executes the first suite".len(string)is a number ... ah – you don't needlen(string)😊 That's my C shining through I'm afraid.
– usr2564301
Nov 20 '18 at 21:01
add a comment |
1
1
1
You forget to explicitly add the very last iteration.
string = "aaabb"
coded = ''
if len(string) == 0:
print('')
else:
count = 1 #start with the first char, not zero!
prev = string[0]
for i in range(1,len(string)):
current = string[i]
if current == prev:
count +=1
else:
coded += prev
coded += str(count)
count = 1
prev = current
coded += prev # these two
coded += str(count) # lines
print(coded)
I would prefer a less complicated loop, though:
string = "aaabbcc"
coded = ''
while string:
i = 0
while i < len(string) and string[0] == string[i]:
i += 1
coded += string[0]+str(i)
string = string[i:]
print(coded)
answered Nov 20 '18 at 20:52
usr2564301usr2564301
17.7k73370
You forget to explicitly add the very last iteration.
string = "aaabb"
coded = ''
if len(string) == 0:
print('')
else:
count = 1 #start with the first char, not zero!
prev = string[0]
for i in range(1,len(string)):
current = string[i]
if current == prev:
count +=1
else:
coded += prev
coded += str(count)
count = 1
prev = current
coded += prev # these two
coded += str(count) # lines
print(coded)
I would prefer a less complicated loop, though:
string = "aaabbcc"
coded = ''
while string:
i = 0
while i < len(string) and string[0] == string[i]:
i += 1
coded += string[0]+str(i)
string = string[i:]
print(coded)
answered Nov 20 '18 at 20:52
usr2564301usr2564301
17.7k73370
edited Nov 20 '18 at 21:01
answered Nov 20 '18 at 20:52
usr2564301usr2564301
17.7k73370
answered Nov 20 '18 at 20:52
usr2564301usr2564301
17.7k73370
answered Nov 20 '18 at 20:52
usr2564301usr2564301
17.7k73370
17.7k73370
how can you write "while len(string)" ?
– ERJAN
Nov 20 '18 at 20:57
1
@ERJAN: uhm. Using my keyboard? Do you see something wrong with it? I did test my code.
– usr2564301
Nov 20 '18 at 20:58
python does not allow this? as condition? len(string) = True? can it be a condtion?
– ERJAN
Nov 20 '18 at 21:00
1
never imagined such a while cond...
– ERJAN
Nov 20 '18 at 21:00
1
while"tests the expression and, if it is true, executes the first suite".len(string)is a number ... ah – you don't needlen(string)😊 That's my C shining through I'm afraid.
– usr2564301
Nov 20 '18 at 21:01
add a comment |
how can you write "while len(string)" ?
– ERJAN
Nov 20 '18 at 20:57
1
@ERJAN: uhm. Using my keyboard? Do you see something wrong with it? I did test my code.
– usr2564301
Nov 20 '18 at 20:58
python does not allow this? as condition? len(string) = True? can it be a condtion?
– ERJAN
Nov 20 '18 at 21:00
1
never imagined such a while cond...
– ERJAN
Nov 20 '18 at 21:00
1
while"tests the expression and, if it is true, executes the first suite".len(string)is a number ... ah – you don't needlen(string)😊 That's my C shining through I'm afraid.
– usr2564301
Nov 20 '18 at 21:01
how can you write "while len(string)" ?
– ERJAN
Nov 20 '18 at 20:57
how can you write "while len(string)" ?
– ERJAN
Nov 20 '18 at 20:57
1
1
@ERJAN: uhm. Using my keyboard? Do you see something wrong with it? I did test my code.
– usr2564301
Nov 20 '18 at 20:58
@ERJAN: uhm. Using my keyboard? Do you see something wrong with it? I did test my code.
– usr2564301
Nov 20 '18 at 20:58
python does not allow this? as condition? len(string) = True? can it be a condtion?
– ERJAN
Nov 20 '18 at 21:00
python does not allow this? as condition? len(string) = True? can it be a condtion?
– ERJAN
Nov 20 '18 at 21:00
1
1
never imagined such a while cond...
– ERJAN
Nov 20 '18 at 21:00
never imagined such a while cond...
– ERJAN
Nov 20 '18 at 21:00
1
1
while "tests the expression and, if it is true, executes the first suite". len(string) is a number ... ah – you don't need len(string) 😊 That's my C shining through I'm afraid.– usr2564301
Nov 20 '18 at 21:01
while "tests the expression and, if it is true, executes the first suite". len(string) is a number ... ah – you don't need len(string) 😊 That's my C shining through I'm afraid.– usr2564301
Nov 20 '18 at 21:01
add a comment |
3
Use itertools.groupby.
>>> from itertools import groupby
>>> s = 'aaabcca'
>>> ''.join('{}{}'.format(c, sum(1 for _ in g)) for c, g in groupby(s))
'a3b1c2a1'
Details on what groupby produces:
>>> groups = groupby(s)
>>> [(char, list(group)) for char, group in groups]
[('a', ['a', 'a', 'a']), ('b', ['b']), ('c', ['c', 'c']), ('a', ['a'])]
answered Nov 20 '18 at 20:46
timgebtimgeb
50.7k116393
add a comment |
3
Use itertools.groupby.
>>> from itertools import groupby
>>> s = 'aaabcca'
>>> ''.join('{}{}'.format(c, sum(1 for _ in g)) for c, g in groupby(s))
'a3b1c2a1'
Details on what groupby produces:
>>> groups = groupby(s)
>>> [(char, list(group)) for char, group in groups]
[('a', ['a', 'a', 'a']), ('b', ['b']), ('c', ['c', 'c']), ('a', ['a'])]
answered Nov 20 '18 at 20:46
timgebtimgeb
50.7k116393
add a comment |
3
3
3
Use itertools.groupby.
>>> from itertools import groupby
>>> s = 'aaabcca'
>>> ''.join('{}{}'.format(c, sum(1 for _ in g)) for c, g in groupby(s))
'a3b1c2a1'
Details on what groupby produces:
>>> groups = groupby(s)
>>> [(char, list(group)) for char, group in groups]
[('a', ['a', 'a', 'a']), ('b', ['b']), ('c', ['c', 'c']), ('a', ['a'])]
answered Nov 20 '18 at 20:46
timgebtimgeb
50.7k116393
Use itertools.groupby.
>>> from itertools import groupby
>>> s = 'aaabcca'
>>> ''.join('{}{}'.format(c, sum(1 for _ in g)) for c, g in groupby(s))
'a3b1c2a1'
Details on what groupby produces:
>>> groups = groupby(s)
>>> [(char, list(group)) for char, group in groups]
[('a', ['a', 'a', 'a']), ('b', ['b']), ('c', ['c', 'c']), ('a', ['a'])]
answered Nov 20 '18 at 20:46
timgebtimgeb
50.7k116393
answered Nov 20 '18 at 20:46
timgebtimgeb
50.7k116393
answered Nov 20 '18 at 20:46
timgebtimgeb
50.7k116393
answered Nov 20 '18 at 20:46
timgebtimgeb
50.7k116393
50.7k116393
add a comment |
add a comment |
3
Some regex magic:
import re
s = 'aaawbbbccddddd'
counts = re.sub(r'(.)1*', lambda m: m.group(1) + str(len(m.group())), s)
print(counts)
The output:
a3w1b3c2d5
Details:
regex pattern:
(.)- capturing a character.(any char) into the 1st captured group
1*- matches zero or more consecutive1which is a reference to the 1st captured group value (matching a potentially sequence of the same character)
replacement:
m.group(1)- contains the 1st matched group value
str(len(m.group()))- get length of the entire character sequence matched
answered Nov 20 '18 at 21:01
RomanPerekhrestRomanPerekhrest
55.6k32253
1
"Daily vote limit reached" – damn. I'll be back in a day. Wicked.
– usr2564301
Nov 20 '18 at 21:03
@usr2564301, no problem )
– RomanPerekhrest
Nov 20 '18 at 21:05
@RomanPerekhrest, can u elaborate on this expression in sub()?
– ERJAN
Nov 20 '18 at 21:17
@ERJAN, yes, see my update
– RomanPerekhrest
Nov 20 '18 at 21:30
add a comment |
3
Some regex magic:
import re
s = 'aaawbbbccddddd'
counts = re.sub(r'(.)1*', lambda m: m.group(1) + str(len(m.group())), s)
print(counts)
The output:
a3w1b3c2d5
Details:
regex pattern:
(.)- capturing a character.(any char) into the 1st captured group
1*- matches zero or more consecutive1which is a reference to the 1st captured group value (matching a potentially sequence of the same character)
replacement:
m.group(1)- contains the 1st matched group value
str(len(m.group()))- get length of the entire character sequence matched
answered Nov 20 '18 at 21:01
RomanPerekhrestRomanPerekhrest
55.6k32253
1
"Daily vote limit reached" – damn. I'll be back in a day. Wicked.
– usr2564301
Nov 20 '18 at 21:03
@usr2564301, no problem )
– RomanPerekhrest
Nov 20 '18 at 21:05
@RomanPerekhrest, can u elaborate on this expression in sub()?
– ERJAN
Nov 20 '18 at 21:17
@ERJAN, yes, see my update
– RomanPerekhrest
Nov 20 '18 at 21:30
add a comment |
3
3
3
Some regex magic:
import re
s = 'aaawbbbccddddd'
counts = re.sub(r'(.)1*', lambda m: m.group(1) + str(len(m.group())), s)
print(counts)
The output:
a3w1b3c2d5
Details:
regex pattern:
(.)- capturing a character.(any char) into the 1st captured group
1*- matches zero or more consecutive1which is a reference to the 1st captured group value (matching a potentially sequence of the same character)
replacement:
m.group(1)- contains the 1st matched group value
str(len(m.group()))- get length of the entire character sequence matched
answered Nov 20 '18 at 21:01
RomanPerekhrestRomanPerekhrest
55.6k32253
Some regex magic:
import re
s = 'aaawbbbccddddd'
counts = re.sub(r'(.)1*', lambda m: m.group(1) + str(len(m.group())), s)
print(counts)
The output:
a3w1b3c2d5
Details:
regex pattern:
(.)- capturing a character.(any char) into the 1st captured group
1*- matches zero or more consecutive1which is a reference to the 1st captured group value (matching a potentially sequence of the same character)
replacement:
m.group(1)- contains the 1st matched group value
str(len(m.group()))- get length of the entire character sequence matched
answered Nov 20 '18 at 21:01
RomanPerekhrestRomanPerekhrest
55.6k32253
edited Nov 21 '18 at 8:40
answered Nov 20 '18 at 21:01
RomanPerekhrestRomanPerekhrest
55.6k32253
answered Nov 20 '18 at 21:01
RomanPerekhrestRomanPerekhrest
55.6k32253
answered Nov 20 '18 at 21:01
RomanPerekhrestRomanPerekhrest
55.6k32253
55.6k32253
1
"Daily vote limit reached" – damn. I'll be back in a day. Wicked.
– usr2564301
Nov 20 '18 at 21:03
@usr2564301, no problem )
– RomanPerekhrest
Nov 20 '18 at 21:05
@RomanPerekhrest, can u elaborate on this expression in sub()?
– ERJAN
Nov 20 '18 at 21:17
@ERJAN, yes, see my update
– RomanPerekhrest
Nov 20 '18 at 21:30
add a comment |
1
"Daily vote limit reached" – damn. I'll be back in a day. Wicked.
– usr2564301
Nov 20 '18 at 21:03
@usr2564301, no problem )
– RomanPerekhrest
Nov 20 '18 at 21:05
@RomanPerekhrest, can u elaborate on this expression in sub()?
– ERJAN
Nov 20 '18 at 21:17
@ERJAN, yes, see my update
– RomanPerekhrest
Nov 20 '18 at 21:30
1
1
"Daily vote limit reached" – damn. I'll be back in a day. Wicked.
– usr2564301
Nov 20 '18 at 21:03
"Daily vote limit reached" – damn. I'll be back in a day. Wicked.
– usr2564301
Nov 20 '18 at 21:03
@usr2564301, no problem )
– RomanPerekhrest
Nov 20 '18 at 21:05
@usr2564301, no problem )
– RomanPerekhrest
Nov 20 '18 at 21:05
@RomanPerekhrest, can u elaborate on this expression in sub()?
– ERJAN
Nov 20 '18 at 21:17
@RomanPerekhrest, can u elaborate on this expression in sub()?
– ERJAN
Nov 20 '18 at 21:17
@ERJAN, yes, see my update
– RomanPerekhrest
Nov 20 '18 at 21:30
@ERJAN, yes, see my update
– RomanPerekhrest
Nov 20 '18 at 21:30
add a comment |
1
If you wonder why your code didn't work or you don't want to use any external libraries here's working version of your code
string = "aaabbcc"
coded = ''
if len(string) == 0:
print('')
else:
count = 0
prev = string[0]
for i in range(1,len(string)):
current = string[i]
count +=1
if current != prev:
coded += prev
coded += str(count)
count = 0
prev = current
coded += current
coded += str(count+1)
print(coded) # -> a3b2c2
answered Nov 20 '18 at 20:53
Filip MłynarskiFilip Młynarski
1,7591413
Fixed, thanks for letting me know.
– Filip Młynarski
Nov 20 '18 at 21:09
add a comment |
1
If you wonder why your code didn't work or you don't want to use any external libraries here's working version of your code
string = "aaabbcc"
coded = ''
if len(string) == 0:
print('')
else:
count = 0
prev = string[0]
for i in range(1,len(string)):
current = string[i]
count +=1
if current != prev:
coded += prev
coded += str(count)
count = 0
prev = current
coded += current
coded += str(count+1)
print(coded) # -> a3b2c2
answered Nov 20 '18 at 20:53
Filip MłynarskiFilip Młynarski
1,7591413
Fixed, thanks for letting me know.
– Filip Młynarski
Nov 20 '18 at 21:09
add a comment |
1
1
1
If you wonder why your code didn't work or you don't want to use any external libraries here's working version of your code
string = "aaabbcc"
coded = ''
if len(string) == 0:
print('')
else:
count = 0
prev = string[0]
for i in range(1,len(string)):
current = string[i]
count +=1
if current != prev:
coded += prev
coded += str(count)
count = 0
prev = current
coded += current
coded += str(count+1)
print(coded) # -> a3b2c2
answered Nov 20 '18 at 20:53
Filip MłynarskiFilip Młynarski
1,7591413
If you wonder why your code didn't work or you don't want to use any external libraries here's working version of your code
string = "aaabbcc"
coded = ''
if len(string) == 0:
print('')
else:
count = 0
prev = string[0]
for i in range(1,len(string)):
current = string[i]
count +=1
if current != prev:
coded += prev
coded += str(count)
count = 0
prev = current
coded += current
coded += str(count+1)
print(coded) # -> a3b2c2
answered Nov 20 '18 at 20:53
Filip MłynarskiFilip Młynarski
1,7591413
edited Nov 20 '18 at 21:08
answered Nov 20 '18 at 20:53
Filip MłynarskiFilip Młynarski
1,7591413
answered Nov 20 '18 at 20:53
Filip MłynarskiFilip Młynarski
1,7591413
answered Nov 20 '18 at 20:53
Filip MłynarskiFilip Młynarski
1,7591413
1,7591413
Fixed, thanks for letting me know.
– Filip Młynarski
Nov 20 '18 at 21:09
add a comment |
Fixed, thanks for letting me know.
– Filip Młynarski
Nov 20 '18 at 21:09
Fixed, thanks for letting me know.
– Filip Młynarski
Nov 20 '18 at 21:09
Fixed, thanks for letting me know.
– Filip Młynarski
Nov 20 '18 at 21:09
add a comment |
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hint: itertoools.groupby
– Chris_Rands
Nov 20 '18 at 20:44
Counter?– Jaba
Nov 20 '18 at 20:44
input letters may be unsorted, right?
wwaaadvc– RomanPerekhrest
Nov 20 '18 at 20:45
@RomanPerekhrest, yes
– ERJAN
Nov 20 '18 at 20:46