How would I count consecutive characters in Python to see the number of times each unique digit repeats before the next unique digit?
At first, I thought I could do something like:
word = '1000'
counter = 0
print range(len(word))
for i in range(len(word) - 1):
while word[i] == word[i + 1]:
counter += 1
print counter * "0"
else:
counter = 1
print counter * "1"
So that in this manner I could see the number of times each unique digit repeats. But this, of course, falls out of range when i
reaches the last value.
In the example above, I would want Python to tell me that 1 repeats 1, and that 0 repeats 3 times. The code above fails, however, because of my while
statement.
How could I do this with just built-in functions?
Ooh nobody's posted itertools.groupby
yet!
s = "111000222334455555"
from itertools import groupby
groups = groupby(s)
result = [(label, sum(1 for _ in group)) for label, group in groups]
After which, result
looks like:
[("1": 3), ("0", 3), ("2", 3), ("3", 2), ("4", 2), ("5", 5)]
And you could format with something like:
", ".join("{}x{}".format(label, count) for label, count in result)
# "1x3, 0x3, 2x3, 3x2, 4x2, 5x5"
Someone in the comments is concerned that you want a total count of numbers so "11100111" -> {"1":6, "0":2}
. In that case you want to use a collections.Counter
:
from collections import Counter
s = "11100111"
result = Counter(s)
# {"1":6, "0":2}
As many have pointed out, your method fails because you're looping through range(len(s))
but addressing s[i+1]
. This leads to an off-by-one error when i
is pointing at the last index of s
, so i+1
raises an IndexError
. One way to fix this would be to loop through range(len(s)-1)
, but it's more pythonic to generate something to iterate over.
For string that's not absolutely huge, zip(s, s[1:])
isn't aa performance issue, so you could do:
counts = []
count = 1
for a, b in zip(s, s[1:]):
if a==b:
count += 1
else:
counts.append((a, count))
count = 1
The only problem being that you'll have to special-case the last character if it's unique. That can be fixed with itertools.zip_longest
import itertools
counts = []
count = 1
for a, b in itertools.zip_longest(s, s[1:], fillvalue=None):
if a==b:
count += 1
else:
counts.append((a, count))
count = 1
If you do have a truly huge string and can't stand to hold two of them in memory at a time, you can use the itertools
recipe pairwise
.
def pairwise(iterable):
"""iterates pairwise without holding an extra copy of iterable in memory"""
a, b = itertools.tee(iterable)
next(b, None)
return itertools.zip_longest(a, b, fillvalue=None)
counts = []
count = 1
for a, b in pairwise(s):
...
A solution "that way", with only basic statements:
word="100011010" #word = "1"
count=1
length=""
if len(word)>1:
for i in range(1,len(word)):
if word[i-1]==word[i]:
count+=1
else :
length += word[i-1]+" repeats "+str(count)+", "
count=1
length += ("and "+word[i]+" repeats "+str(count))
else:
i=0
length += ("and "+word[i]+" repeats "+str(count))
print (length)
Output :
'1 repeats 1, 0 repeats 3, 1 repeats 2, 0 repeats 1, 1 repeats 1, and 0 repeats 1'
#'1 repeats 1'
#!/usr/bin/python3 -B
charseq = 'abbcccdddd'
distros = { c:1 for c in charseq }
for c in range(len(charseq)-1):
if charseq[c] == charseq[c+1]:
distros[charseq[c]] += 1
print(distros)
I'll provide a brief explanation for the interesting lines.
distros = { c:1 for c in charseq }
The line above is a dictionary comprehension, and it basically iterates over the characters in charseq
and creates a key/value pair for a dictionary where the key is the character and the value is the number of times it has been encountered so far.
Then comes the loop:
for c in range(len(charseq)-1):
We go from 0
to length - 1
to avoid going out of bounds with the c+1
indexing in the loop's body.
if charseq[c] == charseq[c+1]:
distros[charseq[c]] += 1
At this point, every match we encounter we know is consecutive, so we simply add 1 to the character key. For example, if we take a snapshot of one iteration, the code could look like this (using direct values instead of variables, for illustrative purposes):
# replacing vars for their values
if charseq[1] == charseq[1+1]:
distros[charseq[1]] += 1
# this is a snapshot of a single comparison here and what happens later
if 'b' == 'b':
distros['b'] += 1
You can see the program output below with the correct counts:
➜ /tmp ./counter.py
{'b': 2, 'a': 1, 'c': 3, 'd': 4}
You only need to change len(word)
to len(word) - 1
. That said, you could also use the fact that False
's value is 0 and True
's value is 1 with sum
:
sum(word[i] == word[i+1] for i in range(len(word)-1))
This produces the sum of (False, True, True, False)
where False
is 0 and True
is 1 - which is what you're after.
If you want this to be safe you need to guard empty words (index -1 access):
sum(word[i] == word[i+1] for i in range(max(0, len(word)-1)))
And this can be improved with zip
:
sum(c1 == c2 for c1, c2 in zip(word[:-1], word[1:]))
If we want to count consecutive characters without looping , we can make use of pandas
:
In [1]: import pandas as pd
In [2]: sample = 'abbcccddddaaaaffaaa'
In [3]: d = pd.Series(list(sample))
In [4]: [(cat[1], grp.shape[0]) for cat, grp in d.groupby([d.ne(d.shift()).cumsum(), d])]
Out[4]: [('a', 1), ('b', 2), ('c', 3), ('d', 4), ('a', 4), ('f', 2), ('a', 3)]
The key is to find the first elements that are different from their previous values and then make proper groupings in pandas
:
In [5]: sample = 'abba'
In [6]: d = pd.Series(list(sample))
In [7]: d.ne(d.shift())
Out[7]:
0 True
1 True
2 False
3 True
dtype: bool
In [8]: d.ne(d.shift()).cumsum()
Out[8]:
0 1
1 2
2 2
3 3
dtype: int32
This is my simple code for finding maximum number of consecutive 1's in binaray string in python 3:
count= 0
maxcount = 0
for i in str(bin(13)):
if i == '1':
count +=1
elif count > maxcount:
maxcount = count;
count = 0
else:
count = 0
if count > maxcount: maxcount = count
maxcount
Unique method:- In case if you are just looking for counting consecutive 1's Using Bit Magic: The idea is based on the concept that if we AND a bit sequence with a shifted version of itself, we're effectively removing the trailing 1 from every sequence of consecutive 1s.
11101111 (x)
& 11011110 (x << 1)
----------
11001110 (x & (x << 1))
^ ^
| |
trailing 1 removed So the operation x = (x & (x << 1)) reduces length of every sequence of 1s by one in binary representation of x. If we keep doing this operation in a loop, we end up with x = 0. The number of iterations required to reach 0 is actually length of the longest consecutive sequence of 1s.
There is no need to count or groupby. Just note the indices where a change occurs and subtract consecutive indicies.
w = "111000222334455555"
iw = [0] + [i+1 for i in range(len(w)-1) if w[i] != w[i+1]] + [len(w)]
dw = [w[i] for i in range(len(w)-1) if w[i] != w[i+1]] + [w[-1]]
cw = [ iw[j] - iw[j-1] for j in range(1, len(iw) ) ]
print(dw) # digits
['1', '0', '2', '3', '4']
print(cw) # counts
[3, 3, 3, 2, 2, 5]
w = 'XXYXYYYXYXXzzzzzYYY'
iw = [0] + [i+1 for i in range(len(w)-1) if w[i] != w[i+1]] + [len(w)]
dw = [w[i] for i in range(len(w)-1) if w[i] != w[i+1]] + [w[-1]]
cw = [ iw[j] - iw[j-1] for j in range(1, len(iw) ) ]
print(dw) # characters
print(cw) # digits
['X', 'Y', 'X', 'Y', 'X', 'Y', 'X', 'z', 'Y']
[2, 1, 1, 3, 1, 1, 2, 5, 3]
A one liner that returns the amount of consecutive characters with no imports:
def f(x):s=x+" ";t=[x[1] for x in zip(s[0:],s[1:],s[2:]) if (x[1]==x[0])or(x[1]==x[2])];return {h: t.count(h) for h in set(t)}
That returns the amount of times any repeated character in a list is in a consecutive run of characters.
alternatively, this accomplishes the same thing, albeit much slower:
def A(m):t=[thing for x,thing in enumerate(m) if thing in [(m[x+1] if x+1<len(m) else None),(m[x-1] if x-1>0 else None)]];return {h: t.count(h) for h in set(t)}
In terms of performance, I ran them with
site = 'https://web.njit.edu/~cm395/theBeeMovieScript/'
s = urllib.request.urlopen(site).read(100_000)
s = str(copy.deepcopy(s))
print(timeit.timeit('A(s)',globals=locals(),number=100))
print(timeit.timeit('f(s)',globals=locals(),number=100))
which resulted in:
12.528256356999918
5.351301653001428
This method can definitely be improved, but without using any external libraries, this was the best I could come up with.
In python
your_string = "wwwwweaaaawwbbbbn"
current = ''
count = 0
for index, loop in enumerate(your_string):
current = loop
count = count + 1
if index == len(your_string)-1:
print(f"{count}{current}", end ='')
break
if your_string[index+1] != current:
print(f"{count}{current}",end ='')
count = 0
continue
This will output
5w1e4a2w4b1n
#I wrote the code using simple loops and if statement
s='feeekksssh' #len(s) =11
count=1 #f:0, e:3, j:2, s:3 h:1
l=[]
for i in range(1,len(s)): #range(1,10)
if s[i-1]==s[i]:
count = count+1
else:
l.append(count)
count=1
if i == len(s)-1: #To check the last character sequence we need loop reverse order
reverse_count=1
for i in range(-1,-(len(s)),-1): #Lopping only for last character
if s[i] == s[i-1]:
reverse_count = reverse_count+1
else:
l.append(reverse_count)
break
print(l)
Today I had an interview and was asked the same question. I was struggling with the original solution in mind:
s = 'abbcccda'
old = ''
cnt = 0
res = ''
for c in s:
cnt += 1
if old != c:
res += f'{old}{cnt}'
old = c
cnt = 0 # default 0 or 1 neither work
print(res)
# 1a1b2c3d1
Sadly this solution always got unexpected edge cases result(is there anyone to fix the code? maybe i need post another question), and finally timeout the interview.
After the interview I calmed down and soon got a stable solution I think(though I like the groupby best).
s = 'abbcccda'
olds = []
for c in s:
if olds and c in olds[-1]:
olds[-1].append(c)
else:
olds.append([c])
print(olds)
res = ''.join([f'{lst[0]}{len(lst)}' for lst in olds])
print(res)
# [['a'], ['b', 'b'], ['c', 'c', 'c'], ['d'], ['a']]
# a1b2c3d1a1
Here is my simple solution:
def count_chars(s):
size = len(s)
count = 1
op = ''
for i in range(1, size):
if s[i] == s[i-1]:
count += 1
else:
op += "{}{}".format(count, s[i-1])
count = 1
if size:
op += "{}{}".format(count, s[size-1])
return op
data_input = 'aabaaaabbaaaaax'
start = 0
end = 0
temp_dict = dict()
while start < len(data_input):
if data_input[start] == data_input[end]:
end = end + 1
if end == len(data_input):
value = data_input[start:end]
temp_dict[value] = len(value)
break
if data_input[start] != data_input[end]:
value = data_input[start:end]
temp_dict[value] = len(value)
start = end
print(temp_dict)
PROBLEM: we need to count consecutive characters and return characters with their count.
def countWithString(input_string:str)-> str:
count = 1
output = ''
for i in range(1,len(input_string)):
if input_string[i]==input_string[i-1]:
count +=1
else:
output += f"{count}{input_string[i-1]}"
count = 1
# Used to add last string count (at last else condition will not run and data will not be inserted to ouput string)
output += f"{count}{input_string[-1]}"
return output
countWithString(input)
input:'aaabbbaabbcc'
output:'3a3b2a2b2c'
Time Complexity: O(n)
Space Complexity: O(1)
temp_str = "aaaajjbbbeeeeewwjjj"
def consecutive_charcounter(input_str):
counter = 0
temp_list = []
for i in range(len(input_str)):
if i==0:
counter+=1
elif input_str[i]== input_str[i-1]:
counter+=1
if i == len(input_str)-1:
temp_list.extend([input_str[i - 1], str(counter)])
else:
temp_list.extend([input_str[i-1],str(counter)])
counter = 1
print("".join(temp_list))
consecutive_charcounter(temp_str)
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