Given a string S, how to print a string containing number of times the character is repeated?
for example: input: aaabbbbccaa
output: a3b4c2a2
my approach:
s = input()
len_string = ''
cur_char = s[0]
cur_counter = 0
for i in range(len(s)):
if s[i] == cur_char:
cur_counter += 1
if s[i] != cur_char or i == len(s) - 1:
len_string += cur_char + str(cur_counter)
cur_char = s[i]
cur_counter = 1
print(len_string)
Because you have shared your code, here is one concise way using groupby
:
from itertools import groupby
s = 'aaabbbbccaa'
print(''.join([k + str(len(list(g))) for k, g in groupby(s)]))
# a3b4c2a2
I would use collections.Counter
https://docs.python.org/2/library/collections.html#counter-objects
Init signature: collections.Counter(*args, **kwds)
Docstring:
Dict subclass for counting hashable items. Sometimes called a bag
or multiset. Elements are stored as dictionary keys and their counts
are stored as dictionary values.
>>> c = Counter('abcdeabcdabcaba') # count elements from a string
>>> c.most_common(3) # three most common elements
[('a', 5), ('b', 4), ('c', 3)]
>>> sorted(c) # list all unique elements
['a', 'b', 'c', 'd', 'e']
>>> ''.join(sorted(c.elements())) # list elements with repetitions
'aaaaabbbbcccdde'
>>> sum(c.values()) # total of all counts
15
>>> c['a'] # count of letter 'a'
5
>>> for elem in 'shazam': # update counts from an iterable
... c[elem] += 1 # by adding 1 to each element's count
>>> c['a'] # now there are seven 'a'
7
>>> del c['b'] # remove all 'b'
>>> c['b'] # now there are zero 'b'
0
>>> d = Counter('simsalabim') # make another counter
>>> c.update(d) # add in the second counter
>>> c['a'] # now there are nine 'a'
9
>>> c.clear() # empty the counter
>>> c
Counter()
Note: If a count is set to zero or reduced to zero, it will remain
in the counter until the entry is deleted or the counter is cleared:
>>> c = Counter('aaabbc')
>>> c['b'] -= 2 # reduce the count of 'b' by two
>>> c.most_common() # 'b' is still in, but its count is zero
[('a', 3), ('c', 1), ('b', 0)]
I think there is a way to do this much more easy:
x='aaabbbbccaa'
noreplist = list(dict.fromkeys(x))
countstring=''
for i in noreplist:
z=z+i+str(x.count(i))
print(countstring)
First, you have x that is your string. Then you make a list with every char from that string, but without repeating any char. And last, just counts how many times is that char repeated on the original string, and concatenate in a 'count string'.
#y is a list that contains every character in the string
#z is a list parallel to y but it contains the number of times each element in list y #has
x=input("Input string: ")
y=[]
z=[]
acc=0
for i in range(len(x)):
if(x[i] not in y):
y.append(x[i])
for i in range(len(y)):
for j in range(len(x)):
if(y[i]==x[j]):
acc=acc+1
z.append(acc)
acc=0
for k in range(len(y)):
print(str(y[k])+str(z[k]))
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