Here in the program how can you find the second repetitive character in the string. for ex:abcdaabdefaggcbd" Output : d (because 'd' occurred 3 times where 'a' occurred 4 times) how can I get the output, please help me.
Given below is my code:
s="abcdaabdefaggcbd"
d={}
for i in s:
d[i] = d.get(i,0)+1
print(d,"ddddd")
max2 = 0
for k,v in d.items():
if(v>max2 and v<max(d.values())):
max2=v
if max2 in d.values():
print k,"kkk"
The magnificent Python Counter and its most_common() method are very handy here.
import collections
my_string = "abcdaabdefaggcbd"
result = collections.Counter(my_string).most_common()
print(result[1])
Output
('b', 3)
In case you need to capture all the second values (if you have more than one entry) you can use the following:
import collections
my_string = "abcdaabdefaggcbd"
result = collections.Counter(my_string).most_common()
second_value = result[1][1]
seconds = []
for item in result:
if item[1] == second_value:
seconds.append(item)
print(seconds)
Output
[('b', 3), ('d', 3)]
I also wanted to add an example of solving the problem using a methodology more similar to the one that you showed in your question:
my_string="abcdaabdefaggcbd"
result={}
for character in my_string:
if character in result:
result[character] = result.get(character) + 1
else:
result[character] = 1
sorted_data = sorted([(value,key) for (key,value) in result.items()])
second_value = sorted_data[-2][0]
result = []
for item in sorted_data:
if item[0] == second_value:
result.append(item)
print(result)
Output
[(3, 'b'), (3, 'd')]
Ps
Please forgive me if I took the freedom to change variable names but I think that in this way my answer will be more readable for a broader audience.
b and d can be considered as the second most repetitive. Would it not be better to compare the values to find the second highest value and then check the keys associated with that value ? This way, both b& dwill be displayed.
Counter.most_common() :facepalm:Sort the dict's items on their values (descending) and get the second item:
>>> from collections import Counter
>>> c = Counter("abcdaabdefaggcbd")
>>> vals = sorted(c.items(), key=lambda item:item[1], reverse=True)
>>> vals
[('a', 4), ('b', 3), ('d', 3), ('c', 2), ('g', 2), ('e', 1), ('f', 1)]
>>> print(vals[1])
('b', 3)
>>>
EDIT:
or just use Counter.most_common():
>>> from collections import Counter
>>> c = Counter("abcdaabdefaggcbd")
>>> print(c.most_common()[1])
Both b and d are second most repetitive. I would think that both should be displayed. This is how I would do it:
Code:
s="abcdaabdefaggcbd"
d={}
for i in s:
ctr=s.count(i)
d[i]=ctr
fir = max(d.values())
sec = 0
for j in d.values():
if(j>sec and j<fir):
sec = j
for k,v in d.items():
if v == sec:
print(k,v)
Output:
b 3
d 3
in order to find the second most repetitive character in string you can very well use collections.Counter()
Here's an example:
import collections
s='abcdaabdefaggcbd'
count=collections.Counter(s)
print(count.most_common(2)[1])
Output: ('b', 3)
You can do a lot with Counter(). Here's a link for a further read: More about Counter()
I hope this answers your question. Cheers!
