Not able to append elements in python list












1















I am writing a program, to find the character that occurs maximum number of odd times, in a given string using python. However i am not able to append characters to a list, if two or more characters occur maximum number of odd times.



input used : AAAbbccc



Error i am getting :



Traceback (most recent call last):
File "./prog.py", line 18, in
AttributeError: 'str' object has no attribute 'append'



inputString = input()



dict = {}



for i in inputString:

  if i in dict:

    dict[i] += 1

  else:

    dict[i] = 1





print(dict)

max = -1

lst = 

for i in dict:

  if(dict[i]%2!=0 and max<=dict[i]):

    if(max == dict[i]):

      lst.append(i)

    else:

      max = dict[i]

      lst = i



print(lst)





python string python-3.x list counter







share|improve this question




















  • 2





    In line 21 you have an assignment 'lst' to 'i' which is not an array, then when when line 18 occurs lst is not an array anymore so it doesn't have an append method

    – szogoon
    Nov 14 '18 at 11:57













  • Thanks but that's not the issue that's working completely fine i'm getting error for line " lst.append(i) " where i am trying to append character into the list but i'm getting the above mentioned error

    – Anand
    Nov 14 '18 at 12:01
















1















I am writing a program, to find the character that occurs maximum number of odd times, in a given string using python. However i am not able to append characters to a list, if two or more characters occur maximum number of odd times.



input used : AAAbbccc



Error i am getting :



Traceback (most recent call last):
File "./prog.py", line 18, in
AttributeError: 'str' object has no attribute 'append'



inputString = input()



dict = {}



for i in inputString:

  if i in dict:

    dict[i] += 1

  else:

    dict[i] = 1





print(dict)

max = -1

lst = 

for i in dict:

  if(dict[i]%2!=0 and max<=dict[i]):

    if(max == dict[i]):

      lst.append(i)

    else:

      max = dict[i]

      lst = i



print(lst)





python string python-3.x list counter







share|improve this question




















  • 2





    In line 21 you have an assignment 'lst' to 'i' which is not an array, then when when line 18 occurs lst is not an array anymore so it doesn't have an append method

    – szogoon
    Nov 14 '18 at 11:57













  • Thanks but that's not the issue that's working completely fine i'm getting error for line " lst.append(i) " where i am trying to append character into the list but i'm getting the above mentioned error

    – Anand
    Nov 14 '18 at 12:01














1












1








1








I am writing a program, to find the character that occurs maximum number of odd times, in a given string using python. However i am not able to append characters to a list, if two or more characters occur maximum number of odd times.



input used : AAAbbccc



Error i am getting :



Traceback (most recent call last):
File "./prog.py", line 18, in
AttributeError: 'str' object has no attribute 'append'



inputString = input()



dict = {}



for i in inputString:

  if i in dict:

    dict[i] += 1

  else:

    dict[i] = 1





print(dict)

max = -1

lst = 

for i in dict:

  if(dict[i]%2!=0 and max<=dict[i]):

    if(max == dict[i]):

      lst.append(i)

    else:

      max = dict[i]

      lst = i



print(lst)





python string python-3.x list counter







share|improve this question
















I am writing a program, to find the character that occurs maximum number of odd times, in a given string using python. However i am not able to append characters to a list, if two or more characters occur maximum number of odd times.



input used : AAAbbccc



Error i am getting :



Traceback (most recent call last):
File "./prog.py", line 18, in
AttributeError: 'str' object has no attribute 'append'



inputString = input()



dict = {}



for i in inputString:

  if i in dict:

    dict[i] += 1

  else:

    dict[i] = 1





print(dict)

max = -1

lst = 

for i in dict:

  if(dict[i]%2!=0 and max<=dict[i]):

    if(max == dict[i]):

      lst.append(i)

    else:

      max = dict[i]

      lst = i



print(lst)





python string python-3.x list counter




python string python-3.x list counter






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 14 '18 at 12:39









ctrl-alt-delor

4,24632444




4,24632444










asked Nov 14 '18 at 11:52









AnandAnand

226




226








  • 2





    In line 21 you have an assignment 'lst' to 'i' which is not an array, then when when line 18 occurs lst is not an array anymore so it doesn't have an append method

    – szogoon
    Nov 14 '18 at 11:57













  • Thanks but that's not the issue that's working completely fine i'm getting error for line " lst.append(i) " where i am trying to append character into the list but i'm getting the above mentioned error

    – Anand
    Nov 14 '18 at 12:01














  • 2





    In line 21 you have an assignment 'lst' to 'i' which is not an array, then when when line 18 occurs lst is not an array anymore so it doesn't have an append method

    – szogoon
    Nov 14 '18 at 11:57













  • Thanks but that's not the issue that's working completely fine i'm getting error for line " lst.append(i) " where i am trying to append character into the list but i'm getting the above mentioned error

    – Anand
    Nov 14 '18 at 12:01








2




2





In line 21 you have an assignment 'lst' to 'i' which is not an array, then when when line 18 occurs lst is not an array anymore so it doesn't have an append method

– szogoon
Nov 14 '18 at 11:57







In line 21 you have an assignment 'lst' to 'i' which is not an array, then when when line 18 occurs lst is not an array anymore so it doesn't have an append method

– szogoon
Nov 14 '18 at 11:57















Thanks but that's not the issue that's working completely fine i'm getting error for line " lst.append(i) " where i am trying to append character into the list but i'm getting the above mentioned error

– Anand
Nov 14 '18 at 12:01





Thanks but that's not the issue that's working completely fine i'm getting error for line " lst.append(i) " where i am trying to append character into the list but i'm getting the above mentioned error

– Anand
Nov 14 '18 at 12:01












1 Answer
1






active

oldest

votes


















1














There are some issues with your code:




  1. Don't name variables after built-ins (even as an example). Use d or dict_ instead of dict. Ditto max.

  2. Your max (fixed at -1) will always be <= dict[i], since counts are always >= 1.

  3. You define lst as a list, then assign a string to it.


Much simpler, use collections.Counter, calculate the maximum value, then use max with a custom function:



from collections import Counter



inputString = input()

c = Counter(inputString)

print(c)



maxval = max(c.values())



def max_logic(x):

    cond1 = x[1] % 2

    cond2 = x[1] - maxval

    return cond1, cond2



key, val = max(c.items(), key=max_logic)


Example run:



print(key, val)



thisisateststring

Counter({'t': 4, 's': 4, 'i': 3, 'h': 1, 'a': 1, 'e': 1, 'r': 1, 'n': 1, 'g': 1})

i 3


The solution assumes a valid odd count does exist in your string. If it doesn't and you need to apply special treatment, you'll need to add additional logic. I leave that as an exercise.






share|improve this answer





















  • 1





    Thanks a lot...

    – Anand
    Nov 14 '18 at 12:16











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1














There are some issues with your code:




  1. Don't name variables after built-ins (even as an example). Use d or dict_ instead of dict. Ditto max.

  2. Your max (fixed at -1) will always be <= dict[i], since counts are always >= 1.

  3. You define lst as a list, then assign a string to it.


Much simpler, use collections.Counter, calculate the maximum value, then use max with a custom function:



from collections import Counter



inputString = input()

c = Counter(inputString)

print(c)



maxval = max(c.values())



def max_logic(x):

    cond1 = x[1] % 2

    cond2 = x[1] - maxval

    return cond1, cond2



key, val = max(c.items(), key=max_logic)


Example run:



print(key, val)



thisisateststring

Counter({'t': 4, 's': 4, 'i': 3, 'h': 1, 'a': 1, 'e': 1, 'r': 1, 'n': 1, 'g': 1})

i 3


The solution assumes a valid odd count does exist in your string. If it doesn't and you need to apply special treatment, you'll need to add additional logic. I leave that as an exercise.






share|improve this answer





















  • 1





    Thanks a lot...

    – Anand
    Nov 14 '18 at 12:16
















1














There are some issues with your code:




  1. Don't name variables after built-ins (even as an example). Use d or dict_ instead of dict. Ditto max.

  2. Your max (fixed at -1) will always be <= dict[i], since counts are always >= 1.

  3. You define lst as a list, then assign a string to it.


Much simpler, use collections.Counter, calculate the maximum value, then use max with a custom function:



from collections import Counter



inputString = input()

c = Counter(inputString)

print(c)



maxval = max(c.values())



def max_logic(x):

    cond1 = x[1] % 2

    cond2 = x[1] - maxval

    return cond1, cond2



key, val = max(c.items(), key=max_logic)


Example run:



print(key, val)



thisisateststring

Counter({'t': 4, 's': 4, 'i': 3, 'h': 1, 'a': 1, 'e': 1, 'r': 1, 'n': 1, 'g': 1})

i 3


The solution assumes a valid odd count does exist in your string. If it doesn't and you need to apply special treatment, you'll need to add additional logic. I leave that as an exercise.






share|improve this answer





















  • 1





    Thanks a lot...

    – Anand
    Nov 14 '18 at 12:16














1












1








1







There are some issues with your code:




  1. Don't name variables after built-ins (even as an example). Use d or dict_ instead of dict. Ditto max.

  2. Your max (fixed at -1) will always be <= dict[i], since counts are always >= 1.

  3. You define lst as a list, then assign a string to it.


Much simpler, use collections.Counter, calculate the maximum value, then use max with a custom function:



from collections import Counter



inputString = input()

c = Counter(inputString)

print(c)



maxval = max(c.values())



def max_logic(x):

    cond1 = x[1] % 2

    cond2 = x[1] - maxval

    return cond1, cond2



key, val = max(c.items(), key=max_logic)


Example run:



print(key, val)



thisisateststring

Counter({'t': 4, 's': 4, 'i': 3, 'h': 1, 'a': 1, 'e': 1, 'r': 1, 'n': 1, 'g': 1})

i 3


The solution assumes a valid odd count does exist in your string. If it doesn't and you need to apply special treatment, you'll need to add additional logic. I leave that as an exercise.






share|improve this answer















There are some issues with your code:




  1. Don't name variables after built-ins (even as an example). Use d or dict_ instead of dict. Ditto max.

  2. Your max (fixed at -1) will always be <= dict[i], since counts are always >= 1.

  3. You define lst as a list, then assign a string to it.


Much simpler, use collections.Counter, calculate the maximum value, then use max with a custom function:



from collections import Counter



inputString = input()

c = Counter(inputString)

print(c)



maxval = max(c.values())



def max_logic(x):

    cond1 = x[1] % 2

    cond2 = x[1] - maxval

    return cond1, cond2



key, val = max(c.items(), key=max_logic)


Example run:



print(key, val)



thisisateststring

Counter({'t': 4, 's': 4, 'i': 3, 'h': 1, 'a': 1, 'e': 1, 'r': 1, 'n': 1, 'g': 1})

i 3


The solution assumes a valid odd count does exist in your string. If it doesn't and you need to apply special treatment, you'll need to add additional logic. I leave that as an exercise.







share|improve this answer














share|improve this answer



share|improve this answer








edited Nov 14 '18 at 12:17

























answered Nov 14 '18 at 12:11









jppjpp

100k2162111




100k2162111








  • 1





    Thanks a lot...

    – Anand
    Nov 14 '18 at 12:16














  • 1





    Thanks a lot...

    – Anand
    Nov 14 '18 at 12:16








1




1





Thanks a lot...

– Anand
Nov 14 '18 at 12:16





Thanks a lot...

– Anand
Nov 14 '18 at 12:16




















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