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How To Fix String Index Out of Range in Python

I'm currently learning python. There is an exercise in the book which outlines code for decoding. I followed along with this code however I keep getting back an error in the middle of the program.
Here is the code that is causing the problem:
def decode(string_for_decoding):
result = ""
for i in range(0, len(string_for_decoding)):
if string_for_decoding[i+1].isdigit():
result += string_for_decoding[i] * int(string_for_decoding[i+1])
elif string_for_decoding[i].isalpha():
result += string_for_decoding[i]
return result
string_for_decoding = input("Enter a string to decode: ")
print(decode(string_for_decoding))

Check if the index from range is larger than the number of chars in the string. It might look like this:
def decode(string_for_decoding: str):
result = ""
for i in range(0, len(string_for_decoding)):
if len(string_for_decoding) > i + 1 and string_for_decoding[i + 1].isdigit():
result += string_for_decoding[i] * int(string_for_decoding[i + 1])
elif string_for_decoding.isalpha():
result += string_for_decoding[i]
return result
print(decode(input("Enter a string to decode: ")))

You are going from 0 to len(string) and inside for loop you are trying to access index: i+1
THis is the root cause! Either:
iterate till one length less e.g. till len(string) - 1
Or use indices
inside appropriately
Moreover it is highly discouraged to use range(len(x)). You can simply iterate over any collective datatype as for i in x: If you want indices too have a look at enumerate(). And you can forget about index errors all together.

You are trying to get index i+1 which cannot be equal or greater than len(string).

Largest substring of non-repeating letters of a string

From the beginning I want to point out, that I am using Python Language. In this question I initially have a string. For example 'abcagfhtgba'. I need to find the length of the largest substring of non-repeating letters. In the case provided above it is 'agfht' (5), because at position [4] the 'a' repeats, so we start the count from the begining.
My idea for this question is to create a dictionary, which stores letters as keys, and numbers of their appearances as values. Whenever any key has corresponding value 2, we append the length of the dictionary to the list named result and completely substitute it with an empty list. For some tests this approach holds, for some not. I will provide the code that I used with brief comments of explanation.
Here I store the input in form of a list
this = list(map(str, input()))
def function(list):
dict = {}
count = 0
result = [1]
Here I start the loop and for every element if it is not in the keys I create a key
with value 1. If the element is in the dictionary I substitute the dict with the empty one. I don't forget to store the first repeating element in a new dictionary and do this. Another important point is at the end to append the count after the loop. Because the tail of the string (if it has the largest non-repeating sequence of letters) should be considered.
for i in range(len(list)):
if list[i] not in dict:
dict[list[i]] = 1
count += 1
elif list[i] in dict:
dict = {}
dict[list[i]] = 1
result.append(count)
count = 1
result.append(count)
print(result)
return max(result)
Here i make my function to choose choose the largest between the string and the inverse of it, to deal with the cases 'adabc', where the largest substring is at the end.
if len(this) != 0:
print(max(function(this), function(this[::-1])))
else:
print('')
I need help of people to tell me where in the approach to the problem I am wrong and edit my code.

Hopefully you might find this a little easier. The idea is to keep track of the seen substrings up to a given point in a set for a faster lookup, and if the current value is contained, build the set anew and append the substring seen up to that point. As you mention you have to check whether the last values have been added or not, hence the final if:
s = 'abcagfhtgba'
seen = set()
out = []
current_out = []
for i in s:
if i not in seen:
current_out += i
seen.update(i)
else:
seen = set(i)
out.append(''.join(current_out))
current_out = [i]
if current_out:
out.append(''.join(current_out))
max(out, key=len)
# 'agfht'
So some key differences:
Iterate over the string itself, not a range
Use sets rather than counts and dictionaries

Remember the last duplicate you have seen, maintain a map of letter to index. If you have already seen then this is duplicate, so we need to reset the index. But index can be this new one or just after the last duplicate character is seen.
s = 'abcagfhtgba'
seen = dict()
longest = ""
start = 0
last_duplicate = 0
for i, c in enumerate(s):
if seen.has_key(c):
if len(longest) < (i - start + 1):
longest = s[start:i]
new_start = seen.get(c) + 1
if last_duplicate > new_start:
start = i
else:
start = new_start
last_duplicate = i
seen[c] = I
if len(longest) < (len(s) - start + 1):
longest = s[start:]
print longest

I want to create a program that finds the longest substring in alphabetical order using python using basics

my code for finding longest substring in alphabetical order using python
what I mean by longest substring in alphabetical order?
if the input was"asdefvbrrfqrstuvwxffvd" the output wil be "qrstuvwx"
#we well use the strings as arrays so don't be confused
s='abcbcd'
#give spaces which will be our deadlines
h=s+' (many spaces) '
#creat outputs
g=''
g2=''
#list of alphapets
abc='abcdefghijklmnopqrstuvwxyz'
#create the location of x"the character the we examine" and its limit
limit=len(s)
#start from 1 becouse we substract one in the rest of the code
x=1
while (x<limit):
#y is the curser that we will move the abc array on it
y=0
#putting our break condition first
if ((h[x]==' ') or (h[x-1]==' ')):
break
for y in range(0,26):
#for the second character x=1
if ((h[x]==abc[y]) and (h[x-1]==abc[y-1]) and (x==1)):
g=g+abc[y-1]+abc[y]
x+=1
#for the third to the last character x>1
if ((h[x]==abc[y]) and (h[x-1]==abc[y-1]) and (x!=1)):
g=g+abc[y]
x+=1
if (h[x]==' '):
break
print ("Longest substring in alphabetical order is:" +g )
it doesn't end,as if it's in infinite loop
what should I do?
I am a beginner so I want some with for loops not functions from libraries
Thanks in advance

To avoid infinite loop add x += 1 in the very end of your while-loop. As a result your code works but works wrong in general case.
The reason why it works wrong is that you use only one variable g to store the result. Use at least two variables to compare previous found substring and new found substring or use list to remember all substrings and then choose the longest one.

s = 'abcbcdiawuhdawpdijsamksndaadhlmwmdnaowdihasoooandalw'
longest = ''
current = ''
for idx, item in enumerate(s):
if idx == 0 or item > s[idx-1]:
current = current + item
if idx > 0 and item <= s[idx-1]:
current = ''
if len(current)>len(longest):
longest = current
print(longest)
Output:
dhlmw
For your understanding 'b'>'a' is True, 'a'>'b' is False etc
Edit:
For longest consecutive substring:
s = 'asdefvbrrfqrstuvwxffvd'
abc = 'abcdefghijklmnopqrstuvwxyz'
longest = ''
current = ''
for idx, item in enumerate(s):
if idx == 0 or abc.index(item) - abc.index(s[idx-1]) == 1:
current = current + item
else:
current = item
if len(current)>len(longest):
longest = current
print(longest)
Output:
qrstuvwx

def sub_strings(string):
substring = ''
string +='\n'
i = 0
string_dict ={}
while i < len(string)-1:
substring += string[i]
if ord(substring[-1])+1 != ord(string[i+1]):
string_dict[substring] = len(substring)
substring = ''
i+=1
return string_dict
s='abcbcd'
sub_strings(s)
{'abc': 3, 'bcd': 3}
To find the longest you can domax(sub_strings(s))
So here which one do you want to be taken as the longest??. Now that is a problem you would need to solve

You can iterate through the string and keep comparing to the last character and append to the potentially longest string if the current character is greater than the last character by one ordinal number:
def longest_substring(s):
last = None
current = longest = ''
for c in s:
if not last or ord(c) - ord(last) == 1:
current += c
else:
if len(current) > len(longest):
longest = current
current = c
last = c
if len(current) > len(longest):
longest = current
return longest
so that:
print(longest_substring('asdefvbrrfqrstuvwxffvd'))
would output:
qrstuvwx

Python: Specific List Output Inconsistency

I believe the output I'm getting is a product of sloppy code somewhere, but I can't seem to get it. What I want my function to do is make a new list of the elements in a list it's being passed, in those elements' Pig-Latin translation. I've got the function working, but it makes all of the new list elements the Pig-Latin translation of only the first element in the list being passed to the function, even though I'm using an index to iterate through the elements of the passed list.
def pigLatin(targetlist):
newlist = ()
listnewlist = list(newlist)
index = 0 # gets incremented
firstletter = targetlist[index][0]
word = targetlist[index][1:]
print('Words made into Pig Latin: ')
while index < len(targetlist):
listnewlist.append(word + firstletter + 'ay')
index += 1
print(listnewlist)
which, after user input is "how now brown cow", displays:
['owhay', 'owhay', 'owhay', 'owhay']

word and firstletter are both computed outside of the loop and keep the value of first word of the targetlist.
IMHO, you should write :
def pigLatin(targetlist):
newlist = ()
listnewlist = list(newlist)
index = 0 # gets incremented
print('Words made into Pig Latin: ')
while index < len(targetlist):
firstletter = targetlist[index][0]
word = targetlist[index][1:]
listnewlist.append(word + firstletter + 'ay')
index += 1
print(listnewlist)

Reference next item in list: python

I'm making a variation of Codecademy's pyglatin.py to make a translator that accepts and translates multiple words. However, I'm having trouble translating more than one word. I've been able to transfer the raw input into a list and translate the first, but I do not know how to reference the next item in the list. Any help would be greatly appreciated.
def piglatin1():
pig = 'ay'
original = raw_input('Enter a phrase:').split(' ')
L = list(original)
print L
i = iter(L)
item = i.next()
for item in L:
if len(item) > 0 and item.isalpha():
word = item.lower()
first = word
if first == "a" or first == "e" or first == "i" or first == "o" or first =="u":
new_word = word + pig
print new_word
else:
new_word = word[1:] + word[0:1] + pig
# first word translated
L = []
M = L[:]
L.append(new_word)
print L # secondary list created.
again = raw_input('Translate again? Y/N')
print again
if len(again) > 0 and again.isalpha():
second_word = again.lower()
if second_word == "y":
return piglatin()
else:
print "Okay Dokey!"
else:
print 'Letters only please!'
return piglatin1()

I was working on this problem recently as well and came up with the following solution (rather than use range, use enumerate to get the index).
for index, item in enumerate(L):
next = index + 1
if next < len(L):
print index, item, next
This example shows how to access the current index, the current item, and then the next item in the list (if it exists in the bounds of the list).

Here are a few things to note that might help.
The lines i = iter(L) and item = i.next() are unnecessary. They have no effect in this method because you are redefining item immediately afterwards in the line for item in L. Go ahead and comment out those two lines to see if it makes any changes in your output.
The looping construct for item in L will go once over every item in the list. Whatever code you write within this loop will be executed once for each item in the list. The variable item is your handle to the list element of an iteration.
If, during any iteration, you really do want to access the "next" element in the list as well, then consider using a looping construct such as for i in range(0,len(L)). Then L[i] will be the current item and L[i+1] will you give the subsequent item.

There are some slight issues with the code but I think there is one main reason why it will not repeat.
In order to process the entire string the
again = raw_input('Translate again? Y/N')
and it's succeeding lines should be brought outside the for statement.
Also you appear to be setting L to an empty string inside the loop:
L = []
The following is a modified version of your code which will loop through the entire sentence and then ask for another one.
def piglatin():
pig = 'ay'
while True:
L = raw_input('Enter a phrase:').split(' ')
M = []
for item in L:
if len(item) > 0 and item.isalpha():
word = item.lower()
first = word
if first == "a" or first == "e" or first == "i" or first == "o" or first =="u":
new_word = word + pig
print new_word
else:
new_word = word[1:] + word[0:1] + pig
M.append(new_word)
else:
print 'Letters only please!'
print M # secondary list created.
again = raw_input('Translate again? Y/N')
print again
if len(again) > 0 and again.isalpha():
second_word = again.lower()
if second_word == "n":
print "Okay Dokey!"
break
Changes made:
You don't need to cast the return of the split to a list. The split
return type is a list.
It isn't necessary to make an iterator, the for loop will do this for you.
I removed the function as the return type. I'm assuming you were attempting some form of recursion but it isn't strictly necessary.
Hope this helps.

Step by step:
If you set variable original in this way:
original = raw_input('Enter a phrase:').split()
it will be already a list, so need to additional assignment.
What is the purpose of these lines?
i = iter(L)
item = i.next()
In a loop, you assign variable to the word, when it is actually only the first letter of the word, so it’ll be better like this: first = word[0]
Then if you want to check if first is a vowel, you can just do:
if first in 'aeuoiy'
Answer to your actual question: do not assign L to an empty list!
If you want to repeat the action of a function, you can just call it again, no need to rewrite the code.