I I've finally found how to make a RLE algorithm by watching a tutorial but This tutorial didn' t explain something in that code I didn't get why we write j = i instead of j = 0 (Knowing that I = 0) it's the same no ?
I didn't get why i = j + 1 either. Why i = j + 1 At the end of the function ? Why not simply i += 1 but if we want to repeat a loop in a loop then we do j + 1 ?
Did the first while loop is supposed to repeat the second while loop until the string is finished ?
And finally why encoded_message is repeated two times ? instead of one. We return encoded_message so that's it ? We can simply do print(encode(text)) instead of
"print('The encoded message is the output ',encoded_message)" (when we put encode(text) into encoded_message)
I know i'm asking a lot of questions but I just can't memorize the code without understanding it, it would be totally useless and unproductive
def encode(message):
encoded_message = ""
i = 0
while(i<len(message)):
count = 1
ch = message[i]
j = i # ???
while(j<len(message)-1): # GET IT -----------------------------------------------------------
if message[j] == message[j+1]: # if the previous and next characters are the same
count = count + 1 # we increase count variable
j += 1 # we increase j position
# GET IT ----------------------------------------------------------------------------
else:
break
encoded_message = encoded_message + str(count) + ch # "" + count converted to string + character (ch)
i = j + 1 # ???
return encoded_message
text = input('enter your charcter chain...')
encoded_message = encode(text)
print('The encoded message is the output ',encoded_message)
When I replaced j = i by j = 0 nothing is displayed in the terminal
see : no result
There is an outer loop and an inner loop. The outer loop with the variable i starts iterating over the message. The inner loop uses the variable j and starts at the current position of i.
That is: when i=0 then j=0. But when i=5 (for example) then j=5 also.
The inner loops task is to check whether 2 or more identical characters follow one another. If they do i is increased accordingly at the end of the inner loop. So that each letter of the message is only looked at once.
That is why j should not be set to a constant value. Setting it to j=0 would cause the inner loop to start at the beginning of the message at every iteration.
I added two simple print() statements to your code to clarify:
def encode(message):
encoded_message = ""
i = 0
while(i<len(message)):
print(f'outer loop: i={i}')
count = 1
ch = message[i]
j = i
while(j<len(message)-1):
print(f'\tinner loop: j={j}')
if message[j] == message[j+1]: # if the previous and next characters are the same
count = count + 1 # we increase count variable
j += 1 # we increase j position
else:
break
encoded_message = encoded_message + str(count) + ch # "" + count converted to string + character (ch)
i = j + 1
return encoded_message
text = 'Hello World'
encoded_message = encode(text)
print('The encoded message is the output ', encoded_message)
(Please note: I do not know the RLE algorithm but just looked at your code.)
The input I gave is like this
Input- abccdddeffg
and the output I want is character and it's occurrence number
Output- a1b1c2d3e1f2g1
my code
uni = []
string = 'abcccdd'
for i in range(0, len(string)):
for j in range(i+1, len(string)):
if (string[i] == string[j]):
uni.append(string[i])
for oc in uni:
cou= uni.count(oc)
print(oc,cou)
Thanks in advance
You can use the Counter from collections to get the count of every character in the list. Then use a forloop to generate your result and use set to make sure no character is repeated in the result.
from collections import Counter
string = "abccdddeffg"
counts = Counter(string)
sets = set()
result = []
for s in string:
if s not in sets:
result.append(f"{s}{counts[s]}")
sets.add(s)
result = ''.join(result)
print(result)
Firstly, your output contains an error, there should be 1 next to e as it was there in first occurring characters.
After clearing this, this is what you need:
import collections
s = "abccdddeffg" # your string
a = dict((letter,s.count(letter)) for letter in set(s))
a = collections.OrderedDict(sorted(a.items()))
answer = "" # to store the result
for i,j in zip(a.keys(),a.values()):
answer+= i + str(j)
print(answer)
answer will return:
'a1b1c2d3e1f2g1'
Here is a simpler approach:
def freq_string(string):
output, buffer = "", ""
for letter in string:
if buffer != letter:
buffer = letter
output += f"{letter}{string.count(letter)}"
return output
Note: this does assume that same characters are in succession rather than spread randomly around the string.
Assuming your input string is sorted and has at least one letter, you can do a simple loop to handle it:
if len(string) == 1:
# print out the only string and 1 as its occurences
print(string + '1')
else:
# initialize first string, its counter, and our result string
prev = string[0]
counter = 1
result = ''
# loop over each letter
for letter in string[1:]:
curr = letter
# if current letter is different from previous letter,
# concat the result and refresh our counter
# else just increase the counter
if curr != prev:
result = result + prev + str(counter)
counter = 1
else:
counter = counter + 1
prev = curr
# don't forget to handle the last case
result = result + prev + str(counter)
print(result)
in the simplest way:
inp = 'abccdddeffg'
l=[]
o=""
for i in inp:
if i in l:
pass
else:
l.append(i)
o+="{}{}".format(i,inp.count(i))
print(o)
output
'a1b1c2d3e1f2g1'
My goal is to write a function which change every even letter into upper letter and odd to lower (space also count as a one element).
This is my code
def to_weird_case(s):
for i in s:
if len(i) % 2 == 0:
s[i] = i.upper() + s(i+1)
else:
s[i] = i.lower() + s(i+2)
return i
I think it should be quite correct, but it gives me error.
line 7, in to_weird_case
s[i] = i.lower() + s(str(i)+2)
TypeError: must be str, not int
EDIT:
I have a sugesstion but I don't know how to make it. I try it for myself and back here.
This needs to definitly explicietly state that the zero indexing uppercase is for each word.
Do you know guys how to make it?
So we can analyze your code and just explain what you typed:
def to_weird_case(s):
for i in s: # s is your string, and i is the actual character
if len(i) % 2 == 0: # if your length of the character can be divided by 2. Hmm this is weird
s[i] = i.upper() + s(i+1) # s[i] change a character in the string but you should provide an index (i) so an integer and not a character. But this is not supported in Python.
else:
s[i] = i.lower() + s(i+2)
return i # This will exit after first iteraction, so to_weird_case("this") will return "t".
So what you need to is first create a output string and fill that. And when iteration over s, you want the index of the char and the char value itself.
def to_weird_case(s):
output = ""
for i, myChar in enumerate(s):
if i % 2 == 0:
output += myChar.upper()
else:
output += myChar.lower()
return output
my_sentence = "abcdef"
print(to_weird_case(my_sentence))
And when you want to ignore spaces, you need to keep track of actual characters (excluding spaces)
def to_weird_case(s):
output = ""
count = 0
for myChar in s:
if myChar.isspace():
output += myChar
else:
if count % 2 == 0:
output += myChar.upper()
else:
output += myChar.lower()
count += 1
return output
my_sentence = "abc def"
print(to_weird_case(my_sentence))
Test this yourself
def to_weird_case(s):
for i in s:
print (i)
After doing this you will find that i gives you characters.
if len(i) % 2 == 0:
This line is incorrect as you are trying to find the length of a single character. len(s) would be much better.
So the code will be like
def to_weird_case(s):
s2 = "" #We create another string as strings are immutable in python
for i in range(len(s)):
if i % 2 == 0:
s2 = s2 + s[i].upper()
else:
s2 = s2 + s[i].lower()
return s2
From #RvdK analysis, you'ld have seen where corrections are needed. In addition to what has been pointed out, I want you to note that s[i] will work fine only if i is an integer, but in your case where (by assumption) i is a string you'll encounter several TypeErrors. From my understanding of what you want to do, it should go this way:
def to_weird_case(s):
for i in s:
if s.index(i) % 2 == 0:
s[s.index(i)] = i.upper() + s[s.index(i)]
elif s.index(i) % 2 == 1:
s[s.index(i)] = i.lower() + s[s.index(i)]
return i # or possibly return s
It is possible to do in a single line using a list comprehension
def funny_case(s):
return "".join([c.upper() if idx%2==0 else c.lower() for idx,c in enumerate(s)])
If you want to treat each word separately then you can split it up in to a list of words and "funny case" each word individually, see below code
original = "hello world"
def funny_case(s):
return "".join([c.upper() if idx%2==0 else c.lower() for idx,c in enumerate(s) ])
def funny_case_by_word(s):
return " ".join((funny_case(word) for word in s.split()))
print(funny_case_by_word(original))
Corrected code is as follows
def case(s):
txt=''
for i in range(len(s)):
if i%2==0:
txt+=s[i].upper()
else:
txt+=s[i].lower()
return txt
String assignment gives error in Python therefore i recommend considering my approach
When looping over elements of s, you get the letter itself, not its index. You can use enumerate to get both index and letter.
def to_weird_case(s):
result = ''
for index, letter in enumerate(s):
if index % 2 == 0:
result += letter.upper()
else:
result += letter.lower()
return result
correct code:
def to_weird_case(s):
str2 = ""
s.split() # through splitting string is converted to list as it is easy to traverse through list
for i in range(0,len(s)):
n = s[i] # storing value in n
if(i % 2 == 0):
str2 = str2 + n.upper()
else:
str2 = str2 + n.lower()
return str2
str1 = "hello world"
r = to_weird_case(str1)
print(r)
I'm writing a random substitution cipher, and it works well when being used to write regular sentences. However, when I try to test 'abbcccddddeeeee', I will get something back like 'G H H I I I J J J J K K K K K'. How can I change my code so that either 'key' or 'vari' changes for each letter in the cycle yet stays the same so that no letter corresponds to more than one other letter?
import random
intext = raw_input("Enter a message to encrypt: ")
intext = intext.upper()
key = random.randint(0,27)
vari = random.randint(0,27)
def randomsub(intext):
outtext = []
intext = intext.upper()
for c in intext:
num = ord(c)
if num >= ord('A') and num <= ord('Z'):
num = num - ord('A')
num = num + (key + vari) % 26
num += ord('A')
outtext.append(chr(num))
return ' '.join(outtext)
print randomsub(intext)
Not sure why you have both key and vari, but the randomness should be inside your function, and at least one of the calls to randint should be inside the loop. I'd recommend making key a user-provided input, and using it to seed random. That way, with the same text and the same key, you get the same output, but duplicate letters in the text will vary in the result. Last, but not least, if you want the output to be letters you need to expand the parentheses where you add key and vari to include num, i.e., num = (num + vari) % 26 if you switch to using key as I suggested.
I have this code that I found on another topic, but it sorts the substring by contiguous characters and not by alphabetical order. How do I correct it for alphabetical order? It prints out lk, and I want to print ccl. Thanks
ps: I'm a beginner in python
s = 'cyqfjhcclkbxpbojgkar'
from itertools import count
def long_alphabet(input_string):
maxsubstr = input_string[0:0] # empty slice (to accept subclasses of str)
for start in range(len(input_string)): # O(n)
for end in count(start + len(maxsubstr) + 1): # O(m)
substr = input_string[start:end] # O(m)
if len(set(substr)) != (end - start): # found duplicates or EOS
break
if (ord(max(sorted(substr))) - ord(min(sorted(substr))) + 1) == len(substr):
maxsubstr = substr
return maxsubstr
bla = (long_alphabet(s))
print "Longest substring in alphabetical order is: %s" %bla
s = 'cyqfjhcclkbxpbojgkar'
r = ''
c = ''
for char in s:
if (c == ''):
c = char
elif (c[-1] <= char):
c += char
elif (c[-1] > char):
if (len(r) < len(c)):
r = c
c = char
else:
c = char
if (len(c) > len(r)):
r = c
print(r)
Try changing this:
if len(set(substr)) != (end - start): # found duplicates or EOS
break
if (ord(max(sorted(substr))) - ord(min(sorted(substr))) + 1) == len(substr):
to this:
if len(substr) != (end - start): # found duplicates or EOS
break
if sorted(substr) == list(substr):
That will display ccl for your example input string. The code is simpler because you're trying to solve a simpler problem :-)
You can improve your algorithm by noticing that the string can be broken into runs of ordered substrings of maximal length. Any ordered substring must be contained in one of these runs
This allows you to just iterate once through the string O(n)
def longest_substring(string):
curr, subs = '', ''
for char in string:
if not curr or char >= curr[-1]:
curr += char
else:
curr, subs = '', max(curr, subs, key=len)
return max(curr, subs, key=len)
s = 'cyqfjhcclkbxpbojgkar'
longest = ""
max =""
for i in range(len(s) -1):
if(s[i] <= s[i+1] ):
longest = longest + s[i]
if(i==len(s) -2):
longest = longest + s[i+1]
else:
longest = longest + s[i]
if(len(longest) > len(max)):
max = longest
longest = ""
if(len(s) == 1):
longest = s
if(len(longest) > len(max)):
print("Longest substring in alphabetical order is: " + longest)
else:
print("Longest substring in alphabetical order is: " + max)
In a recursive way, you can import count from itertools
Or define a same method:
def loops( I=0, S=1 ):
n = I
while True:
yield n
n += S
With this method, you can obtain the value of an endpoint, when you create any substring in your anallitic process.
Now looks the anallize method (based on spacegame issue and Mr. Tim Petters suggestion)
def anallize(inStr):
# empty slice (maxStr) to implement
# str native methods
# in the anallize search execution
maxStr = inStr[0:0]
# loop to read the input string (inStr)
for i in range(len(inStr)):
# loop to sort and compare each new substring
# the loop uses the loops method of past
# I = sum of:
# (i) current read index
# (len(maxStr)) current answer length
# and 1
for o in loops(i + len(maxStr) + 1):
# create a new substring (newStr)
# the substring is taked:
# from: index of read loop (i)
# to: index of sort and compare loop (o)
newStr = inStr[i:o]
if len(newStr) != (o - i):# detect and found duplicates
break
if sorted(newStr) == list(newStr):# compares if sorted string is equal to listed string
# if success, the substring of sort and compare is assigned as answer
maxStr = newStr
# return the string recovered as longest substring
return maxStr
Finally, for test or execution pourposes:
# for execution pourposes of the exercise:
s = "azcbobobegghakl"
print "Longest substring in alphabetical order is: " + anallize( s )
The great piece of this job started by: spacegame and attended by Mr. Tim Petters, is in the use of the native str methods and the reusability of the code.
The answer is:
Longest substring in alphabetical order is: ccl
In Python character comparison is easy compared to java script where the ASCII values have to be compared. According to python
a>b gives a Boolean False and b>a gives a Boolean True
Using this the longest sub string in alphabetical order can be found by using the following algorithm :
def comp(a,b):
if a<=b:
return True
else:
return False
s = raw_input("Enter the required sting: ")
final = []
nIndex = 0
temp = []
for i in range(nIndex, len(s)-1):
res = comp(s[i], s[i+1])
if res == True:
if temp == []:
#print i
temp.append(s[i])
temp.append(s[i+1])
else:
temp.append(s[i+1])
final.append(temp)
else:
if temp == []:
#print i
temp.append(s[i])
final.append(temp)
temp = []
lengths = []
for el in final:
lengths.append(len(el))
print lengths
print final
lngStr = ''.join(final[lengths.index(max(lengths))])
print "Longest substring in alphabetical order is: " + lngStr
Use list and max function to reduce the code drastically.
actual_string = 'azcbobobegghakl'
strlist = []
i = 0
while i < len(actual_string)-1:
substr = ''
while actial_string[i + 1] > actual_string[i] :
substr += actual_string[i]
i += 1
if i > len(actual_string)-2:
break
substr += actual-string[i]
i += 1
strlist.append(subst)
print(max(strlist, key=len))
Wow, some really impressing code snippets here...
I want to add my solution, as I think it's quite clean:
s = 'cyqfjhcclkbxpbojgkar'
res = ''
tmp = ''
for i in range(len(s)):
tmp += s[i]
if len(tmp) > len(res):
res = tmp
if i > len(s)-2:
break
if s[i] > s[i+1]:
tmp = ''
print("Longest substring in alphabetical order is: {}".format(res))
Without using a library, but using a function ord() which returns ascii value for a character.
Assumption: input will be in lowercase, and no special characters are used
s = 'azcbobobegghakl'
longest = ''
for i in range(len(s)):
temp_longest=s[i]
for j in range(i+1,len(s)):
if ord(s[i])<=ord(s[j]):
temp_longest+=s[j]
i+=1
else:
break
if len(temp_longest)>len(longest):
longest = temp_longest
print(longest)
Slightly different implementation, building up a list of all substrings in alphabetical order and returning the longest one:
def longest_substring(s):
in_orders = ['' for i in range(len(s))]
index = 0
for i in range(len(s)):
if (i == len(s) - 1 and s[i] >= s[i - 1]) or s[i] <= s[i + 1]:
in_orders[index] += s[i]
else:
in_orders[index] += s[i]
index += 1
return max(in_orders, key=len)
s = "azcbobobegghakl"
ls = ""
for i in range(0, len(s)-1):
b = ""
ss = ""
j = 2
while j < len(s):
ss = s[i:i+j]
b = sorted(ss)
str1 = ''.join(b)
j += 1
if str1 == ss:
ks = ss
else:
break
if len(ks) > len(ls):
ls = ks
print("The Longest substring in alphabetical order is "+ls)
This worked for me
s = 'cyqfjhcclkbxpbojgkar'
lstring = s[0]
slen = 1
for i in range(len(s)):
for j in range(i,len(s)-1):
if s[j+1] >= s[j]:
if (j+1)-i+1 > slen:
lstring = s[i:(j+1)+1]
slen = (j+1)-i+1
else:
break
print("Longest substring in alphabetical order is: " + lstring)
Output: Longest substring in alphabetical order is: ccl
input_str = "cyqfjhcclkbxpbojgkar"
length = len(input_str) # length of the input string
iter = 0
result_str = '' # contains latest processed sub string
longest = '' # contains longest sub string alphabetic order
while length > 1: # loop till all char processed from string
count = 1
key = input_str[iter] #set last checked char as key
result_str += key # start of the new sub string
for i in range(iter+1, len(input_str)): # discard processed char to set new range
length -= 1
if(key <= input_str[i]): # check the char is in alphabetic order
key = input_str[i]
result_str += key # concatenate the char to result_str
count += 1
else:
if(len(longest) < len(result_str)): # check result and longest str length
longest = result_str # if yes set longest to result
result_str = '' # re initiate result_str for new sub string
iter += count # update iter value to point the index of last processed char
break
if length is 1: # check for the last iteration of while loop
if(len(longest) < len(result_str)):
longest = result_str
print(longest);
finding the longest substring in alphabetical order in Python
in python shell 'a' < 'b' or 'a' <= 'a' is True
result = ''
temp = ''
for char in s:
if (not temp or temp[-1] <= char):
temp += char
elif (temp[-1] > char):
if (len(result) < len(temp)):
result = temp
temp = char
if (len(temp) > len(result)):
result = temp
print('Longest substring in alphabetical order is:', result)
s=input()
temp=s[0]
output=s[0]
for i in range(len(s)-1):
if s[i]<=s[i+1]:
temp=temp+s[i+1]
if len(temp)>len(output):
output=temp
else:
temp=s[i+1]
print('Longest substring in alphabetic order is:' + output)
I had similar question on one of the tests on EDX online something. Spent 20 minutes brainstorming and couldn't find solution. But the answer got to me. And it is very simple. The thing that stopped me on other solutions - the cursor should not stop or have unique value so to say if we have the edx string s = 'azcbobobegghakl' it should output - 'beggh' not 'begh'(unique set) or 'kl'(as per the longest identical to alphabet string). Here is my answer and it works
n=0
for i in range(1,len(s)):
if s[n:i]==''.join(sorted(s[n:i])):
longest_try=s[n:i]
else:
n+=1
In some cases, input is in mixed characters like "Hello" or "HelloWorld"
**Condition 1:**order determination is case insensitive, i.e. the string "Ab" is considered to be in alphabetical order.
**Condition 2:**You can assume that the input will not have a string where the number of possible consecutive sub-strings in alphabetical order is 0. i.e. the input will not have a string like " zxec ".
string ="HelloWorld"
s=string.lower()
r = ''
c = ''
last=''
for char in s:
if (c == ''):
c = char
elif (c[-1] <= char):
c += char
elif (c[-1] > char):
if (len(r) < len(c)):
r = c
c = char
else:
c = char
if (len(c) > len(r)):
r = c
for i in r:
if i in string:
last=last+i
else:
last=last+i.upper()
if len(r)==1:
print(0)
else:
print(last)
Out:elloW
```python
s = "cyqfjhcclkbxpbojgkar" # change this to any word
word, temp = "", s[0] # temp = s[0] for fence post problem
for i in range(1, len(s)): # starting from 1 not zero because we already add first char
x = temp[-1] # last word in var temp
y = s[i] # index in for-loop
if x <= y:
temp += s[i]
elif x > y:
if len(temp) > len(word): #storing longest substring so we can use temp for make another substring
word = temp
temp = s[i] #reseting var temp with last char in loop
if len(temp) > len(word):
word = temp
print("Longest substring in alphabetical order is:", word)
```
My code store longest substring at the moment in variable temp, then compare every string index in for-loop with last char in temp (temp[-1]) if index higher or same with (temp[-1]) then add that char from index in temp. If index lower than (temp[-1]) checking variable word and temp which one have longest substring, after that reset variable temp so we can make another substring until last char in strings.
s = 'cyqfjhcclkbxpbojgkar'
long_sub = '' #longest substring
temp = '' # temporarily hold current substr
if len(s) == 1: # if only one character
long_sub = s
else:
for i in range(len(s) - 1):
index = i
temp = s[index]
while index < len(s) - 1:
if s[index] <= s[index + 1]:
temp += s[index + 1]
else:
break
index += 1
if len(temp) > len(long_sub):
long_sub = temp
temp = ''
print(long_sub)
For comprehensibility, I also add this code snippet based on regular expressions. It's hard-coded and seems clunky. On the other hand, it seems to be the shortest and easiest answer to this problem. And it's also among the most efficient in terms of runtime complexity (see graph).
import re
def longest_substring(s):
substrings = re.findall('a*b*c*d*e*f*g*h*i*j*k*l*m*n*o*p*q*r*s*t*u*v*w*x*y*z*', s)
return max(substrings, key=len)
(Unfortunately, I'm not allowed to paste a graph here as a "newbie".)
Source + Explanation + Graph: https://blog.finxter.com/python-how-to-find-the-longest-substring-in-alphabetical-order/
Another way:
s = input("Please enter a sentence: ")
count = 0
maxcount = 0
result = 0
for char in range(len(s)-1):
if(s[char]<=s[char+1]):
count += 1
if(count > maxcount):
maxcount = count
result = char + 1
else:
count = 0
startposition = result - maxcount
print("Longest substring in alphabetical order is: ", s[startposition:result+1])