def encode(message):
encoded_message = ""
i = 0
while (i <= len(message)-1):
count = 1
ch = message[i]
j = i
while (j < len(message)-1):
if (message[j] == message[j+1]):
count = count+1
j = j+1
else:
break
encoded_message=encoded_message+str(count)+ch
i = j+1
return encoded_message
#Provide different values for message and test your program
encoded_message=encode("ABBBBCCCCCCCCAB")
print(encoded_message)
This code generates the following output: 1A4B8C1A1B
But if the value is 1 it should just display the letter like this: A4B8CAB
Simply replace this line:
encoded_message=encoded_message+str(count)+ch
with this:
encoded_message += (str(count) if count > 1 else "") + ch
It should do the trick. Now it only appends the count to the string if said count is bigger than one.
Related
Using Python, how to print output string as -> aaa3bb2c1ddddd5 when Input string is aaabbcddddd
I want to concatenate actual character value and number of times a character is repeated in a string
def mycode(myString):
lenstr = len(myString)
print('length of string is '+str(lenstr));
for ele in myString:
count=0
for character in myString:
if character == ele:
count = count+1
totalstr = ele+str(count)
return totalstr
If the string is always sorted and grouped together like that, then you can use a collections.Counter to do it.
from collections import Counter
inp = "aaabbcddddd"
counter = Counter(inp)
out = "".join(k * v + str(v) for k,v in counter.items())
Or in one line:
print(''.join(k * v + str(v) for k,v in Counter(inp).items()))
Output:
aaa3bb2c1ddddd5
Or you can do it manually:
inp = "aaabbcddddd"
last = inp[0]
out = inp[0]
count = 1
for i in inp[1:]:
if i == last:
count += 1
else:
out += str(count)
count = 1
last = i
out += i
out += str(count)
print(out)
Here is a one line solution using a regex replacement with callback:
inp = "aaabbcddddd"
output = re.sub(r'((\w)\2*)', lambda m: m.group(1) + str(len(m.group(1))), inp)
print(output) # aaa3bb2c1ddddd5
Another one-liner:
import itertools
test = 'aaabbcddddd'
out = ''.join(f"{(g := ''.join(ig))}{len(g)}" for _, ig in itertools.groupby(test))
assert out == 'aaa3bb2c1ddddd5'
def char_counter_string(string):
prev_char = None
char_counter = 0
output = ''
for char_index in range(len(string)+1):
if char_index == len(string):
output += str(char_counter)
break
if string[char_index] != prev_char and prev_char is not None:
output += str(char_counter)
char_counter = 0
output += string[char_index]
char_counter += 1
prev_char = string[char_index]
return output
if __name__ == '__main__':
print(char_counter_string('aaabbcddddd'))
you can do like..
Code:
Time Complexity: O(n)
input_string="aaabbcddddd"
res=""
count=1
for i in range(1, len(input_string)):
if input_string[i] == input_string[i-1]:
count += 1
else:
res+=input_string[i-1]*count + str(count)
count = 1
res+=input_string[-1]*count + str(count)
print(res) #aaa3bb2c1ddddd5
Here's another way, ...
Full disclosure: ... as long as the run of characters is 10 or less, it will work. I.e., if there are 11 of anything in row, this won't work (the count will be wrong).
It's just a function wrapping a reduce.
from functools import reduce
def char_rep_count(in_string):
return reduce(
lambda acc, inp:
(acc[:-1]+inp+str(int(acc[-1])+1))
if (inp==acc[-2])
else (acc+inp+"1"),
in_string[1:],
in_string[0]+"1"
)
And here's some sample output:
print(char_rep_count("aaabbcdddd"))
aaa3bb2c1dddd4
I think this fulfils the brief and is also very fast:
s = 'aaabbcddddd'
def mycode(myString):
if myString:
count = 1
rs = [prev := myString[0]]
for c in myString[1:]:
if c != prev:
rs.append(f'{count}')
count = 1
else:
count += 1
rs.append(prev := c)
rs.append(f'{count}')
return ''.join(rs)
return myString
I have a school project question (for Python) that goes like this:
Given a string_input such as "abcd&1-4efg", the function must remove the "&1-4" and insert the string slice from 1 to 4 where the "&1-4" was.
eg. if string_input = "abcd&1-4efg",
"&1-4" is removed.
The remaining characters are indexed as follows: a=0, b=1, c=2, d=3, e=4, f=5, g=6
The new string becomes:
"abcdbcdeefg"
I've managed to write a long chunk of code to do this, but I'm wondering if anyone has any more efficient solutions?
Things to note:
The instructions can include double digits (eg. &10-15)
If the index isn't found, the returned string should print "?" for every missing index
(eg. "abcd&5-10efgh" would return "abcdfgh???efgh")
Intructions can be back-to-back (eg. "&10-15abcdef&1-5&4-5pqrs")
The code I've written is:
def expand(text):
text += "|"
import string
digits_dash = string.digits + "-"
idx_ref_str = ""
replace_list = []
record_val = False
output_to_list = []
instruct = ""
and_idx_mark = 0
#builds replace_list & idx_ref_list
for idx in range(len(text)):
if text[idx] == "&" and record_val==True:
output_to_list.append(instruct)
output_to_list.append(and_idx_mark)
replace_list.append(output_to_list)
output_to_list, instruct, inst_idx, and_idx_mark = [],"",0,0
and_idx_mark = idx
continue
elif text[idx] == "&":
record_val = True
and_idx_mark = idx
continue
#executes if currently in instruction part
if record_val == True:
#adds to instruct
if text[idx] in digits_dash:
instruct += text[idx]
#take info, add to replace list
else:
output_to_list.append(instruct)
output_to_list.append(and_idx_mark)
replace_list.append(output_to_list)
output_to_list, instruct, inst_idx, and_idx_mark, record_val = [],"",0,0,False
#executes otherwise
if record_val == False:
idx_ref_str += text[idx]
idx_ref_str = idx_ref_str[:-1]
text = text[:-1]
#converts str to int indexes in replace list[x][2]
for item in replace_list:
start_idx = ""
end_idx = ""
#find start idx
for char in item[0]:
if char in string.digits:
start_idx += char
elif char == "-":
start_idx = int(start_idx)
break
#find end idx
for char in item[0][::-1]:
if char in string.digits:
end_idx = char + end_idx
elif char == "-":
end_idx = int(end_idx)
break
start_end_list = [start_idx,end_idx]
item+=start_end_list
#split text into parts in list
count = 0
text_block = ""
text_block_list = []
idx_replace = 0
for char in text:
if char == "&":
text_block_list.append(text_block)
text_block = ""
count += len(replace_list[idx_replace][0])
idx_replace +=1
elif count > 0:
count -= 1
else:
text_block += char
text_block_list.append(text_block)
#creates output str
output_str = ""
for idx in range(len(text_block_list)-1):
output_str += text_block_list[idx]
#creates to_add var to add to output_str
start_repl = replace_list[idx][1]
end_repl = replace_list[idx][1] + len(replace_list[idx][0])
find_start = replace_list[idx][2]
find_end = replace_list[idx][3]
if end_idx >= len(idx_ref_str):
gap = end_idx + 1 - len(idx_ref_str)
to_add = idx_ref_str[find_start:] + "?" * gap
else:
to_add = idx_ref_str[find_start:find_end+1]
output_str += to_add
output_str += text_block_list[-1]
return output_str
Here's how I would do it. Always open to criticism.
import re
s = 'abcd&1-4efg'
c = re.compile('&[0-9]+-[0-9]+')
if (m := c.search(s)):
a, b = m.span()
left = s[:a]
right = s[b:]
o = [int(x) for x in m.group(0)[1:].split('-')]
mid = (left+right)[o[0]:o[1]+1]
print(left + mid + right)
I'm currently learning Python and I'm stuck on this specific question.
Image
Here is my current code:
word = input()
text = 0
wordch = 0
positions = 0
repeated = 0
while repeated != 2:
for i in range(0, len(tablet)):
if tablet[i] == word[wordch]:
text += 1
wordch += 1
if text == len(word):
positions += 1
text = 0
wordch = 0
elif repeated == 1 and text == len(word):
positions += 1
text = 0
wordch = 0
break
elif i == len(tablet)-1:
repeated += 1
break
elif tablet[i] != word[wordch]:
text == 0
wordch == 0
print(positions)
I would hope for a code that is really basic using the same concepts but please do answer.
Thank you!
I have tried to solve the problem by using a different approach. As we know that we can only use (len(fav_word)) - 1 letters if we tried to create the substring in a cyclic manner from the end since if we took any more characters, we would have created them from the start itself without the cycle.
So, I just created a new string from the original string by appending the starting (len(fav_word)) - 1 to the original string and then find all occurrences of the fav_string in the new string.
def find_all(a_str, sub):
start = 0
while True:
start = a_str.find(sub, start)
if start == -1: return
yield start
start += 1
x = "cabccabcab"
fav = "abc"
y = x + x[0:len(fav)-1]
print(len(list(find_all(y, fav)))) # Output: 3
x = "ababa"
fav = "aba"
y = x + x[0:len(fav)-1]
print(len(list(find_all(y, fav)))) # Output: 2
x = "aaaaaa"
fav = "aa"
y = x + x[0:len(fav)-1]
print(len(list(find_all(y, fav)))) # Output: 6
x = "abaaba"
fav = "aaba"
y = x + x[0:len(fav)-1]
print(len(list(find_all(y, fav)))) # Output: 2
def find_str(g,find):
lg = len(g)
lf = len(find)
x=0
s=""
for index, i in enumerate(g):
if i == find[0]:
if index+lf <= lg:
s = "".join(g[index:index+lf])
if s == find:
x+=1
else:
rem = "".join(g[index:])
lr = len(rem)
for index,i in enumerate(g):
rem+=i
lr+=1
if lr == lf:
if rem == find:
x+=1
break
return x
print(find_str("abaaba","aaba"))
def split(word):
return [char for char in word]
x = "aaaaaa"
pattern = "aa"
mylist=split(x)
ok=True
occurrences=0
buffer=""
while ok:
char=mylist.pop(0)
buffer+=char
if buffer==pattern:
occurrences+=1
buffer=""
if len(mylist)==0:
ok=False
print(occurrences)
output:3
I want to inset new key,value pair into a variable that was not previously defined
Here is what i want to do:
def drawboard(nextmove):
result = nextmove
pos = 1
print result
for i in range(7):
for j in range(7):
if (j == 0 or i % 2 == 0 or j % 2 == 0):
print '*',
if (i % 2 != 0 and j % 2 != 0):
print pos,
pos += 1
if (j == 6):
print '\n'
move = raw_input("Input you move(i.e. x,3[position, your symbol]: ")
if move == 'q':
exit()
calcres(move)
def calcres(move):
nextmove = dict([move.split(',')])
drawboard(nextmove)
drawboard({0: 0})
Inside drawboard function i want to concatenate nextmove with result, and save all the moves inside the result finally. I tried initializing result = {} but as expected that removes all items from result and re-initializes it resulting only one item inside that dictionary.
Use setdefault to initialize the result dictionary value to an empty list whenever the key is not present
def drawboard(nextmove):
result.setdefault(nextmove[0],[]).append(nextmove[1])
pos = 1
#print result
for i in range(7):
for j in range(7):
if (j == 0 or i % 2 == 0 or j % 2 == 0):
print '*',
if (i % 2 != 0 and j % 2 != 0):
print pos,
pos += 1
if (j == 6):
print '\n'
move = raw_input("Input you move(i.e. x,3[position, your symbol]: ")
if move == 'q':
exit()
calcres(move)
def calcres(move):
nextmove = move.split(',')
drawboard(nextmove)
result = {}
drawboard([0,0])
I am fairly new to python, I am not sure on how to fix a index string out of range. it happens right after the while loop when I want to send mylist[i][0] to formatting function. Any pointer on my code in general would be awesome!
def formatting(str1):
if str1 == '?':
return True
else:
return False
while(i <= len(mylist)):
val = formatting(mylist[i][0])
if val == True:
str1 = mylist[i]
str2 = mylist[i+1]
i = i + 2
format_set(str1, str2)
else:
if format == True:
if (margin + count + len(mylist[i])) <= width:
if (i == (len(mylist)-1)):
list2.append(mylist[i])
print(" " * margin + " ".join(list2))
break
list2.append(mylist[i])
count += len(mylist[i])
i += 1
else:
print(" " * margin + " ".join(list2))
list2 = []
count = 0
else:
temp_margin = margin
temp_width = width
width = 60
margin = 0
if (margin + count + len(mylist[i])) <= width:
if (i == (len(mylist)-1)):
list2.append(mylist[i])
print(" " * margin + " ".join(list2))
margin = temp_margin
width = temp_width
break
list2.append(mylist[i])
count += len(mylist[i])
i += 1
else:
print(" " * margin + " ".join(list2))
list2 = []
count = 0
change
i <= len(mylist)
to
i < len(mylist)
In the last iteration of the while loop, i is referring to the last value. Hence,
str2 = mylist[i+1]
is trying to reference a string outside the allowed range and you get an error.
EDIT: Also, as Wcrousse mentioned, the while (i <= len(...)) should be changed to i < len(...) because indexes go from 0 - (length-1).