How to replace characters in a string in python - python

How to replace characters in a string which we know the exact indexes in python?
Ex : name = "ABCDEFGH"
I need to change all odd index positions characters into '$' character.
name = "A$C$E$G$"
(Considered indexes bigin from 0 )

Also '$'.join(s[::2])
Just takes even letters, casts them to a list of chars and then interleaves $
''.join(['$' if i in idx else s[i] for i in range(len(s))])
works for any index array idx

You can use enumerate to loop over the string and get the indices in each iteration then based your logic you can keep the proper elements :
>>> ''.join([j if i%2==0 else '$' for i,j in enumerate(name)])
'A$C$E$G$'

name = "ABCDEFGH"
nameL = list(name)
for i in range(len(nameL)):
if i%2==1:
nameL[i] = '$'
name = ''.join(nameL)
print(name)

You can reference string elements by index and form a new string. Something like this should work:
startingstring = 'mylittlestring'
nstr = ''
for i in range(0,len(startingstring)):
if i % 2 == 0:
nstr += startingstring[i]
else:
nstr += '$'
Then do with nstr as you like.

Related

how to add commas to a integer string

I have a task to create a comma separated list. What I have now works except I'm trying to figure out how to remove the last comma. What I mean is it looks like this with my code 1,2,3, but I need to have it as 1,2,3 is there any way to change what I have now to do this.
new_string = new_string+str(numbers[index])+comma
The easiest way to render a list as a string of comma separated values is str.join. If your list items aren't strings, use str() on each individual item as you join them:
>>> numbers = [1, 2, 3]
>>> ",".join(str(n) for n in numbers)
'1,2,3'
if you want to remove the last char of a string in python you can do this:
new_string = new_string[:-1] or just print(new_string[:-1])
However I'd recommend you check to see if it's the last element and not add the comma that time:
new_string = ''
my_list = [1,2,3,4]
for i, item in enumerate(my_list):
if i + 1 == len(my_list):
new_string += f'{item}'
else:
new_string += f'{item},'
try this:
#your ,ed output
mystr = "1,2,3,4,"
#removes the last item of the string
mylist = list(mystr)
mystr = ""
mylist[len(mylist)-1] = "";
for x in mylist:
mystr += x
#prints output(1,2,3,4)
print(mystr)
the first part sets 'myvar' to "1,2,3,4,"
then it turns 'myvar' into a list and removes the last element
replace 'myvar' width your output

How to find the largest repeating substring given character in Python?

Given some string say 'aabaaab', how would I go about finding the largest substring of a. So it should return 'aaa'. Any help would be greatly appreciated.
def sub_string(s):
best_run = 0
current_run = 0
for char in s:
if char == 'a'
current_run += 1
else:
current_letter = char
return(best_run)
I have something like the one above. Not sure where I can fix it up.
not the most efficient, but a straightforward solution:
word = "aasfgaaassaasdsddaaaaaafff"
substr_count = 0
substr_counts = []
character = "f"
for i, letter in enumerate(word):
if (letter == character):
substr_count += 1
else:
substr_counts.append(substr_count)
substr_count = 0
if (i == len(word) - 1):
substr_counts.append(substr_count)
print(max(substr_counts))
If you want a short method using standard python tools (and avoid writing loops to reconstruct the string as you iterate), you can use regex to split the string by any non-a characters than get the max() according to len:
import re
test_string = 'aabaaab'
split_string_list = re.split( '[^a]', test_string )
longest_string_subset = max( split_string_list, key=len )
print( longest_string_subset )
The re library is for regex, the '[^a]' is a regex statement for any non-a character. Basically, the 'aabaaab' is being split into a list according to any matches on the regex statement, so that it becomes [ 'aa' 'aaa' '' ]. Then, the max() statement looks for the longest string based on len (aka length).
You can read more about functions like re.split() in the docs: https://docs.python.org/2/library/re.html

Split a string, loop through it character by character, and replace specific ones?

I'm working on an assignment and have gotten stuck on a particular task. I need to write two functions that do similar things. The first needs to correct capitalization at the beginning of a sentence, and count when this is done. I've tried the below code:
def fix_capitalization(usrStr):
count = 0
fixStr = usrStr.split('.')
for sentence in fixStr:
if sentence[0].islower():
sentence[0].upper()
count += 1
print('Number of letters capitalized: %d' % count)
print('Edited text: %s' % fixStr)
Bu receive an out of range error. I'm getting an "Index out of range error" and am not sure why. Should't sentence[0] simply reference the first character in that particular string in the list?
I also need to replace certain characters with others, as shown below:
def replace_punctuation(usrStr):
s = list(usrStr)
exclamationCount = 0
semicolonCount = 0
for sentence in s:
for i in sentence:
if i == '!':
sentence[i] = '.'
exclamationCount += 1
if i == ';':
sentence[i] = ','
semicolonCount += 1
newStr = ''.join(s)
print(newStr)
print(semicolonCount)
print(exclamationCount)
But I'm struggling to figure out how to actually do the replacing once the character is found. Where am I going wrong here?
Thank you in advance for any help!
I would use str.capitalize over str.upper on one character. It also works correctly on empty strings. The other major improvement would be to use enumerate to also track the index as you iterate over the list:
def fix_capitalization(s):
sentences = [sentence.strip() for sentence in s.split('.')]
count = 0
for index, sentence in enumerate(sentences):
capitalized = sentence.capitalize()
if capitalized != sentence:
count += 1
sentences[index] = capitalized
result = '. '.join(sentences)
return result, count
You can take a similar approach to replacing punctuation:
replacements = {'!': '.', ';': ','}
def replace_punctuation(s):
l = list(s)
counts = dict.fromkeys(replacements, 0)
for index, item in enumerate(l):
if item in replacements:
l[index] = replacements[item]
counts[item] += 1
print("Replacement counts:")
for k, v in counts.items():
print("{} {:>5}".format(k, v))
return ''.join(l)
There are better ways to do these things but I'll try to change your code minimally so you will learn something.
The first function's issue is that when you split the sentence like "Hello." there will be two sentences in your fixStr list that the last one is an empty string; so the first index of an empty string is out of range. fix it by doing this.
def fix_capitalization(usrStr):
count = 0
fixStr = usrStr.split('.')
for sentence in fixStr:
# changed line
if sentence != "":
sentence[0].upper()
count += 1
print('Number of letters capitalized: %d' % count)
print('Edited text: %s' % fixStr)
In second snippet you are trying to write, when you pass a string to list() you get a list of characters of that string. So all you need to do is to iterate over the elements of the list and replace them and after that get string from the list.
def replace_punctuation(usrStr):
newStr = ""
s = list(usrStr)
exclamationCount = 0
semicolonCount = 0
for c in s:
if c == '!':
c = '.'
exclamationCount += 1
if c == ';':
c = ','
semicolonCount += 1
newStr = newStr + c
print(newStr)
print(semicolonCount)
print(exclamationCount)
Hope I helped!
Python has a nice build in function for this
for str in list:
new_str = str.replace('!', '.').replace(';', ',')
You can write a oneliner to get a new list
new_list = [str.replace('!', '.').replace(';', ',') for str in list]
You also could go for the split/join method
new_str = '.'.join(str.split('!'))
new_str = ','.join(str.split(';'))
To count capitalized letters you could do
result = len([cap for cap in str if str(cap).isupper()])
And to capitalize them words just use the
str.capitalize()
Hope this works out for you

Find out word at specific index

I have a string with multiple words separated by underscores like this:
string = 'this_is_my_string'
And let's for example take string[n] which will return a letter.
Now for this index I want to get the whole word between the underscores.
So for string[12] I'd want to get back the word 'string' and for string[1] I'd get back 'this'
Very simple approach using string slicing is to:
slice the list in two parts based on position
split() each part based on _.
concatenate last item from part 1 and first item from part 2
Sample code:
>>> my_string = 'this_is_my_sample_string'
# ^ index 14
>>> pos = 14
>>> my_string[:pos].split('_')[-1] + my_string[pos:].split('_')[0]
'sample'
This shuld work:
string = 'this_is_my_string'
words = string.split('_')
idx = 0
indexes = {}
for word in words:
for i in range(len(word)):
idx += 1
indexes[idx] = word
print(indexes[1]) # this
print(indexes[12]) #string
The following code works. You can change the index and string variables and adapt to new strings. You can also define a new function with the code to generalize it.
string = 'this_is_my_string'
sp = string.split('_')
index = 12
total_len = 0
for word in sp:
total_len += (len(word) + 1) #The '+1' accounts for the underscore
if index < total_len:
result = word
break
print result
A little bit of regular expression magic does the job:
import re
def wordAtIndex(text, pos):
p = re.compile(r'(_|$)')
beg = 0
for m in p.finditer(text):
#(end, sym) = (m.start(), m.group())
#print (end, sym)
end = m.start()
if pos < end: # 'pos' is within current split piece
break
beg = end+1 # advance to next split piece
if pos == beg-1: # handle case where 'pos' is index of split character
return ""
else:
return text[beg:end]
text = 'this_is_my_string'
for i in range(0, len(text)+1):
print ("Text["+str(i)+"]: ", wordAtIndex(text, i))
It splits the input string at '_' characters or at end-of-string, and then iteratively compares the given position index with the actual split position.

python - ordinal value - list indices must be integers not str

what i want to do is take a string and for each character make the ordinal value 1 more from the value it has.
myinput=input("Message : ")
mylist =list(myinput) #convert to list in order to take each character
for character in mylist:
mylist[character]+=ord(mylist[character])+1
print(character)
The problem is with the "ord(mylist[character])+1"
Thank you!
Probably you are looking for the next:
>>> m = raw_input('Message:')
Message:asdf
>>> ''.join(chr(ord(c) + 1) for c in m)
'bteg'
Notes:
use raw_input when you need to get string input from a user;
ord convert character to integer, chr - vise versa;
... for c in m syntax is a generator expression. It is also used for list comprehension.
Three problems here. First, you're mixing up list indices and list elements. Second, you didn't convert back to a character (I'm assuming you want characters, not numbers). Third, you're adding to the existing value.
One way:
for i range(len(mylist)):
mylist[i] = chr(ord(mylist[i])+1)
Another way:
for i, character in enumerate(mylist):
mylist[i] = chr(ord(character)+1)
Instead of
for character in mylist:
mylist[character]+=ord(mylist[character])+1
(where character is a list index and therefore invalid), you probably want:
mylist = [ord(character) + 1 for character in mylist]
Or a Counter.
You can do like this
def ordsum(astring, tablesize):
sum = 0
for num in range(len(astring)):
sum = sum + ord(astring[num])
return sum
myinput = input() # use raw_input() in Python 2
myinput = map(lambda ch: chr(ord(ch) + 1), myinput)
# or list comp.
myinput = [chr(ord(ch) + 1) for ch in myinput]
You can iterate directly over a string, you do not have to make it a list first. If your end goal is to have a new string, you can do this:
myinput=input("Message : ")
result = []
for character in myinput:
result.append( chr( ord( character ) + 1 )
mynewstring = ' '.join(result)

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