We Keep Coding

How to replace characters of string from a list entry in Python?

I have a string in which I want to replace certain characters with "*". But replace() function of python doesn't replace the characters. I understand that the strings in python are immutable and I am creating a new variable to store the replaced string. But still the function doesn't provide the replaced strings.
This is the following code that I have written. I have tried the process in two ways but still don't get the desired output:
1st way:
a = "AGGCFTFGADFADTRFCAGFADARTRADFACDGFLKLIAP"
rep = ['A','C','P']
for char in rep:
new = a.replace(char, "*")
print(new)
Output:
AGGCFTFGADFADTRFCAGFADARTRADFACDGFLKLIA*
2nd way:
a = "AGGCFTFGADFADTRFCAGFADARTRADFACDGFLKLIAP"
rep = ['A','C','P']
for i in a:
if(i in rep):
new = a.replace(i, "*")
print(new)
Output:
AGGCFTFGADFADTRFCAGFADARTRADFACDGFLKLIA*
Any help would be much appreciated. Thanks

You assign the result of a.replace(char, "*") to new, but then on the next iteration of the for loop, you again replace parts of a, not new. Instead of assigning to new, just assign the result to a, replacing the original string.
a = "AGGCFTFGADFADTRFCAGFADARTRADFACDGFLKLIAP"
rep = ['A','C','P']
for char in rep:
a = a.replace(char, "*")
print(a)

In addition to the answers offered, I would suggest that regular expressions make this perhaps more straightforward, accomplishing all of the substitutions with a single function call.
>>> import re
>>> a = "AGGCFTFGADFADTRFCAGFADARTRADFACDGFLKLIAP"
>>> rep = ['A','C','P']
>>> r = re.compile('|'.join(rep))
>>> r.sub('*', a)
'*GG*FTFG*DF*DTRF**GF*D*RTR*DF**DGFLKLI**'
Just in case someone decides to be clever and puts something regex significant in rep, you could escape those when compiling your regex.
r = re.compile('|'.join(re.escape(x) for x in rep))

Others have explained errors in posted code. An alternative using generator expression:
new = ''.join("*" if char in ['A','C','P'] else char for char in a)
print(new)
>>> '*GG*FTFG*DF*DTRF**GF*D*RTR*DF**DGFLKLI**'

A simple loop is easy to understand and efficient. The crucial part of the looping approach is to re-assign the string reference to the output of replace()
I've taken the liberty of plagiarising two pieces of code from other contributors in order to demonstrate the performance differences (in case that's important).
import re
from timeit import timeit
a = "AGGCFTFGADFADTRFCAGFADARTRADFACDGFLKLIAP"
rep = 'A', 'C', 'P'
p = re.compile('|'.join(rep))
def v1(s):
for c in rep:
s = s.replace(c, '*')
return s
def v2(s):
return p.sub('*', s)
def v3(s):
return ''.join("*" if char in rep else char for char in s)
for func in v1, v2, v3:
print(func.__name__, timeit(lambda: func(a)))
assert v1(a) == v2(a)
assert v1(a) == v3(a)
Output:
v1 0.3363962830003402
v2 1.8725565750000897
v3 3.3800653280000006
Platform:
macOS 13.0.1
Python 3.11.0
3 GHz 10-Core Intel Xeon W

As already mentioned, you should write a = a.replace(i, "*") because you are looping through rep and you want to do the replacement in the string a. Strings are immutable, and replace gives back a copy of the string.
The variable new only gives you the replacement over the last iteration of rep which is a P char and will result in AGGCFTFGADFADTRFCAGFADARTRADFACDGFLKLIA* because there is only a single P at the end of the string and you are never actually changing the value of rep.
If you have single characters, you can use a character class [ACP] with a single call to re.sub
import re
a = "AGGCFTFGADFADTRFCAGFADARTRADFACDGFLKLIAP"
print(re.sub("[ACP]", "*", a))
Output
*GG*FTFG*DF*DTRF**GF*D*RTR*DF**DGFLKLI**

Why isn't my alternate string replacement method working?

I got a working solution to this question with my second attempt. But I'm wondering why the first option doesn't work as well.
The question is:
Complete the function to replace the word test with cat and print the new string
my code (working solution):
def replaceTest(mystring):
answer = mystring.replace("test", "cat")
print(answer)
replaceTest('This is a test')
# output:
# This is a cat
my code (solution that doesn't work):
def replaceTest(mystring):
stringSplit = mystring.split()
for i in stringSplit:
if i == 'test':
i = 'cat'
answer = " ".join(stringSplit)
print(answer)
print(replaceTest('This is a test'))
# output:
# This is a test
# None
I can't see why it doesn't work. I think I'm iterating over each item in the list and saying if that item is test, make it cat instead. And then making the list into a string. Where did i go wrong?

With i = 'cat' you are just assigning to the loop variable, which is then discarded. You need to assign to the original strinSplit list (stringSplit[i] = 'cat') where by i is the index of stringSplit, obtained from enumerate:
def replaceTest(mystring):
stringSplit = mystring.split()
for i, word in enumerate(stringSplit):
if word == 'test':
stringSplit[i] = 'cat'
answer = " ".join(stringSplit)
print(answer)
replaceTest('This is a test')
Some improvements while still using your original logic: Having an index in a loop used to access list elements is awkward and error-prone. In this case (and many others) you can avoid it using a list comprehension instead. Also, I assume you wouldn't really want to print inside the function but instead return the modified string so you can use it further:
def replaceTest(mystring):
string_split_replace = ["cat" if word == "test" else word
for word in mystring.split()]
return " ".join(string_split_replace)
print(replaceTest('This is a test'))

python - in-place string reversal [duplicate]

There is no built in reverse function for Python's str object. What is the best way of implementing this method?
If supplying a very concise answer, please elaborate on its efficiency. For example, whether the str object is converted to a different object, etc.

Using slicing:
>>> 'hello world'[::-1]
'dlrow olleh'
Slice notation takes the form [start:stop:step]. In this case, we omit the start and stop positions since we want the whole string. We also use step = -1, which means, "repeatedly step from right to left by 1 character".

#Paolo's s[::-1] is fastest; a slower approach (maybe more readable, but that's debatable) is ''.join(reversed(s)).

What is the best way of implementing a reverse function for strings?
My own experience with this question is academic. However, if you're a pro looking for the quick answer, use a slice that steps by -1:
>>> 'a string'[::-1]
'gnirts a'
or more readably (but slower due to the method name lookups and the fact that join forms a list when given an iterator), str.join:
>>> ''.join(reversed('a string'))
'gnirts a'
or for readability and reusability, put the slice in a function
def reversed_string(a_string):
return a_string[::-1]
and then:
>>> reversed_string('a_string')
'gnirts_a'
Longer explanation
If you're interested in the academic exposition, please keep reading.
There is no built-in reverse function in Python's str object.
Here is a couple of things about Python's strings you should know:
In Python, strings are immutable. Changing a string does not modify the string. It creates a new one.
Strings are sliceable. Slicing a string gives you a new string from one point in the string, backwards or forwards, to another point, by given increments. They take slice notation or a slice object in a subscript:
string[subscript]
The subscript creates a slice by including a colon within the braces:
string[start:stop:step]
To create a slice outside of the braces, you'll need to create a slice object:
slice_obj = slice(start, stop, step)
string[slice_obj]
A readable approach:
While ''.join(reversed('foo')) is readable, it requires calling a string method, str.join, on another called function, which can be rather relatively slow. Let's put this in a function - we'll come back to it:
def reverse_string_readable_answer(string):
return ''.join(reversed(string))
Most performant approach:
Much faster is using a reverse slice:
'foo'[::-1]
But how can we make this more readable and understandable to someone less familiar with slices or the intent of the original author? Let's create a slice object outside of the subscript notation, give it a descriptive name, and pass it to the subscript notation.
start = stop = None
step = -1
reverse_slice = slice(start, stop, step)
'foo'[reverse_slice]
Implement as Function
To actually implement this as a function, I think it is semantically clear enough to simply use a descriptive name:
def reversed_string(a_string):
return a_string[::-1]
And usage is simply:
reversed_string('foo')
What your teacher probably wants:
If you have an instructor, they probably want you to start with an empty string, and build up a new string from the old one. You can do this with pure syntax and literals using a while loop:
def reverse_a_string_slowly(a_string):
new_string = ''
index = len(a_string)
while index:
index -= 1 # index = index - 1
new_string += a_string[index] # new_string = new_string + character
return new_string
This is theoretically bad because, remember, strings are immutable - so every time where it looks like you're appending a character onto your new_string, it's theoretically creating a new string every time! However, CPython knows how to optimize this in certain cases, of which this trivial case is one.
Best Practice
Theoretically better is to collect your substrings in a list, and join them later:
def reverse_a_string_more_slowly(a_string):
new_strings = []
index = len(a_string)
while index:
index -= 1
new_strings.append(a_string[index])
return ''.join(new_strings)
However, as we will see in the timings below for CPython, this actually takes longer, because CPython can optimize the string concatenation.
Timings
Here are the timings:
>>> a_string = 'amanaplanacanalpanama' * 10
>>> min(timeit.repeat(lambda: reverse_string_readable_answer(a_string)))
10.38789987564087
>>> min(timeit.repeat(lambda: reversed_string(a_string)))
0.6622700691223145
>>> min(timeit.repeat(lambda: reverse_a_string_slowly(a_string)))
25.756799936294556
>>> min(timeit.repeat(lambda: reverse_a_string_more_slowly(a_string)))
38.73570013046265
CPython optimizes string concatenation, whereas other implementations may not:
... do not rely on CPython's efficient implementation of in-place string concatenation for statements in the form a += b or a = a + b . This optimization is fragile even in CPython (it only works for some types) and isn't present at all in implementations that don't use refcounting. In performance sensitive parts of the library, the ''.join() form should be used instead. This will ensure that concatenation occurs in linear time across various implementations.

Quick Answer (TL;DR)
Example
### example01 -------------------
mystring = 'coup_ate_grouping'
backwards = mystring[::-1]
print(backwards)
### ... or even ...
mystring = 'coup_ate_grouping'[::-1]
print(mystring)
### result01 -------------------
'''
gnipuorg_eta_puoc
'''
Detailed Answer
Background
This answer is provided to address the following concern from #odigity:
Wow. I was horrified at first by the solution Paolo proposed, but that
took a back seat to the horror I felt upon reading the first
comment: "That's very pythonic. Good job!" I'm so disturbed that such
a bright community thinks using such cryptic methods for something so
basic is a good idea. Why isn't it just s.reverse()?
Problem
Context
Python 2.x
Python 3.x
Scenario:
Developer wants to transform a string
Transformation is to reverse order of all the characters
Solution
example01 produces the desired result, using extended slice notation.
Pitfalls
Developer might expect something like string.reverse()
The native idiomatic (aka "pythonic") solution may not be readable to newer developers
Developer may be tempted to implement his or her own version of string.reverse() to avoid slice notation.
The output of slice notation may be counter-intuitive in some cases:
see e.g., example02
print 'coup_ate_grouping'[-4:] ## => 'ping'
compared to
print 'coup_ate_grouping'[-4:-1] ## => 'pin'
compared to
print 'coup_ate_grouping'[-1] ## => 'g'
the different outcomes of indexing on [-1] may throw some developers off
Rationale
Python has a special circumstance to be aware of: a string is an iterable type.
One rationale for excluding a string.reverse() method is to give python developers incentive to leverage the power of this special circumstance.
In simplified terms, this simply means each individual character in a string can be easily operated on as a part of a sequential arrangement of elements, just like arrays in other programming languages.
To understand how this works, reviewing example02 can provide a good overview.
Example02
### example02 -------------------
## start (with positive integers)
print 'coup_ate_grouping'[0] ## => 'c'
print 'coup_ate_grouping'[1] ## => 'o'
print 'coup_ate_grouping'[2] ## => 'u'
## start (with negative integers)
print 'coup_ate_grouping'[-1] ## => 'g'
print 'coup_ate_grouping'[-2] ## => 'n'
print 'coup_ate_grouping'[-3] ## => 'i'
## start:end
print 'coup_ate_grouping'[0:4] ## => 'coup'
print 'coup_ate_grouping'[4:8] ## => '_ate'
print 'coup_ate_grouping'[8:12] ## => '_gro'
## start:end
print 'coup_ate_grouping'[-4:] ## => 'ping' (counter-intuitive)
print 'coup_ate_grouping'[-4:-1] ## => 'pin'
print 'coup_ate_grouping'[-4:-2] ## => 'pi'
print 'coup_ate_grouping'[-4:-3] ## => 'p'
print 'coup_ate_grouping'[-4:-4] ## => ''
print 'coup_ate_grouping'[0:-1] ## => 'coup_ate_groupin'
print 'coup_ate_grouping'[0:] ## => 'coup_ate_grouping' (counter-intuitive)
## start:end:step (or start:end:stride)
print 'coup_ate_grouping'[-1::1] ## => 'g'
print 'coup_ate_grouping'[-1::-1] ## => 'gnipuorg_eta_puoc'
## combinations
print 'coup_ate_grouping'[-1::-1][-4:] ## => 'puoc'
Conclusion
The cognitive load associated with understanding how slice notation works in python may indeed be too much for some adopters and developers who do not wish to invest much time in learning the language.
Nevertheless, once the basic principles are understood, the power of this approach over fixed string manipulation methods can be quite favorable.
For those who think otherwise, there are alternate approaches, such as lambda functions, iterators, or simple one-off function declarations.
If desired, a developer can implement her own string.reverse() method, however it is good to understand the rationale behind this aspect of python.
See also
alternate simple approach
alternate simple approach
alternate explanation of slice notation

This answer is a bit longer and contains 3 sections: Benchmarks of existing solutions, why most solutions here are wrong, my solution.
The existing answers are only correct if Unicode Modifiers / grapheme clusters are ignored. I'll deal with that later, but first have a look at the speed of some reversal algorithms:
NOTE: I've what I called list_comprehension should be called slicing
list_comprehension : min: 0.6μs, mean: 0.6μs, max: 2.2μs
reverse_func : min: 1.9μs, mean: 2.0μs, max: 7.9μs
reverse_reduce : min: 5.7μs, mean: 5.9μs, max: 10.2μs
reverse_loop : min: 3.0μs, mean: 3.1μs, max: 6.8μs
list_comprehension : min: 4.2μs, mean: 4.5μs, max: 31.7μs
reverse_func : min: 75.4μs, mean: 76.6μs, max: 109.5μs
reverse_reduce : min: 749.2μs, mean: 882.4μs, max: 2310.4μs
reverse_loop : min: 469.7μs, mean: 577.2μs, max: 1227.6μs
You can see that the time for the list comprehension (reversed = string[::-1]) is in all cases by far the lowest (even after fixing my typo).
String Reversal
If you really want to reverse a string in the common sense, it is WAY more complicated. For example, take the following string (brown finger pointing left, yellow finger pointing up). Those are two graphemes, but 3 unicode code points. The additional one is a skin modifier.
example = "👈🏾👆"
But if you reverse it with any of the given methods, you get brown finger pointing up, yellow finger pointing left. The reason for this is that the "brown" color modifier is still in the middle and gets applied to whatever is before it. So we have
U: finger pointing up
M: brown modifier
L: finger pointing left
and
original: LMU 👈🏾👆
reversed: UML (above solutions) ☝🏾👈
reversed: ULM (correct reversal) 👆👈🏾
Unicode Grapheme Clusters are a bit more complicated than just modifier code points. Luckily, there is a library for handling graphemes:
>>> import grapheme
>>> g = grapheme.graphemes("👈🏾👆")
>>> list(g)
['👈🏾', '👆']
and hence the correct answer would be
def reverse_graphemes(string):
g = list(grapheme.graphemes(string))
return ''.join(g[::-1])
which also is by far the slowest:
list_comprehension : min: 0.5μs, mean: 0.5μs, max: 2.1μs
reverse_func : min: 68.9μs, mean: 70.3μs, max: 111.4μs
reverse_reduce : min: 742.7μs, mean: 810.1μs, max: 1821.9μs
reverse_loop : min: 513.7μs, mean: 552.6μs, max: 1125.8μs
reverse_graphemes : min: 3882.4μs, mean: 4130.9μs, max: 6416.2μs
The Code
#!/usr/bin/env python3
import numpy as np
import random
import timeit
from functools import reduce
random.seed(0)
def main():
longstring = ''.join(random.choices("ABCDEFGHIJKLM", k=2000))
functions = [(list_comprehension, 'list_comprehension', longstring),
(reverse_func, 'reverse_func', longstring),
(reverse_reduce, 'reverse_reduce', longstring),
(reverse_loop, 'reverse_loop', longstring)
]
duration_list = {}
for func, name, params in functions:
durations = timeit.repeat(lambda: func(params), repeat=100, number=3)
duration_list[name] = list(np.array(durations) * 1000)
print('{func:<20}: '
'min: {min:5.1f}μs, mean: {mean:5.1f}μs, max: {max:6.1f}μs'
.format(func=name,
min=min(durations) * 10**6,
mean=np.mean(durations) * 10**6,
max=max(durations) * 10**6,
))
create_boxplot('Reversing a string of length {}'.format(len(longstring)),
duration_list)
def list_comprehension(string):
return string[::-1]
def reverse_func(string):
return ''.join(reversed(string))
def reverse_reduce(string):
return reduce(lambda x, y: y + x, string)
def reverse_loop(string):
reversed_str = ""
for i in string:
reversed_str = i + reversed_str
return reversed_str
def create_boxplot(title, duration_list, showfliers=False):
import seaborn as sns
import matplotlib.pyplot as plt
import operator
plt.figure(num=None, figsize=(8, 4), dpi=300,
facecolor='w', edgecolor='k')
sns.set(style="whitegrid")
sorted_keys, sorted_vals = zip(*sorted(duration_list.items(),
key=operator.itemgetter(1)))
flierprops = dict(markerfacecolor='0.75', markersize=1,
linestyle='none')
ax = sns.boxplot(data=sorted_vals, width=.3, orient='h',
flierprops=flierprops,
showfliers=showfliers)
ax.set(xlabel="Time in ms", ylabel="")
plt.yticks(plt.yticks()[0], sorted_keys)
ax.set_title(title)
plt.tight_layout()
plt.savefig("output-string.png")
if __name__ == '__main__':
main()

1. using slice notation
def rev_string(s):
return s[::-1]
2. using reversed() function
def rev_string(s):
return ''.join(reversed(s))
3. using recursion
def rev_string(s):
if len(s) == 1:
return s
return s[-1] + rev_string(s[:-1])

A lesser perplexing way to look at it would be:
string = 'happy'
print(string)
'happy'
string_reversed = string[-1::-1]
print(string_reversed)
'yppah'
In English [-1::-1] reads as:
"Starting at -1, go all the way, taking steps of -1"

Reverse a string in python without using reversed() or [::-1]
def reverse(test):
n = len(test)
x=""
for i in range(n-1,-1,-1):
x += test[i]
return x

This is also an interesting way:
def reverse_words_1(s):
rev = ''
for i in range(len(s)):
j = ~i # equivalent to j = -(i + 1)
rev += s[j]
return rev
or similar:
def reverse_words_2(s):
rev = ''
for i in reversed(range(len(s)):
rev += s[i]
return rev
Another more 'exotic' way using bytearray which supports .reverse()
b = bytearray('Reverse this!', 'UTF-8')
b.reverse()
b.decode('UTF-8')`
will produce:
'!siht esreveR'

def reverse(input):
return reduce(lambda x,y : y+x, input)

Here is a no fancy one:
def reverse(text):
r_text = ''
index = len(text) - 1
while index >= 0:
r_text += text[index] #string canbe concatenated
index -= 1
return r_text
print reverse("hello, world!")

There are multiple ways to reverse a string in Python
Slicing Method
string = "python"
rev_string = string[::-1]
print(rev_string)
using reversed function
string = "python"
rev= reversed(string)
rev_string = "".join(rev)
print(rev_string)
Using Recursion
string = "python"
def reverse(string):
if len(string)==0:
return string
else:
return reverse(string[1:])+string[0]
print(reverse(string))
Using for Loop
string = "python"
rev_string =""
for s in string:
rev_string = s+ rev_string
print(rev_string)
Using while Loop
string = "python"
rev_str =""
length = len(string)-1
while length >=0:
rev_str += string[length]
length -= 1
print(rev_str)

original = "string"
rev_index = original[::-1]
rev_func = list(reversed(list(original))) #nsfw
print(original)
print(rev_index)
print(''.join(rev_func))

To solve this in programing way for interview
def reverse_a_string(string: str) -> str:
"""
This method is used to reverse a string.
Args:
string: a string to reverse
Returns: a reversed string
"""
if type(string) != str:
raise TypeError("{0} This not a string, Please provide a string!".format(type(string)))
string_place_holder = ""
start = 0
end = len(string) - 1
if end >= 1:
while start <= end:
string_place_holder = string_place_holder + string[end]
end -= 1
return string_place_holder
else:
return string
a = "hello world"
rev = reverse_a_string(a)
print(rev)
Output:
dlrow olleh

Recursive method:
def reverse(s): return s[0] if len(s)==1 else s[len(s)-1] + reverse(s[0:len(s)-1])
example:
print(reverse("Hello!")) #!olleH

def reverse_string(string):
length = len(string)
temp = ''
for i in range(length):
temp += string[length - i - 1]
return temp
print(reverse_string('foo')) #prints "oof"
This works by looping through a string and assigning its values in reverse order to another string.

a=input()
print(a[::-1])
The above code recieves the input from the user and prints an output that is equal to the reverse of the input by adding [::-1].
OUTPUT:
>>> Happy
>>> yppaH
But when it comes to the case of sentences, view the code output below:
>>> Have a happy day
>>> yad yppah a evaH
But if you want only the characters of the string to be reversed and not the sequence of string, try this:
a=input().split() #Splits the input on the basis of space (" ")
for b in a: #declares that var (b) is any value in the list (a)
print(b[::-1], end=" ") #End declares to print the character in its quotes (" ") without a new line.
In the above code in line 2 in I said that ** variable b is any value in the list (a)** I said var a to be a list because when you use split in an input the variable of the input becomes a list. Also remember that split can't be used in the case of int(input())
OUTPUT:
>>> Have a happy day
>>> evaH a yppah yad
If we don't add end(" ") in the above code then it will print like the following:
>>> Have a happy day
>>> evaH
>>> a
>>> yppah
>>> yad
Below is an example to understand end():
CODE:
for i in range(1,6):
print(i) #Without end()
OUTPUT:
>>> 1
>>> 2
>>> 3
>>> 4
>>> 5
Now code with end():
for i in range(1,6):
print(i, end=" || ")
OUTPUT:
>>> 1 || 2 || 3 || 4 || 5 ||

Here is how we can reverse a string using for loop:
string = "hello,world"
for i in range(-1,-len(string)-1,-1):
print (string[i], end=(" "))

Just as a different solution(because it's asked in interviews):
def reverse_checker(string):
ns = ""
for h in range(1,len(string)+1):
ns += string[-h]
print(ns)
if ns == string:
return True
else:
return False

How to delete repeating letters in a string?

I am trying to write a function which will return me the string of unique characters present in the passed string. Here's my code:
def repeating_letters(given_string):
counts = {}
for char in given_string:
if char in counts:
return char
else:
counts[char] = 1
if counts[char] > 1:
del(char)
else:
return char
I am not getting expected results with it. How can I get the desired result.
Here when I am passing this string as input:
sample_input = "abcadb"
I am expecting the result to be:
"abcd"
However my code is returning me just:
nothing

def repeating_letters(given_string):
seen = set()
ret = []
for c in given_string:
if c not in seen:
ret.append(c)
seen.add(c)
return ''.join(ret)
Here we add each letter to the set seen the first time we see it, at the same time adding it to a list ret. Then we return the joined list.

Here's the one-liner to achieve this if the order in the resultant string matters via using set with sorted as:
>>> my_str = 'abcadbgeg'
>>> ''.join(sorted(set(my_str),key=my_str.index))
'abcdge'
Here sorted will sort the characters in the set based on the first index of each in the original string, resulting in ordered list of characters.
However if the order in the resultant string doesn't matter, then you may simply do:
>>> ''.join(set(my_str))
'acbedg'

Rearranging a string based on specific requirements

Hi there so I am looking to build this python function with simple things like def, find etc. so far I know how to get the first part of the code.
Given a string such as "HELLODOGMEMEDOGPAPA", I will need to return a list that gives me three things:
Everything before the word dog which i will denote as before_dog
The word dog until dog appears again dog_todog
Everything after the second time dog appears will be denoted by after_todog
The list will be in the form [before_dog,dog_todog,after_todog].
so for example given ("HELLODOGMEMEDOGPAPADD") this will return the list
("HELLO","DOGMEME","DOGPAPADD")
another example would be ("HEYHELLOMANDOGYDOGDADDY") this would return the list
("HEYHELLOMAN","DOGY","DOGDADDY")
but if I have ("HEYHELLODOGDADDY")
the output will be ("HEYHELLO","DOGDADDY","")
also if dog never appears ("HEYHELLOYO") then the output will be ("HEYHELLOYO,"","")
This is what I have so far:
def split_list(words):
# declare the list
lst = []
# find the first position
first_pos=words.find("DOG")
# find the first_pos
before_dog = words [0:first_pos]
lst.append(before_dog)
return lst

Funny function split_2_dogs() with re.findall() function:
import re
def split_2_dogs(s):
if s.count('DOG') == 2: # assuring 2 dogs are "walking" there
return list(re.findall(r'^(.*)(DOG.*)(DOG.*)$', s)[0])
print(split_2_dogs("HELLODOGMEMEDOGPAPADD"))
print(split_2_dogs("HEYHELLOMANDOGYDOGDADDY"))
The output:
['HELLO', 'DOGMEME', 'DOGPAPADD']
['HEYHELLOMAN', 'DOGY', 'DOGDADDY']
Alternative solution with str.index() and str.rfind() functions:
def split_2_dogs(s):
if 'DOG' not in s: return [s,'']
pos1, pos2 = s.index('DOG'), s.rfind('DOG')
return [s[0:pos1], s[pos1:pos2], s[pos2:]]

This is pretty easy to do using the split function. For example, you can split any string by a delimiter, like dog, as so:
>>> chunks = 'HELLODOGMEMEDOGPAPA'.split('DOG')
>>> print(chunks)
['HELLO', 'MEME', 'PAPA']
You could then use the output of that in a list comprehension, like so:
>>> dog_chunks = chunks[:1] + ["DOG" + chunk for chunk in chunks[1:]]
>>> print(dog_chunks)
['HELLO', 'DOGMEME', 'DOGPAPA']
The only slightly tricky bit is making sure you don't prepend dog to the first string in the list, hence the little bits of slicing.

Split the string at 'DOG' and use conditions to get the desired result
s = 'HELLODOGMEMEDOGPAPADD'
l = s.split('DOG')
dl = ['DOG'+i for i in l[1:]]
[l[0]]+dl if l[0] else dl
Output:
['HELLO', 'DOGMEME', 'DOGPAPADD']

Splitting at DOG is the key!! This code will for all the cases that you have mentioned.
from itertools import izip_longest
words = 'HEYHELLODOGDADDY'
words = words.split("DOG")
words = ['DOG'+j if i>0 else j for i,j in enumerate(words)]
# words = ['HEYHELLO', 'DOGDADDY']
ans = ['','','']
# stitch words and ans together
ans = [m+n for m,n in izip_longest(words,ans,fillvalue='')]
print ans
Output :
['HEYHELLO', 'DOGDADDY', '']