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Why doesn't replace () change all occurrences?

I have the following code:
dna = "TGCGAGAAGGGGCGATCATGGAGATCTACTATCCTCTCGGGGTATGGTGGGGTTGAGA"
print(dna.count("GAGA"))
dna = dna.replace("GAGA", "AGAG")
print(dna.count("GAGA"))
Replace does not replace all occurrences. Could somebody help my in understanding why it happened?

It replaces all occurences. That might lead to new occurences (look at your replacement string!).
I'd say, logically, all is fine.
You could repeat this replace while dna.count("GAGA") > 0 , but: that sounds not like what you should be doing. (I bet you really just want to do one round of replacement to simulate something specific happening. Not a genetics expert at all though.)

It did make all replacements (that's what .replace() does in Python unless specified otherwise), but some of these replacements inadvertently introduced new instances of GAGA. Take the beginning of your string:
TGCGAGAA
There's GAGA at indices 3-6. If you replace that with AGAG, you get
TGCAGAGA
So the last G from that AGAG, together with the subsequent A that was already there before, forms a new GAGA.

Replacements does not occur "until exhausted"; they occur when a substring is matched in your original string.
Consider the following from your string:
>>> a = "TGCGAGAA"
>>> a.replace("GAGA", "AGAG")
'TGCAGAGA'
>>>
The replacement does not happen again, since the original string did not match GAGA in that location.
If you want to do the replacement until no match is found, you can wrap it in a loop:
>>> while a.count("GAGA") > 0: # you probably don't want to use count here if the string is long because of performance considerations
... a = a.replace("GAGA", "AGAG")
...
>>> a
'TGCAAGAG'

How to remove a substrings from a list of strings?

I have a list of strings, all of which have a common property, they all go like this "pp:actual_string". I do not know for sure what the substring "pp:" will be, basically : acts as a delimiter; everything before : shouldn't be included in the result.
I have solved the problem using the brute force approach, but I would like to see a clever method, maybe something like regex.
Note : Some strings might not have this "pp:string" format, and could be already a perfect string, i.e. without the delimiter.
This is my current solution:
ll = ["pp17:gaurav","pp17:sauarv","pp17:there","pp17:someone"]
res=[]
for i in ll:
g=""
for j in range(len(i)):
if i[j] == ':':
index=j+1
res.append(i[index:len(i)])
print(res)
Is there a way that I can do it without creating an extra list ?

Whilst regex is an incredibly powerful tool with a lot of capabilities, using a "clever method" is not necessarily the best idea you are unfamiliar with its principles.
Your problem is one that can be solved without regex by splitting on the : character using the str.split() method, and just returning the last part by using the [-1] index value to represent the last (or only) string that results from the split. This will work even if there isn't a :.
list_with_prefixes = ["pp:actual_string", "perfect_string", "frog:actual_string"]
cleaned_list = [x.split(':')[-1] for x in list_with_prefixes]
print(cleaned_list)
This is a list comprehension that takes each of the strings in turn (x), splits the string on the : character, this returns a list containing the prefix (if it exists) and the suffix, and builds a new list with only the suffix (i.e. item [-1] in the list that results from the split. In this example, it returns:
['actual_string', 'perfect_string', 'actual_string']

Here are a few options, based upon different assumptions.
Most explicit
if s.startswith('pp:'):
s = s[len('pp:'):] # aka 3
If you want to remove anything before the first :
s = s.split(':', 1)[-1]
Regular expressions:
Same as startswith
s = re.sub('^pp:', '', s)
Same as split, but more careful with 'pp:' and slower
s = re.match('(?:^pp:)?(.*)', s).group(1)

Getting Substring after certain char in python

I want to cut a String such as "0011165.jpg_Fish" to get only Fish, so everything after the "_", how do i do that in python?
Thank you very much!

Please use str.partition instead of str.split. This is robust, since you can always expect 3 items, unlike, split which maybe tricky to handle if the input string doesn't have the split character,
>>> word = '0011165.jpg_Fish'
>>> not_required, split_char, required = word.partition('_')
>>> required
'Fish'

Try
"0011165.jpg_Fish".split("_")[1]
And in case of a Dataframe
train['Label'] = train.Image_Labels.str.split("_").str[1]

Searching for duplicates and remove them

sometimes I have a string like this
string = "Hett, Agva,"
and sometimes I will have duplicates in it.
string = "Hett, Agva, Delf, Agva, Hett,"
how can I check if my string has duplicates and then if it does remove them?
UPDATE.
So in the second string i need to remove Agva, and Hett, because there is 2x of them in the string

Iterate over the parts (words) and add each part to a set of seen parts and to a list of parts if it is not already in that set. Finally. reconstruct the string:
seen = set()
parts = []
for part in string.split(','):
if part.strip() not in seen:
seen.add(part.strip())
parts.append(part)
no_dups = ','.join(parts)
(note that I had to add some calls to .strip() as there are spaces at the start of some of the words which this method removes)
which gives:
'Hett, Agva, Delf,'
Why use a set?
To query whether an element is in a set, it is O(1) average case - since they are stored by a hash which makes lookup constant time. On the other hand, lookup in a list is O(n) as Python must iterate over the list until the element is found. This means that it is much more efficient for this task to use a set since, for each new word, you can instantly check to see if you have seen in before whereas you'd have to iterate over a list of seen elements otherwise which would take much longer for a large list.
Oh and to just check if there are duplicates, query whether the length of the split list is the same as the set of that list (which removes the duplicates but looses the order).
I.e.
def has_dups(string):
parts = string.split(',')
return len(parts) != len(set(parts))
which works as expected:
>>> has_dups('Hett, Agva,')
False
>>> has_dups('Hett, Agva, Delf, Agva, Hett,')
True

You can use toolz.unique, or equivalently the unique_everseen recipe in the itertools docs, or equivalently #JoeIddon's explicit solution.
Here's the solution using 3rd party toolz:
x = "Hett, Agva, Delf, Agva, Hett,"
from toolz import unique
res = ', '.join(filter(None, unique(x.replace(' ', '').split(','))))
print(res)
'Hett, Agva, Delf'
I've removed whitespace and used filter to clean up a trailing , which may not be required.

if you will receive a string in only this format then you can do the following:
import numpy as np
string_words=string.split(',')
uniq_words=np.unique(string_words)
string=""
for word in uniq_words:
string+=word+", "
string=string[:-1]
what this code does is that it splits words into a list, finds unique items, and then merges them into a string like before

If order of words id important then you can make a list of words in the string and then iterate over the list to make a new list of unique words.
string = "Hett, Agva, Delf, Agva, Hett,"
words_list = string.split()
unique_words = []
[unique_words.append(w) for w in words_list if w not in unique_words]
new_string = ' '.join(unique_words)
print (new_String)
Output:
'Hett, Agva, Delf,'

Quick and easy approach:
', '.join(
set(
filter( None, [ i.strip() for i in string.split(',') ] )
)
)
Hope it helps. Please feel free to ask if anything is not clear :)

How can I make multiple replacements in a string using a dictionary?

Suppose we have:
d = {
'Спорт':'Досуг',
'russianA':'englishA'
}
s = 'Спорт russianA'
How can I replace each appearance within s of any of d's keys, with the corresponding value (in this case, the result would be 'Досуг englishA')?

Using re:
import re
s = 'Спорт not russianA'
d = {
'Спорт':'Досуг',
'russianA':'englishA'
}
keys = (re.escape(k) for k in d.keys())
pattern = re.compile(r'\b(' + '|'.join(keys) + r')\b')
result = pattern.sub(lambda x: d[x.group()], s)
# Output: 'Досуг not englishA'
This will match whole words only. If you don't need that, use the pattern:
pattern = re.compile('|'.join(re.escape(k) for k in d.keys()))
Note that in this case you should sort the words descending by length if some of your dictionary entries are substrings of others.

You could use the reduce function:
reduce(lambda x, y: x.replace(y, dict[y]), dict, s)

Solution found here (I like its simplicity):
def multipleReplace(text, wordDict):
for key in wordDict:
text = text.replace(key, wordDict[key])
return text

one way, without re
d = {
'Спорт':'Досуг',
'russianA':'englishA'
}
s = 'Спорт russianA'.split()
for n,i in enumerate(s):
if i in d:
s[n]=d[i]
print ' '.join(s)

Almost the same as ghostdog74, though independently created. One difference,
using d.get() in stead of d[] can handle items not in the dict.
>>> d = {'a':'b', 'c':'d'}
>>> s = "a c x"
>>> foo = s.split()
>>> ret = []
>>> for item in foo:
... ret.append(d.get(item,item)) # Try to get from dict, otherwise keep value
...
>>> " ".join(ret)
'b d x'

With the warning that it fails if key has space, this is a compressed solution similar to ghostdog74 and extaneons answers:
d = {
'Спорт':'Досуг',
'russianA':'englishA'
}
s = 'Спорт russianA'
' '.join(d.get(i,i) for i in s.split())

I used this in a similar situation (my string was all in uppercase):
def translate(string, wdict):
for key in wdict:
string = string.replace(key, wdict[key].lower())
return string.upper()
hope that helps in some way... :)

Using regex
We can build a regular expression that matches any of the lookup dictionary's keys, by creating regexes to match each individual key and combine them with |. We use re.sub to do the substitution, by giving it a function to do the replacement (this function, of course, will do the dict lookup). Putting it together:
import re
# assuming global `d` and `s` as in the question
# a function that does the dict lookup with the global `d`.
def lookup(match):
return d[match.group()]
# Make the regex.
joined = '|'.join(re.escape(key) for key in d.keys())
pattern = re.compile(joined)
result = pattern.sub(lookup, s)
Here, re.escape is used to escape any characters with special meaning in the replacements (so that they don't interfere with building the regex, and are matched literally).
This regex pattern will match the substrings anywhere they appear, even if they are part of a word or span across multiple words. To avoid this, modify the regex so that it checks for word boundaries:
# pattern = re.compile(joined)
pattern = re.compile(rf'\b({joined})\b')
Using str.replace iteratively
Simply iterate over the .items() of the lookup dictionary, and call .replace with each. Since this method returns a new string, and does not (cannot) modify the string in place, we must reassign the results inside the loop:
for to_replace, replacement in d.items():
s = s.replace(to_replace, replacement)
This approach is simple to write and easy to understand, but it comes with multiple caveats.
First, it has the disadvantage that it works sequentially, in a specific order. That is, each replacement has the potential to interfere with other replacements. Consider:
s = 'one two'
s = s.replace('one', 'two')
s = s.replace('two', 'three')
This will produce 'three three', not 'two three', because the 'two' from the first replacement will itself be replaced in the second step. This is normally not desirable; however, in the rare case when it should work this way, this approach is the only practical one.
This approach also cannot easily be fixed to respect word boundaries, because it must match literal text, and a "word boundary" can be marked in multiple different ways - by varying kinds of whitespace, but also without text at the beginning and end of the string.
Finally, keep in mind that a dict is not an ideal data structure for this approach. If we will iterate over the dict, then its ability to do key lookup is useless; and in Python 3.5 and below, the order of dicts is not guaranteed (making the sequential replacement problem worse). Instead, it would be better to specify a list of tuples for the replacements:
d = [('Спорт', 'Досуг'), ('russianA', 'englishA')]
s = 'Спорт russianA'
for to_replace, replacement in d: # no more `.items()` call
s = s.replace(to_replace, replacement)
By tokenization
The problem becomes much simpler if the string is first cut into pieces (tokenized), in such a way that anything that should be replaced is now an exact match for a dict key. That would allow for using the dict's lookup directly, and processing the entire string in one go, while also not building a custom regex.
Suppose that we want to match complete words. We can use a simpler, hard-coded regex that will match whitespace, and which uses a capturing group; by passing this to re.split, we split the string into whitespace and non-whitespace sections. Thus:
import re
tokenizer = re.compile('([ \t\n]+)')
tokenized = tokenizer.split(s)
Now we look up each of the tokens in the dictionary: if present, it should be replaced with the corresponding value, and otherwise it should be left alone (equivalent to replacing it with itself). The dictionary .get method is a natural fit for this task. Finally, we join the pieces back up. Thus:
s = ''.join(d.get(token, token) for token in tokenized)
More generally, for example if the strings to replace could have spaces in them, a different tokenization rule will be needed. However, it will usually be possible to come up with a tokenization rule that is simpler than the regex from the first section (that matches all the keys by brute force).
Special case: replacing single characters
If the keys of the dict are all one character (technically, Unicode code point) each, there are more specific techniques that can be used. See Best way to replace multiple characters in a string? for details.