I have short strings (tweets) in which I must extract all instances of mentions from the text and return a list of these instances including repeats.
extract_mentions('.#AndreaTantaros-supersleuth! You are a true journalistic professional. Keep up the great work! #MakeAmericaGreatAgain')
[AndreaTantaros]
How do I make it so that I remove all text after the first instance of punctuation after '#'? (In this case it would be '-') Note, punctuation can be varied. Please no use of regex.
I have used the following:
tweet_list = tweet.split()
mention_list = []
for word in tweet_list:
if '#' in word:
x = word.index('#')
y = word[x+1:len(word)]
if y.isalnum() == False:
y = word[x+1:-1]
mention_list.append(y)
else:
mention_list.append(y)
return mention_list
This would only work for instances with one extra character
import string
def extract_mentions(s, delimeters = string.punctuation + string.whitespace):
mentions = []
begin = s.find('#')
while begin >= 0:
end = begin + 1
while end < len(s) and s[end] not in delimeters:
end += 1
mentions.append(s[begin+1:end])
begin = s.find('#', end)
return mentions
>>> print(extract_mentions('.#AndreaTantaros-supersleuth! You are a true journalistic professional. Keep up the great work! #MakeAmericaGreatAgain'))
['AndreaTantaros']
Use string.punctuation module to get all punctuation chars.
Remove the first characters while they are punctuation (else the answer would be empty string all the time). Then find the first punctuation char.
This uses 2 loops with opposite conditions and a set for better speed.
z =".#AndreaTantaros-supersleuth! You are a true journalistic professional. Keep up the great work! #MakeAmericaGreatAgain') [AndreaTantaros]"
import string
# skip leading punctuation: find position of first non-punctuation
spun=set(string.punctuation) # faster if searched from a set
start_pos = 0
while z[start_pos] in spun:
start_pos+=1
end_pos = start_pos
while z[end_pos] not in spun:
end_pos+=1
print(z[start_pos:end_pos])
Just use regexp to match and extract part of the text.
Related
The problem is that มาก technically is in มาก็. Because มาก็ is มาก + ็.
So when I do
"แชมพูมาก็เยอะ".replace("มาก", " X ")
I end up with
แชมพู X ็เยอะ
And what I want
แชมพู X เยอะ
What I really want is to force the last character ก็ to count as a single character, so that มาก no longer matches มาก็.
While I haven't found a proper solution, I was able to find a solution. I split each string into separate (combined) characters via regex. Then I compare those lists to each other.
# Check is list is inside other list
def is_slice_in_list(s,l):
len_s = len(s) #so we don't recompute length of s on every iteration
return any(s == l[i:len_s+i] for i in range(len(l) - len_s+1))
def is_word_in_string(w, s):
a = regex.findall(u'\X', w)
b = regex.findall(u'\X', s)
return is_slice_in_list(a, b)
assert is_word_in_string("มาก็", "พูมาก็เยอะ") == True
assert is_word_in_string("มาก", "พูมาก็เยอะ") == False
The regex will split like this:
พู ม า ก็ เ ย อ ะ
ม า ก
And as it compares ก็ to ก the function figures the words are not the same.
I will mark as answered but if there is a nice or "proper" solution I will chose that one.
My question is more or less similar to:
Is there a way to substring a string in Python?
but it's more specifically oriented.
How can I get a par of a string which is located between two known words in the initial string.
Example:
mySrting = "this is the initial string"
Substring = "initial"
knowing that "the" and "string" are the two known words in the string that can be used to get the substring.
Thank you!
You can start with simple string manipulation here. str.index is your best friend there, as it will tell you the position of a substring within a string; and you can also start searching somewhere later in the string:
>>> myString = "this is the initial string"
>>> myString.index('the')
8
>>> myString.index('string', 8)
20
Looking at the slice [8:20], we already get close to what we want:
>>> myString[8:20]
'the initial '
Of course, since we found the beginning position of 'the', we need to account for its length. And finally, we might want to strip whitespace:
>>> myString[8 + 3:20]
' initial '
>>> myString[8 + 3:20].strip()
'initial'
Combined, you would do this:
startIndex = myString.index('the')
substring = myString[startIndex + 3 : myString.index('string', startIndex)].strip()
If you want to look for matches multiple times, then you just need to repeat doing this while looking only at the rest of the string. Since str.index will only ever find the first match, you can use this to scan the string very efficiently:
searchString = 'this is the initial string but I added the relevant string pair a few more times into the search string.'
startWord = 'the'
endWord = 'string'
results = []
index = 0
while True:
try:
startIndex = searchString.index(startWord, index)
endIndex = searchString.index(endWord, startIndex)
results.append(searchString[startIndex + len(startWord):endIndex].strip())
# move the index to the end
index = endIndex + len(endWord)
except ValueError:
# str.index raises a ValueError if there is no match; in that
# case we know that we’re done looking at the string, so we can
# break out of the loop
break
print(results)
# ['initial', 'relevant', 'search']
You can also try something like this:
mystring = "this is the initial string"
mystring = mystring.strip().split(" ")
for i in range(1,len(mystring)-1):
if(mystring[i-1] == "the" and mystring[i+1] == "string"):
print(mystring[i])
I suggest using a combination of list, split and join methods.
This should help if you are looking for more than 1 word in the substring.
Turn the string into array:
words = list(string.split())
Get the index of your opening and closing markers then return the substring:
open = words.index('the')
close = words.index('string')
substring = ''.join(words[open+1:close])
You may want to improve a bit with the checking for the validity before proceeding.
If your problem gets more complex, i.e multiple occurrences of the pair values, I suggest using regular expression.
import re
substring = ''.join(re.findall(r'the (.+?) string', string))
The re should store substrings separately if you view them in list.
I am using the spaces between the description to rule out the spaces between words, you can modify to your needs as well.
Input:
to-camel-case
to_camel_case
Desired output:
toCamelCase
My code:
def to_camel_case(text):
lst =['_', '-']
if text is None:
return ''
else:
for char in text:
if text in lst:
text = text.replace(char, '').title()
return text
Issues:
1) The input could be an empty string - the above code does not return '' but None;
2) I am not sure that the title()method could help me obtaining the desired output(only the first letter of each word before the '-' or the '_' in caps except for the first.
I prefer not to use regex if possible.
A better way to do this would be using a list comprehension. The problem with a for loop is that when you remove characters from text, the loop changes (since you're supposed to iterate over every item originally in the loop). It's also hard to capitalize the next letter after replacing a _ or - because you don't have any context about what came before or after.
def to_camel_case(text):
# Split also removes the characters
# Start by converting - to _, then splitting on _
l = text.replace('-','_').split('_')
# No text left after splitting
if not len(l):
return ""
# Break the list into two parts
first = l[0]
rest = l[1:]
return first + ''.join(word.capitalize() for word in rest)
And our result:
print to_camel_case("hello-world")
Gives helloWorld
This method is quite flexible, and can even handle cases like "hello_world-how_are--you--", which could be difficult using regex if you're new to it.
If my string is this: 'this is a string', how can I produce all possible combinations by joining each word with its neighboring word?
What this output would look like:
this is a string
thisis a string
thisisa string
thisisastring
thisis astring
this isa string
this isastring
this is astring
What I have tried:
s = 'this is a string'.split()
for i, l in enumerate(s):
''.join(s[0:i])+' '.join(s[i:])
This produces:
'this is a string'
'thisis a string'
'thisisa string'
'thisisastring'
I realize I need to change the s[0:i] part because it's statically anchored at 0 but I don't know how to move to the next word is while still including this in the output.
A simpler (and 3x faster than the accepted answer) way to use itertools product:
s = 'this is a string'
s2 = s.replace('%', '%%').replace(' ', '%s')
for i in itertools.product((' ', ''), repeat=s.count(' ')):
print(s2 % i)
You can also use itertools.product():
import itertools
s = 'this is a string'
words = s.split()
for t in itertools.product(range(len('01')), repeat=len(words)-1):
print(''.join([words[i]+t[i]*' ' for i in range(len(t))])+words[-1])
Well, it took me a little longer than I expected... this is actually tricker than I thought :)
The main idea:
The number of spaces when you split the string is the length or the split array - 1. In our example there are 3 spaces:
'this is a string'
^ ^ ^
We'll take a binary representation of all the options to have/not have either one of the spaces, so in our case it'll be:
000
001
011
100
101
...
and for each option we'll generate the sentence respectively, where 111 represents all 3 spaces: 'this is a string' and 000 represents no-space at all: 'thisisastring'
def binaries(n):
res = []
for x in range(n ** 2 - 1):
tmp = bin(x)
res.append(tmp.replace('0b', '').zfill(n))
return res
def generate(arr, bins):
res = []
for bin in bins:
tmp = arr[0]
i = 1
for digit in list(bin):
if digit == '1':
tmp = tmp + " " + arr[i]
else:
tmp = tmp + arr[i]
i += 1
res.append(tmp)
return res
def combinations(string):
s = string.split(' ')
bins = binaries(len(s) - 1)
res = generate(s, bins)
return res
print combinations('this is a string')
# ['thisisastring', 'thisisa string', 'thisis astring', 'thisis a string', 'this isastring', 'this isa string', 'this is astring', 'this is a string']
UPDATE:
I now see that Amadan thought of the same idea - kudos for being quicker than me to think about! Great minds think alike ;)
The easiest is to do it recursively.
Terminating condition: Schrödinger join of a single element list is that word.
Recurring condition: say that L is the Schrödinger join of all the words but the first. Then the Schrödinger join of the list consists of all elements from L with the first word directly prepended, and all elements from L with the first word prepended with an intervening space.
(Assuming you are missing thisis astring by accident. If it is deliberately, I am sure I have no idea what the question is :P )
Another, non-recursive way you can do it is to enumerate all numbers from 0 to 2^(number of words - 1) - 1, then use the binary representation of each number as a selector whether or not a space needs to be present. So, for example, the abovementioned thisis astring corresponds to 0b010, for "nospace, space, nospace".
I understand that str = str.replace('x', '') will eliminate all the x's.
But let's say I have a string jxjrxxtzxz and I only want to delete the first and last x making the string jjrxxtzz. This is not string specific. I want to be able to handle all strings, and not just that specific example.
edit: assume that x is the only letter I want to remove. Thank you!
One fairly straight forward way is to just use find and rfind to locate the characters to remove;
s = 'jxjrxxtzxz'
# Remove first occurrence of 'x'
ix = s.find('x')
if ix > -1:
s = s[:ix]+s[ix+1:]
# Remove last occurrence of 'x'
ix = s.rfind('x')
if ix > -1:
s = s[:ix]+s[ix+1:]
Not pretty but this will work:
def remove_first_last(c, s):
return s.replace(c,'', 1)[::-1].replace(c,'',1)[::-1]
Usage:
In [1]: remove_first_last('x', 'jxjrxxtzxz')
Out[1]: 'jjrxxtzz'