I have the following code:
output = requests.get(url=url, auth=oauth, headers=headers, data=payload)
output_data = output.content
type(output_date)
<class 'bytes'>
output_data
Squeezed Text (3632 Lines)
When looking at the squeezed text, I have some values that look like this:
Steve likes to walk his dog. Steve says to John "I like \n Pineapple, oranges, \n and pizza.\n" and then he went to bed \n.
John likes his beer cold.\n
Sally likes her teeth brushed with a bottle of jack.\n
How can I remove the \n characters, but ONLY if it is contained within double quotes, so that my results look like this:
Steve likes to walk his dog. Steve says to John "I like Pineapple, oranges, and pizza." and then he went to bed \n.
John likes his beer cold.\n
Sally likes her teeth brushed with a bottle of jack.\n
I know how to remove \n characters, but I am not sure how to do this if I only want to remove the values if they are contained within double quotes.
Here is what I have tries:
I found this, and used this code:
my_text = re.sub(r'"\\n"','',my_text)
But it doesn't seem to be working.
I might be complicating it a bit, but something like this might work
parts = content.split("\"")
for i, part in enumerate(parts):
if i % 2:
parts[i] = part.replace("\n", "")
content = "\"".join(parts)
Figured it out.
Steps:
Convert bytes to String
Create the pattern for Regex
Use regex to format the values.
Step 1:
my_text = my_text.decode("utf-8")
Step 2:
pattern = re.compile(r'".*?"',re.DOTALL)
Step 3:
my_text = pattern.sub(lambda x:x.group().replace('\n',''),my_text)
This solves my problem.
Would it be possible to look at a specific n-grams from the whole list of them and look for it in a list of sentences?
For example:
I have the following sentences (from a dataframe column):
example = ['Mary had a little lamb. Jack went up the hill' ,
'Jack went to the beach' ,
'i woke up suddenly' ,
'it was a really bad dream...']
and n-grams (bigrams) got from
word_v = CountVectorizer(ngram_range=(2,2), analyzer='word')
mat = word_v r.fit_transform(df['Example'])
frequencies = sum(mat).toarray()[0]
which generates the output of the n-grams frequency.
I would like to select
the most frequent bi-grams
a bi-gram selected manually
within the list above example.
So, let's say that the most frequent bi-gram is Jack went, how could I look for it in the example above?
Also, if I want to look, not at the most frequent bi-grams but at the hill/beach in the example, how could I do it?
To select the rows that have the most frequent ngrams in it, you can do:
df.loc[mat.toarray()[:, frequencies==frequencies.max()].astype(bool)]
Example
0 Mary had a little lamb. Jack went up the hill
1 Jack went to the beach
but if two ngrams have the max frequency, you would get all the rows where both are present.
If you want the top/hill 3 and all the rows that have any of them:
top = 3
print (df.loc[mat.toarray()[:, np.argsort(frequencies)][:, -top:].any(axis=1)])
Example
0 Mary had a little lamb. Jack went up the hill
1 Jack went to the beach
2 i woke up suddenly
3 it was a really bad dream...
#here it is all the rows with the example
hill = 3
print (df.loc[mat.toarray()[:, np.argsort(frequencies)][:, :hill].any(axis=1)])
Example
1 Jack went to the beach
3 it was a really bad dream...
Finally if you want a specific ngrams:
ng = 'it was'
df.loc[mat.toarray()[:, np.array(word_v.get_feature_names())==ng].astype(bool)]
Example
3 it was a really bad dream...
I would like reconstruct full names from photo captions using Regex in Python, by appending last name back to the first name in patterns "FirstName1 and FirstName2 LastName". We can rely on names starting with capital letter.
For example,
'John and Albert McDonald' becomes 'John McDonald' and 'Albert McDonald'
'Stephen Stewart, John and Albert Diamond' becomes 'John Diamond' and 'Albert Diamond'
I would need to avoid matching patterns like this: 'Jay Smith and Albert Diamond' and generate a non-existent name 'Smith Diamond'
The photo captions may or may not have more words before this pattern, for example, 'It was a great day hanging out with John and Stephen Diamond.'
This is the code I have so far:
s = 'John and Albert McDonald'
so = re.search('([A-Z][a-z\-]+)\sand\s([A-Z][a-z\-]+\s[A-Z][a-z\-]+(?:[A-Z][a-z]+)?)', s)
if so:
print so.group(1) + ' ' + so.group(2).split()[1]
print so.group(2)
This returns 'John McDonald' and 'Albert McDonald', but 'Jay Smith and Albert Diamond' will result in a non-existent name 'Smith Diamond'.
An idea would be to check whether the pattern is preceded by a capitalized word, something like (?<![A-Z][a-z\-]+)\s([A-Z][a-z\-]+)\sand\s([A-Z][a-z\-]+\s[A-Z][a-z\-]+(?:[A-Z][a-z]+)?) but unfortunately negative lookbehind only works if we know the exact length of the preceding word, which I don't.
Could you please let me know how I can correct my regex epression? Or is there a better way to do what I want? Thanks!
As you can rely on names starting with a capital letter, then you could do something like:
((?:[A-Z]\w+\s+)+)and\s+((?:[A-Z]\w+(?:\s+|\b))+)
Live preview
Swapping out your current pattern, with this pattern should work with your current Python code. You do need to strip() the captured results though.
Which for your examples and current code would yield:
Input
First print
Second print
John and Albert McDonald
John McDonald
Albert McDonald
Stephen Stewart, John and Albert Diamond
John Diamond
Albert Diamond
It was a great day hanging out with John and Stephen Diamond.
John Diamond
Stephen Diamond
The goals of the function is to split one single string into multiple lines to make it more readable. The goal is to replace the first space found after at least n characters (since the beginning of the string, or since the last "\n" dropped in the string)
Hp:
you can assume no \n in the string
Example
Marcus plays soccer in the afternoon
f(10) should result in
Marcus plays\nsoccer in\nthe afternoon
The first space in Marcus plays soccer in the afternoonis skipped because Marcus is only 5 chars long. We put then a \n after plays and we start counting again. The space after soccer is therefore skipped, etc.
So far tried
def replace_space_w_newline_every_n_chars(n,s):
return re.sub("(?=.{"+str(n)+",})(\s)", "\\1\n", s, 0, re.DOTALL)
inspired by this
Try replacing
(.{10}.*?)\s
with
$1\n
Check it out here.
Example:
>>> import re
>>> s = 'Marcus plays soccer in the afternoo
>>> re.sub(r'(.{9}.*?)\s', r'\1\n', s)
'Marcus plays\nsoccer in\nthe afternoon'
I got a string of such format:
"Wilbur Smith (Billy, son of John), Eddie Murphy (John), Elvis Presley, Jane Doe (Jane Doe)"
so basicly it's list of actor's names (optionally followed by their role in parenthesis). The role itself can contain comma (actor's name can not, I strongly hope so).
My goal is to split this string into a list of pairs - (actor name, actor role).
One obvious solution would be to go through each character, check for occurances of '(', ')' and ',' and split it whenever a comma outside occures. But this seems a bit heavy...
I was thinking about spliting it using a regexp: first split the string by parenthesis:
import re
x = "Wilbur Smith (Billy, son of John), Eddie Murphy (John), Elvis Presley, Jane Doe (Jane Doe)"
s = re.split(r'[()]', x)
# ['Wilbur Smith ', 'Billy, son of John', ', Eddie Murphy ', 'John', ', Elvis Presley, Jane Doe ', 'Jane Doe', '']
The odd elements here are actor names, even are the roles. Then I could split the names by commas and somehow extract the name-role pairs. But this seems even worse then my 1st approach.
Are there any easier / nicer ways to do this, either with a single regexp or a nice piece of code?
One way to do it is to use findall with a regex that greedily matches things that can go between separators. eg:
>>> s = "Wilbur Smith (Billy, son of John), Eddie Murphy (John), Elvis Presley, Jane Doe (Jane Doe)"
>>> r = re.compile(r'(?:[^,(]|\([^)]*\))+')
>>> r.findall(s)
['Wilbur Smith (Billy, son of John)', ' Eddie Murphy (John)', ' Elvis Presley', ' Jane Doe (Jane Doe)']
The regex above matches one or more:
non-comma, non-open-paren characters
strings that start with an open paren, contain 0 or more non-close-parens, and then a close paren
One quirk about this approach is that adjacent separators are treated as a single separator. That is, you won't see an empty string. That may be a bug or a feature depending on your use-case.
Also note that regexes are not suitable for cases where nesting is a possibility. So for example, this would split incorrectly:
"Wilbur Smith (son of John (Johnny, son of James), aka Billy), Eddie Murphy (John)"
If you need to deal with nesting your best bet would be to partition the string into parens, commas, and everthing else (essentially tokenizing it -- this part could still be done with regexes) and then walk through those tokens reassembling the fields, keeping track of your nesting level as you go (this keeping track of the nesting level is what regexes are incapable of doing on their own).
s = re.split(r',\s*(?=[^)]*(?:\(|$))', x)
The lookahead matches everything up to the next open-parenthesis or to the end of the string, iff there's no close-parenthesis in between. That ensures that the comma is not inside a set of parentheses.
I think the best way to approach this would be to use python's built-in csv module.
Because the csv module only allows a one character quotechar, you would need to do a replace on your inputs to convert () to something like | or ". Then make sure you are using an appropriate dialect and off you go.
An attempt on human-readable regex:
import re
regex = re.compile(r"""
# name starts and ends on word boundary
# no '(' or commas in the name
(?P<name>\b[^(,]+\b)
\s*
# everything inside parentheses is a role
(?:\(
(?P<role>[^)]+)
\))? # role is optional
""", re.VERBOSE)
s = ("Wilbur Smith (Billy, son of John), Eddie Murphy (John), Elvis Presley,"
"Jane Doe (Jane Doe)")
print re.findall(regex, s)
Output:
[('Wilbur Smith', 'Billy, son of John'), ('Eddie Murphy', 'John'),
('Elvis Presley', ''), ('Jane Doe', 'Jane Doe')]
My answer will not use regex.
I think simple character scanner with state "in_actor_name" should work. Remember then state "in_actor_name" is terminated either by ')' or by comma in this state.
My try:
s = 'Wilbur Smith (Billy, son of John), Eddie Murphy (John), Elvis Presley, Jane Doe (Jane Doe)'
in_actor_name = 1
role = ''
name = ''
for c in s:
if c == ')' or (c == ',' and in_actor_name):
in_actor_name = 1
name = name.strip()
if name:
print "%s: %s" % (name, role)
name = ''
role = ''
elif c == '(':
in_actor_name = 0
else:
if in_actor_name:
name += c
else:
role += c
if name:
print "%s: %s" % (name, role)
Output:
Wilbur Smith: Billy, son of John
Eddie Murphy: John
Elvis Presley:
Jane Doe: Jane Doe
Here's a general technique I've used in the past for such cases:
Use the sub function of the re module with a function as replacement argument. The function keeps track of opening and closing parens, brackets and braces, as well as single and double quotes, and performs a replacement only outside of such bracketed and quoted substrings. You can then replace the non-bracketed/quoted commas with another character which you're sure doesn't appear in the string (I use the ASCII/Unicode group-separator: chr(29) code), then do a simple string.split on that character. Here's the code:
import re
def srchrepl(srch, repl, string):
"""Replace non-bracketed/quoted occurrences of srch with repl in string"""
resrchrepl = re.compile(r"""(?P<lbrkt>[([{])|(?P<quote>['"])|(?P<sep>["""
+ srch + """])|(?P<rbrkt>[)\]}])""")
return resrchrepl.sub(_subfact(repl), string)
def _subfact(repl):
"""Replacement function factory for regex sub method in srchrepl."""
level = 0
qtflags = 0
def subf(mo):
nonlocal level, qtflags
sepfound = mo.group('sep')
if sepfound:
if level == 0 and qtflags == 0:
return repl
else:
return mo.group(0)
elif mo.group('lbrkt'):
level += 1
return mo.group(0)
elif mo.group('quote') == "'":
qtflags ^= 1 # toggle bit 1
return "'"
elif mo.group('quote') == '"':
qtflags ^= 2 # toggle bit 2
return '"'
elif mo.group('rbrkt'):
level -= 1
return mo.group(0)
return subf
If you don't have nonlocal in your version of Python, just change it to global and define level and qtflags at the module level.
Here's how it's used:
>>> GRPSEP = chr(29)
>>> string = "Wilbur Smith (Billy, son of John), Eddie Murphy (John), Elvis Presley, Jane Doe (Jane Doe)"
>>> lst = srchrepl(',', GRPSEP, string).split(GRPSEP)
>>> lst
['Wilbur Smith (Billy, son of John)', ' Eddie Murphy (John)', ' Elvis Presley', ' Jane Doe (Jane Doe)']
This post helped me a lot. I was looking to split a string by commas positioned outside quotes. I used this as a starter. My final line of code was regEx = re.compile(r'(?:[^,"]|"[^"]*")+') This did the trick. Thanks a ton.
I certainly agree with #Wogan above, that using the CSV moudle is a good approach. Having said that if you still want to try a regex solution give this a try, but you will have to adapt it to Python dialect
string.split(/,(?=(?:[^\"]*\"[^\"]*\")*(?![^\"]*\"))/)
HTH
split by ")"
>>> s="Wilbur Smith (Billy, son of John), Eddie Murphy (John), Elvis Presley, Jane Doe (Jane Doe)"
>>> s.split(")")
['Wilbur Smith (Billy, son of John', ', Eddie Murphy (John', ', Elvis Presley, Jane Doe (Jane Doe', '']
>>> for i in s.split(")"):
... print i.split("(")
...
['Wilbur Smith ', 'Billy, son of John']
[', Eddie Murphy ', 'John']
[', Elvis Presley, Jane Doe ', 'Jane Doe']
['']
you can do further checking to get those names that doesn't come with ().
None of the answers above are correct if there are any errors or noise in your data.
It's easy to come up with a good solution if you know the data is right every time. But what happens if there are formatting errors? What do you want to have happen?
Suppose there are nesting parentheses? Suppose there are unmatched parentheses? Suppose the string ends with or begins with a comma, or has two in a row?
All of the above solutions will produce more or less garbage and not report it to you.
Were it up to me, I'd start with a pretty strict restriction on what "correct" data was - no nesting parentheses, no unmatched parentheses, and no empty segments before, between or after comments - validate as I went, and then raise an exception if I wasn't able to validate.