Simple example:
from sympy import *
n = symbols('n',integer=True)
x,y = symbols('x,y')
expression = exp(I*n*x)
expression.subs(exp(I*n),y)
#>> exp(I*n*x)
Why doesn't SymPy replace the exp(I*x) to give y**n? It works perfectly fine the other way around (replacing y in y**n with exp(I*x) to give exp(I*n*x)). Is there a hack to get this done?
subs might not work in general if the expression you are replacing doesn't appear exactly in the expression.
One workaround is to use replace with pattern matching to replace a more general pattern
>>> a = Wild('a')
>>> expression.replace(exp(I*n*a), y**a)
y**x
Related
While matching an email address, after I match something like yasar#webmail, I want to capture one or more of (\.\w+)(what I am doing is a little bit more complicated, this is just an example), I tried adding (.\w+)+ , but it only captures last match. For example, yasar#webmail.something.edu.tr matches but only include .tr after yasar#webmail part, so I lost .something and .edu groups. Can I do this in Python regular expressions, or would you suggest matching everything at first, and split the subpatterns later?
re module doesn't support repeated captures (regex supports it):
>>> m = regex.match(r'([.\w]+)#((\w+)(\.\w+)+)', 'yasar#webmail.something.edu.tr')
>>> m.groups()
('yasar', 'webmail.something.edu.tr', 'webmail', '.tr')
>>> m.captures(4)
['.something', '.edu', '.tr']
In your case I'd go with splitting the repeated subpatterns later. It leads to a simple and readable code e.g., see the code in #Li-aung Yip's answer.
You can fix the problem of (\.\w+)+ only capturing the last match by doing this instead: ((?:\.\w+)+)
This will work:
>>> regexp = r"[\w\.]+#(\w+)(\.\w+)?(\.\w+)?(\.\w+)?(\.\w+)?(\.\w+)?"
>>> email_address = "william.adama#galactica.caprica.fleet.mil"
>>> m = re.match(regexp, email_address)
>>> m.groups()
('galactica', '.caprica', '.fleet', '.mil', None, None)
But it's limited to a maximum of six subgroups. A better way to do this would be:
>>> m = re.match(r"[\w\.]+#(.+)", email_address)
>>> m.groups()
('galactica.caprica.fleet.mil',)
>>> m.group(1).split('.')
['galactica', 'caprica', 'fleet', 'mil']
Note that regexps are fine so long as the email addresses are simple - but there are all kinds of things that this will break for. See this question for a detailed treatment of email address regexes.
This is what you are looking for:
>>> import re
>>> s="yasar#webmail.something.edu.tr"
>>> r=re.compile("\.\w+")
>>> m=r.findall(s)
>>> m
['.something', '.edu', '.tr']
I'd like to know how to search using regular expression.
this is my code
import re
data = "python one test code"
p = re.compile("^python\s\w+")
result
print(p.findall(data))
['python one']
The result I want to get is as below
print(p.findall(data))
['python one test code']
I can get the above result if as below
p = re.compile("^python\s\w+\s\w+\s\w+")
but I don't want to repeat "\s\w+" like as "^python\s\w+\s\w+\s\w+"
How can I get the result using * or + in "\s\w+" ?
You can try this:
^python(?:\s\w+)+
Explanation
^python string starts with python
?: the () group will not be captured
\s\w+ will match a space and the immediate word
(?:\s\w+)+ the outer plus will match all occurrence of no.3 where + means one or more
You can try following
p = re.compile("^python[\w\s]*")
I'm the worst for regex in general, but in python... I need help in fixing my regex for parsing filenames, e.g:
>>> from re import search, I, M
>>> x="/almac/data/vectors_puces_T12_C1_00_d2v_H50_corr_m10_70.mtx"
>>> for i in range(6):
... print search(r"[vectors|pairs]+_(\w+[\-\w+]*[0-9]{0,4})([_T[0-9]{2,3}_C[1-9]_[0-9]{2}]?)(_[d2v|w2v|coocc\w*|doc\w*]*)(_H[0-9]{1,4})(_[sub|co[nvs{0,2}|rr|nc]+]?)(_m[0-9]{1,3}[_[0-9]{0,3}]?)",x, M|I).group(i)
...
It gives the following output:
vectors_puces_T12_C1_00_d2v_H50_corr_m10_70
puces_T
12_C1_00
_d2v
_H50
_corr
However, what I need is
vectors_puces_T12_C1_00_d2v_H50_corr_m10_70
puces
T12_C1_00
_d2v
_H50
_corr
I don't know what exactly is wrong. Thank you
One problem is that \w would also match underscore which you want to be a delimiter between puces and T12_C1_00 in this case. Replace the \w with A-Za-z\-. Also, you should put the underscore between the appropriate saving groups:
(?:vectors|pairs)_([A-Za-z\-]+[0-9]{0,4})_([T[0-9]{2,3}_C[1-9]_[0-9]{2}]?)...
HERE^
Works for me:
>>> import re
>>> re.search(r"(?:vectors|pairs)_([A-Za-z\-]+[0-9]{0,4})_([T[0-9]{2,3}_C[1-9]_[0-9]{2}]?)(_[d2v|w2v|coocc\w*|doc\w*]*)(_H[0-9]{1,4})(_[sub|co[nvs{0,2}|rr|nc]+]?)(_m[0-9]{1,3}[_[0-9]{0,3}]?)",x, re.M|re.I).groups()
('puces', 'T12_C1_00', '_d2v', '_H50', '_corr', '_m10_70')
I've also replaced the [vectors|pairs] with (?:vectors|pairs) which is, I think, what you've actually meant - match either vectors or pairs literal strings, (?:...) is a syntax for a non-capturing group.
I'm not sure what your goal is, but you seem to be interested in what's between each underscore, so it may be simpler to split by it:
path, filename = os.path.split(x)
filename = filename.split('.')
fileparts = filename.split('_')
fileparts will then be this list:
vectors
puces
T12
C1
00
d2v
H50
corr
m10
70
And you can validate / inspect any part, e.g. if fileparts[0] == 'vectors' or tpart = fileparts[2:4]...
I'd like your opinion as you might be more experienced on Python as I do.
I came from C++ and I'm still not used to the Pythonic way to do things.
I want to loop under a string, between 2 capital letters. For example, I could do that this way:
i = 0
str = "PythonIsFun"
for i, z in enumerate(str):
if(z.isupper()):
small = ''
x = i + 1
while(not str[x].isupper()):
small += str[x]
I wrote this on my phone, so I don't know if this even works but you caught the idea, I presume.
I need you to help me get the best results on this, not just in a non-forced way to the cpu but clean code too. Thank you very much
This is one of those times when regexes are the best bet.
(And don't call a string str, by the way: it shadows the built-in function.)
s = 'PythonIsFun'
result = re.search('[A-Z]([a-z]+)[A-Z]', s)
if result is not None:
print result.groups()[0]
you could use regular expressions:
import re
re.findall ( r'[A-Z]([^A-Z]+)[A-Z]', txt )
outputs ['ython'], and
re.findall ( r'(?=[A-Z]([^A-Z]+)[A-Z])', txt )
outputs ['ython', 's']; and if you just need the first match,
re.search ( r'[A-Z]([^A-Z]+)[A-Z]', txt ).group( 1 )
You can use a list comprehension to do this easily.
>>> s = "PythonIsFun"
>>> u = [i for i,x in enumerate(s) if x.isupper()]
>>> s[u[0]+1:u[1]]
'ython'
If you can't guarantee that there are two upper case characters you can check the length of u to make sure it is at least 2. This does iterate over the entire string, which could be a problem if the two upper case characters occur at the start of a lengthy string.
There are many ways to tackle this, but I'd use regular expressions.
This example will take "PythonIsFun" and return "ythonsun"
import re
text = "PythonIsFun"
pattern = re.compile(r'[a-z]') #look for all lower-case characters
matches = re.findall(pattern, text) #returns a list of lower-chase characters
lower_string = ''.join(matches) #turns the list into a string
print lower_string
outputs:
ythonsun
I'm trying to extract set of data from a string that can match one of three patterns. I have a list of compiled regexes. I want to run through them (in order) and go with the first match.
regexes = [
compiled_regex_1,
compiled_regex_2,
compiled_regex_3,
]
m = None
for reg in regexes:
m = reg.match(name)
if m: break
if not m:
print 'ARGL NOTHING MATCHES THIS!!!'
This should work (haven't tested yet) but it's pretty fugly. Is there a better way of boiling down a loop that breaks when it succeeds or explodes when it doesn't?
There might be something specific to re that I don't know about that allows you to test multiple patterns too.
You can use the else clause of the for loop:
for reg in regexes:
m = reg.match(name)
if m: break
else:
print 'ARGL NOTHING MATCHES THIS!!!'
If you just want to know if any of the regex match then you could use the builtin any function:
if any(reg.match(name) for reg in regexes):
....
however this will not tell you which regex matched.
Alternatively you can combine multiple patterns into a single regex with |:
regex = re.compile(r"(regex1)|(regex2)|...")
Again this will not tell you which regex matched, but you will have a match object that you can use for further information. For example you can find out which of the regex succeeded from the group that is not None:
>>> match = re.match("(a)|(b)|(c)|(d)", "c")
>>> match.groups()
(None, None, 'c', None)
However this can get complicated however if any of the sub-regex have groups in them as well, since the numbering will be changed.
This is probably faster than matching each regex individually since the regex engine has more scope for optimising the regex.
Since you have a finite set in this case, you could use short ciruit evaluation:
m = compiled_regex_1.match(name) or
compiled_regex_2.match(name) or
compiled_regex_3.match(name) or
print("ARGHHHH!")
In Python 2.6 or better:
import itertools as it
m = next(it.ifilter(None, (r.match(name) for r in regexes)), None)
The ifilter call could be made into a genexp, but only a bit awkwardly, i.e., with the usual trick for name binding in a genexp (aka the "phantom nested for clause idiom"):
m = next((m for r in regexes for m in (r.match(name),) if m), None)
but itertools is generally preferable where applicable.
The bit needing 2.6 is the next built-in, which lets you specify a default value if the iterator is exhausted. If you have to simulate it in 2.5 or earlier,
def next(itr, deft):
try: return itr.next()
except StopIteration: return deft
I use something like Dave Kirby suggested, but add named groups to the regexps, so that I know which one matched.
regexps = {
'first': r'...',
'second': r'...',
}
compiled = re.compile('|'.join('(?P<%s>%s)' % item for item in regexps.iteritems()))
match = compiled.match(my_string)
print match.lastgroup
Eric is in better track in taking bigger picture of what OP is aiming, I would use if else though. I would also think that using print function in or expression is little questionable. +1 for Nathon of correcting OP to use proper else statement.
Then my alternative:
# alternative to any builtin that returns useful result,
# the first considered True value
def first(seq):
for item in seq:
if item: return item
regexes = [
compiled_regex_1,
compiled_regex_2,
compiled_regex_3,
]
m = first(reg.match(name) for reg in regexes)
print(m if m else 'ARGL NOTHING MATCHES THIS!!!')