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regex in python using OR for single or double quotation marks

I am trying to write regex in python for either single or double quotation marks from these examples:
animal="cat"
animal="horse"
animal='dog'
animal='cow'
It comes up empty when trying with |
re.compile("animal=\"|'(.+?)\"|'").findall
Please help. Thanks

You can take advantage of back-reference:
r = re.compile(r"""animal=(["'])(.+?)\1""")
This guarantees that the opening and closing characters are the same.
It's time to test it:
assert r.search('animal="cat"').group(2) == "cat"
assert r.search('animal="horse"').group(2) == "horse"
assert r.search("animal='dog'").group(2) == "dog"
assert r.search("animal='cow'").group(2) == "cow"

Your logical OR doesn't works on ' and " instead use a character class :
>>> s="""animal="cat"
...
... animal="horse"
...
... animal='dog'
...
... animal='cow'"""
>>>
>>> re.findall(r"""animal=["'](.+?)["']""",s)
['cat', 'horse', 'dog', 'cow']
>>>

Checking two string in python?

let two strings
s='chayote'
d='aceihkjouty'
the characters in string s is present in d Is there any built-in python function to accomplish this ?
Thanks In advance

Using sets:
>>> set("chayote").issubset("aceihkjouty")
True
Or, equivalently:
>>> set("chayote") <= set("aceihkjouty")
True

I believe you are looking for all and a generator expression:
>>> s='chayote'
>>> d='aceihkjouty'
>>> all(x in d for x in s)
True
>>>
The code will return True if all characters in string s can be found in string d.
Also, if string s contains duplicate characters, it would be more efficient to make it a set using set:
>>> s='chayote'
>>> d='aceihkjouty'
>>> all(x in d for x in set(s))
True
>>>

Try this
for i in s:
if i in d:
print i

How to get the first 2 letters of a string in Python?

Let's say I have a string
str1 = "TN 81 NZ 0025"
two = first2(str1)
print(two) # -> TN
How do I get the first two letters of this string? I need the first2 function for this.

It is as simple as string[:2]. A function can be easily written to do it, if you need.
Even this, is as simple as
def first2(s):
return s[:2]

In general, you can get the characters of a string from i until j with string[i:j].
string[:2] is shorthand for string[0:2]. This works for lists as well.
Learn about Python's slice notation at the official tutorial

t = "your string"
Play with the first N characters of a string with
def firstN(s, n=2):
return s[:n]
which is by default equivalent to
t[:2]

Heres what the simple function would look like:
def firstTwo(string):
return string[:2]

In python strings are list of characters, but they are not explicitly list type, just list-like (i.e. it can be treated like a list). More formally, they're known as sequence (see http://docs.python.org/2/library/stdtypes.html#sequence-types-str-unicode-list-tuple-bytearray-buffer-xrange):
>>> a = 'foo bar'
>>> isinstance(a, list)
False
>>> isinstance(a, str)
True
Since strings are sequence, you can use slicing to access parts of the list, denoted by list[start_index:end_index] see Explain Python's slice notation . For example:
>>> a = [1,2,3,4]
>>> a[0]
1 # first element, NOT a sequence.
>>> a[0:1]
[1] # a slice from first to second, a list, i.e. a sequence.
>>> a[0:2]
[1, 2]
>>> a[:2]
[1, 2]
>>> x = "foo bar"
>>> x[0:2]
'fo'
>>> x[:2]
'fo'
When undefined, the slice notation takes the starting position as the 0, and end position as len(sequence).
In the olden C days, it's an array of characters, the whole issue of dynamic vs static list sounds like legend now, see Python List vs. Array - when to use?

All previous examples will raise an exception in case your string is not long enough.
Another approach is to use
'yourstring'.ljust(100)[:100].strip().
This will give you first 100 chars.
You might get a shorter string in case your string last chars are spaces.

For completeness: Instead of using def you could give a name to a lambda function:
first2 = lambda s: s[:2]

python convert unicode to string

I got my results from sqlite by python, it's like this kind of tuples: (u'PR:000017512',)
However, I wanna print it as 'PR:000017512'. At first, I tried to select the first one in tuple by using index [0]. But the print out results is still u'PR:000017512'. Then I used str() to convert and nothing changed. How can I print this without u''?

You're confusing the string representation with its value. When you print a unicode string the u doesn't get printed:
>>> foo=u'abc'
>>> foo
u'abc'
>>> print foo
abc
Update:
Since you're dealing with a tuple, you don't get off this easy: You have to print the members of the tuple:
>>> foo=(u'abc',)
>>> print foo
(u'abc',)
>>> # If the tuple really only has one member, you can just subscript it:
>>> print foo[0]
abc
>>> # Join is a more realistic approach when dealing with iterables:
>>> print '\n'.join(foo)
abc

Don't see the problem:
>>> x = (u'PR:000017512',)
>>> print x
(u'PR:000017512',)
>>> print x[0]
PR:000017512
>>>
You the string is in unicode format, but it still means PR:000017512
Check out the docs on String literals
http://docs.python.org/2/reference/lexical_analysis.html#string-literals

In [22]: unicode('foo').encode('ascii','replace')
Out[22]: 'foo'

Alternative to python string item assignment

What is the best / correct way to use item assignment for python string ?
i.e s = "ABCDEFGH" s[1] = 'a' s[-1]='b' ?
Normal way will throw : 'str' object does not support item assignment

Strings are immutable. That means you can't assign to them at all. You could use formatting:
>>> s = 'abc{0}efg'.format('d')
>>> s
'abcdefg'
Or concatenation:
>>> s = 'abc' + 'd' + 'efg'
>>> s
'abcdefg'
Or replacement (thanks Odomontois for reminding me):
>>> s = 'abc0efg'
>>> s.replace('0', 'd')
'abcdefg'
But keep in mind that all of these methods create copies of the string, rather than modifying it in-place. If you want in-place modification, you could use a bytearray -- though that will only work for plain ascii strings, as alexis points out.
>>> b = bytearray('abc0efg')
>>> b[3] = 'd'
>>> b
bytearray(b'abcdefg')
Or you could create a list of characters and manipulate that. This is probably the most efficient and correct way to do frequent, large-scale string manipulation:
>>> l = list('abc0efg')
>>> l[3] = 'd'
>>> l
['a', 'b', 'c', 'd', 'e', 'f', 'g']
>>> ''.join(l)
'abcdefg'
And consider the re module for more complex operations.
String formatting and list manipulation are the two methods that are most likely to be correct and efficient IMO -- string formatting when only a few insertions are required, and list manipulation when you need to frequently update your string.

Since strings are "immutable", you get the effect of editing by constructing a modified version of the string and assigning it over the old value. If you want to replace or insert to a specific position in the string, the most array-like syntax is to use slices:
s = "ABCDEFGH"
s = s[:3] + 'd' + s[4:] # Change D to d at position 3
It's more likely that you want to replace a particular character or string with another. Do that with re, again collecting the result rather than modifying in place:
import re
s = "ABCDEFGH"
s = re.sub("DE", "--", s)

I guess this Object could help:
class Charray(list):
def __init__(self, mapping=[]):
"A character array."
if type(mapping) in [int, float, long]:
mapping = str(mapping)
list.__init__(self, mapping)
def __getslice__(self,i,j):
return Charray(list.__getslice__(self,i,j))
def __setitem__(self,i,x):
if type(x) <> str or len(x) > 1:
raise TypeError
else:
list.__setitem__(self,i,x)
def __repr__(self):
return "charray['%s']" % self
def __str__(self):
return "".join(self)
For example:
>>> carray = Charray("Stack Overflow")
>>> carray
charray['Stack Overflow']
>>> carray[:5]
charray['Stack']
>>> carray[-8:]
charray['Overflow']
>>> str(carray)
'Stack Overflow'
>>> carray[6] = 'z'
>>> carray
charray['Stack zverflow']

s = "ABCDEFGH" s[1] = 'a' s[-1]='b'
you can use like this
s=s[0:1]+'a'+s[2:]
this is very simple than other complex ways