Python Snippets （4）

#!/usr/bin/env python

eval_code = compile("100 + 200", "", "eval")

print eval(eval_code)

single_code = compile("print 'Hello World'", '', "single")

exec single_code

exec_code = compile("""

req = input('Count how many numbers?')

for eachNum in range(req):

    print eachNum

            """, '','exec')

exec exec_code

#! /usr/bin/env python

def foo():

    return True

def bar():

    'bar() does not much'

    return True

foo.__doc__ = 'foo() does not do much'

foo.tester = '''

if foo():

    print 'PASSED'

else:

    print 'FAILED'

'''

for eachAttr in dir():

    obj = eval(eachAttr) #convert the string to the object

    if isinstance(obj, type(foo)):

        if hasattr(obj, '__doc__'):

            print '\n Function "%s" has a doc string:\n\t%s' %(eachAttr, obj.__doc__)

        if hasattr(obj, 'tester'):

            print 'Function "%s" has a tester...executing ' %eachAttr

            exec obj.tester #to execute the strings

        else:

            print 'Funciton "%s" has no tester...skipping' %eachAttr

    else:

        print '"%s" is not a Function' %eachAttr

print '________________dir()__________________'

print dir()

try:

    _count = int(open("counter").read())

except IOError:

    _count = 0

def incrcounter(n):

    global _count

    _count = _count + n

def savecounter():

    open("counter", "w").write("%d" % _count)

import atexit

atexit.register(savecounter)

>>> m = re.findall(r'(the)(dog)' , 'bite thedog thedog')

>>> m

[('the', 'dog'), ('the', 'dog')]

>>>

>>> patt = '^((\w){3}(\d+))(BB)'

>>> m = re.match(patt, 'THU34324BB')

>>> m.groups()

('THU34324', 'U', '34324', 'BB')

>>>

一个解决办法是用“非贪婪”操作符，“?”. 这个操作符可以用在 “*”, “+”, 或 “?” 的

后面。它的作用是要求正则表达式引擎匹配的字符越少越好。因此，如果我们把“?”放在“.+”

的后面，我们就得到了想要的结果，

>>> patt = '.+?(\d+-\d+-\d+)'

>>> re.match(patt, data).group(1) # subgroup 1 # 子组 1

'1171590364-6-8'

#coding:utf-8

def foo(arg1,arg2="OK",*tupleArg,**dictArg):

    print "arg1=",arg1

    print "arg2=",arg2

    for i,element in enumerate(tupleArg):

        print "tupleArg %d-->%s" % (i,str(element))

    for  key in dictArg:

        print "dictArg %s-->%s" %(key,dictArg[key])

myList=["my1","my2"]

myDict={"name":"Tom","age":22}

foo("formal_args",arg2="argSecond",a = 2) #a = 2 is assigned to dictArg 

print "*"*40

foo(123,myList,myDict) # because there is no * or **, so this is the common assignment.

print "*"*40

foo(123,rt=123,*myList,**myDict) #this is no rt, so rt belongs to dictArg

-------------------

output:

arg1= formal_args

arg2= argSecond

dictArg a-->2

****************************************

arg1= 123

arg2= ['my1', 'my2']

tupleArg 0-->{'age': 22, 'name': 'Tom'}

****************************************

arg1= 123

arg2= my1

tupleArg 0-->my2

dictArg rt-->123

dictArg age-->22

dictArg name-->Tom

[Finished in 0.1s]

def f(a, L=[]):

    L.append(a)

    return L

print f(1)

print f(2)

print f(3)

This will print

[1]

[1, 2]

[1, 2, 3]

>>> def make_incrementor(n):

...     return lambda x: x + n

...

>>> f = make_incrementor(42)

>>> f(0)

42

>>> f(1)

43