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from package.sub import m2

instead. Relative imports can lead to a module

being initialized twice, leading to confusing bugs. See PEP 328 for details.

It is sometimes necessary to move imports to a function or class to avoid problems with

circular imports. Gordon McMillan says:

Circular imports are fine where both modules use the “import <module>” form of

import. They fail when the 2nd module wants to grab a name out of the first (“from

module import name”) and the import is at the top level. That’s because names in

the 1st are not yet available, because the first module is busy importing the 2nd.

In this case, if the second module is only used in one function, then the import can easily

be moved into that function. By the time the import is called, the first module will have

finished initializing, and the second module can do its import.

It may also be necessary to move imports out of the top level of code if some of the

modules are platform-specific. In that case, it may not even be possible to import all of

the modules at the top of the file. In this case, importing the correct modules in the

corresponding platform-specific code is a good option.

Only move imports into a local scope, such as inside a function definition, if it’s necessary

to solve a problem such as avoiding a circular import or are trying to reduce the

initialization time of a module. This technique is especially helpful if many of the imports

are unnecessary depending on how the program executes. You may also want to move

imports into a function if the modules are only ever used in that function. Note that

loading a module the first time may be expensive because of the one time initialization of

the module, but loading a module multiple times is virtually free, costing only a couple of

dictionary lookups. Even if the module name has gone out of scope, the module is

probably available in

sys.modules

If only instances of a specific class use a module, then it is reasonable to import the

module in the class’s

__init__

method and then assign the module to an instance

variable so that the module is always available (via that instance variable) during the life

of the object. Note that to delay an import until the class is instantiated, the import must

be inside a method. Putting the import inside the class but outside of any method still

causes the import to occur when the module is initialized.

How can I pass optional or keyword parameters from one function

to another?

Collect the arguments using the

and

specifiers in the function’s parameter list; this

gives you the positional arguments as a tuple and the keyword arguments as a dictionary.

You can then pass these arguments when calling another function by using

and

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def f(x, *args, **kwargs):

...

kwargs['width'] = '14.3c'

...

g(x, *args, **kwargs)

In the unlikely case that you care about Python versions older than 2.0, use

apply()

def f(x, *args, **kwargs):

...

kwargs['width'] = '14.3c'

...

apply(g, (x,)+args, kwargs)

What is the difference between arguments and parameters?

Parameters are defined by the names that appear in a function definition, whereas

arguments are the values actually passed to a function when calling it. Parameters define

what types of arguments a function can accept. For example, given the function

definition:

def func(foo, bar=None, **kwargs):

pass

foo, bar and kwargs are parameters of

func

. However, when calling

func

, for example:

func(42, bar=314, extra=somevar)

the values

314

, and

somevar

are arguments.

How do I write a function with output parameters (call by

reference)?

Remember that arguments are passed by assignment in Python. Since assignment just

creates references to objects, there’s no alias between an argument name in the caller

and callee, and so no call-by-reference per se. You can achieve the desired effect in a

number of ways.

1. By returning a tuple of the results:

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def func2(a, b):

a = 'new-value' # a and b are local names

b = b + 1 # assigned to new objects

return a, b # return new values

x, y = 'old-value', 99

x, y = func2(x, y)

print x, y # output: new-value 100

This is almost always the clearest solution.

2. By using global variables. This isn’t thread-safe, and is not recommended.

3. By passing a mutable (changeable in-place) object:

def func1(a):

a[0] = 'new-value' # 'a' references a mutable list

a[1] = a[1] + 1 # changes a shared object

args = ['old-value', 99]

func1(args)

print args[0], args[1] # output: new-value 100

4. By passing in a dictionary that gets mutated:

def func3(args):

args['a'] = 'new-value' # args is a mutable dictionary

args['b'] = args['b'] + 1 # change it in-place

args = {'a':' old-value', 'b': 99}

func3(args)

print args['a'], args['b']

5. Or bundle up values in a class instance:

class callByRef:

def __init__(self, **args):

for (key, value) in args.items():

setattr(self, key, value)

def func4(args):

args.a = 'new-value' # args is a mutable callByRef

args.b = args.b + 1 # change object in-place

args = callByRef(a='old-value', b=99)

func4(args)

print args.a, args.b

There’s almost never a good reason to get this complicated.

Your best choice is to return a tuple containing the multiple results.

How do you make a higher order function in Python?

You have two choices: you can use nested scopes or you can use callable objects. For

example, suppose you wanted to define

linear(a,b)

which returns a function

f(x)

that

computes the value

a*x+b

. Using nested scopes:

def linear(a, b):

def result(x):

return a * x + b

return result

Or using a callable object:

class linear:

def __init__(self, a, b):

self.a, self.b = a, b

def __call__(self, x):

return self.a * x + self.b

In both cases,

taxes = linear(0.3, 2)

gives a callable object where

taxes(10e6) == 0.3 * 10e6 + 2

The callable object approach has the disadvantage that it is a bit slower and results in

slightly longer code. However, note that a collection of callables can share their signature

via inheritance:

class exponential(linear):

# __init__ inherited

def __call__(self, x):

return self.a * (x ** self.b)

Object can encapsulate state for several methods:

class counter:

value = 0

def set(self, x):

self.value = x

def up(self):

self.value = self.value + 1

def down(self):

self.value = self.value - 1

count = counter()

inc, dec, reset = count.up, count.down, count.set

Here

inc()

dec()

and

reset()

act like functions which share the same counting

variable.

How do I copy an object in Python?

In general, try

copy.copy()

copy.deepcopy()

for the general case. Not all objects can

be copied, but most can.

Some objects can be copied more easily. Dictionaries have a

copy()

method:

newdict = olddict.copy()

Sequences can be copied by slicing:

new_l = l[:]

How can I find the methods or attributes of an object?

For an instance x of a user-defined class,

dir(x)

returns an alphabetized list of the

names containing the instance attributes and methods and attributes defined by its class.

How can my code discover the name of an object?

Generally speaking, it can’t, because objects don’t really have names. Essentially,

assignment always binds a name to a value; The same is true of

def

and

class

statements, but in that case the value is a callable. Consider the following code:

class A:

pass

B = A

a = B()

b = a

print b

<__main__.A instance at 0x16D07CC>

print a

<__main__.A instance at 0x16D07CC>

Arguably the class has a name: even though it is bound to two names and invoked

through the name B the created instance is still reported as an instance of class A.

However, it is impossible to say whether the instance’s name is a or b, since both names

are bound to the same value.

Generally speaking it should not be necessary for your code to “know the names” of

particular values. Unless you are deliberately writing introspective programs, this is

usually an indication that a change of approach might be beneficial.

In comp.lang.python, Fredrik Lundh once gave an excellent analogy in answer to this

question:

The same way as you get the name of that cat you found on your porch: the cat

(object) itself cannot tell you its name, and it doesn’t really care – so the only way to

find out what it’s called is to ask all your neighbours (namespaces) if it’s their cat

(object)...

....and don’t be surprised if you’ll find that it’s known by many names, or no name at

all!

What’s up with the comma operator’s precedence?

Comma is not an operator in Python. Consider this session:

>>> "a" in "b", "a"

(False, 'a')

Since the comma is not an operator, but a separator between expressions the above is

evaluated as if you had entered:

("a" in "b"), "a"

not:

"a" in ("b", "a")

The same is true of the various assignment operators (

etc). They are not truly

operators but syntactic delimiters in assignment statements.

Is there an equivalent of C’s ”?:” ternary operator?

Yes, this feature was added in Python 2.5. The syntax would be as follows:

[on_true] if [expression] else [on_false]

x, y = 50, 25

small = x if x < y else y

For versions previous to 2.5 the answer would be ‘No’.

Is it possible to write obfuscated one-liners in Python?

Yes. Usually this is done by nesting

lambda

within

lambda

. See the following three

examples, due to Ulf Bartelt:

# Primes < 1000

print filter(None,map(lambda y:y*reduce(lambda x,y:x*y!=0,

map(lambda x,y=y:y%x,range(2,int(pow(y,0.5)+1))),1),range(2,1000)))

# First 10 Fibonacci numbers

print map(lambda x,f=lambda x,f:(f(x-1,f)+f(x-2,f)) if x>1 else 1: f(x,f),

range(10))

# Mandelbrot set

print (lambda Ru,Ro,Iu,Io,IM,Sx,Sy:reduce(lambda x,y:x+y,map(lambda y,

Iu=Iu,Io=Io,Ru=Ru,Ro=Ro,Sy=Sy,L=lambda yc,Iu=Iu,Io=Io,Ru=Ru,Ro=Ro,i=IM,

Sx=Sx,Sy=Sy:reduce(lambda x,y:x+y,map(lambda x,xc=Ru,yc=yc,Ru=Ru,Ro=Ro,

i=i,Sx=Sx,F=lambda xc,yc,x,y,k,f=lambda xc,yc,x,y,k,f:(k<=0)or (x*x+y*y

>=4.0) or 1+f(xc,yc,x*x-y*y+xc,2.0*x*y+yc,k-1,f):f(xc,yc,x,y,k,f):chr(

64+F(Ru+x*(Ro-Ru)/Sx,yc,0,0,i)),range(Sx))):L(Iu+y*(Io-Iu)/Sy),range(Sy

))))(-2.1, 0.7, -1.2, 1.2, 30, 80, 24)

# \___ ___/ \___ ___/ | | |__ lines on screen

# V V | |______ columns on screen

# | | |__________ maximum of "iterations"

# | |_________________ range on y axis

# |____________________________ range on x axis

Don’t try this at home, kids!

Numbers and strings

How do I specify hexadecimal and octal integers?

To specify an octal digit, precede the octal value with a zero, and then a lower or

uppercase “o”. For example, to set the variable “a” to the octal value “10” (8 in decimal),

type:

>>> a = 0o10

>>> a

Hexadecimal is just as easy. Simply precede the hexadecimal number with a zero, and

then a lower or uppercase “x”. Hexadecimal digits can be specified in lower or uppercase.

For example, in the Python interpreter:

>>> a = 0xa5

>>> a

165

>>> b = 0XB2

>>> b

178

Why does -22 // 10 return -3?

It’s primarily driven by the desire that

i % j

have the same sign as

. If you want that,

and also want:

i == (i // j) * j + (i % j)

then integer division has to return the floor. C also requires that identity to hold, and then

compilers that truncate

i // j

need to make

i % j

have the same sign as

There are few real use cases for

i % j

when

is negative. When

is positive, there are

many, and in virtually all of them it’s more useful for

i % j

to be

>= 0

. If the clock says

10 now, what did it say 200 hours ago?

-190 % 12 == 2

is useful;

-190 % 12 == -10

is a

bug waiting to bite.

Note: On Python 2,

a / b

returns the same as

a // b

__future__.division

is not

in effect. This is also known as “classic” division.

How do I convert a string to a number?

For integers, use the built-in

int()

type constructor, e.g.

int('144') == 144

. Similarly,

float()

converts to floating-point, e.g.

float('144') == 144.0

By default, these interpret the number as decimal, so that

int('0144') == 144

and

int('0x144')

raises

ValueError

int(string, base)

takes the base to convert from as a

second optional argument, so

int('0x144', 16) == 324

. If the base is specified as 0, the

number is interpreted using Python’s rules: a leading ‘0’ indicates octal, and ‘0x’ indicates

a hex number.

Do not use the built-in function

eval()

if all you need is to convert strings to numbers.

eval()

will be significantly slower and it presents a security risk: someone could pass you

a Python expression that might have unwanted side effects. For example, someone could

pass

__import__('os').system("rm -rf $HOME")

which would erase your home

directory.

eval()

also has the effect of interpreting numbers as Python expressions, so that e.g.

eval('09')

gives a syntax error because Python regards numbers starting with ‘0’ as

octal (base 8).

How do I convert a number to a string?

To convert, e.g., the number 144 to the string ‘144’, use the built-in type constructor

str()

. If you want a hexadecimal or octal representation, use the built-in functions

hex()

o r

oct()

. For fancy formatting, see the Format String Syntax section, e.g.

{:04d}".format(144)

yields

'0144'

and

"{:.3f}".format(1/3)

yields

'0.333'

. You may

also use the % operator on strings. See the library reference manual for details.

How do I modify a string in place?

You can’t, because strings are immutable. If you need an object with this ability, try

converting the string to a list or use the array module:

>>> import io

>>> s = "Hello, world"

>>> a = list(s)

>>> print a

['H', 'e', 'l', 'l', 'o', ',', ' ', 'w', 'o', 'r', 'l', 'd']

>>> a[7:] = list("there!")

>>> ''.join(a)

'Hello, there!'

>>> import array

>>> a = array.array('c', s)

>>> print a

array('c', 'Hello, world')

>>> a[0] = 'y' ; print a

array('c', 'yello, world')

>>> a.tostring()

'yello, world'

How do I use strings to call functions/methods?

There are various techniques.

The best is to use a dictionary that maps strings to functions. The primary advantage

of this technique is that the strings do not need to match the names of the functions.

This is also the primary technique used to emulate a case construct:

def a():

pass

def b():

pass

dispatch = {'go': a, 'stop': b} # Note lack of parens for funcs

dispatch[get_input()]() # Note trailing parens to call function

Use the built-in function

getattr()

import foo

getattr(foo, 'bar')()

Note that

getattr()

works on any object, including classes, class instances,

modules, and so on.

This is used in several places in the standard library, like this:

class Foo:

def do_foo(self):

...

def do_bar(self):

...

f = getattr(foo_instance, 'do_' + opname)

f()

Use

locals()

eval()

to resolve the function name:

def myFunc():

print "hello"

fname = "myFunc"

f = locals()[fname]

f()

f = eval(fname)

f()

Note: Using

eval()

is slow and dangerous. If you don’t have absolute control over

the contents of the string, someone could pass a string that resulted in an arbitrary

function being executed.

Is there an equivalent to Perl’s chomp() for removing trailing

newlines from strings?

Starting with Python 2.2, you can use

S.rstrip("\r\n")

to remove all occurrences of

any line terminator from the end of the string

without removing other trailing

whitespace. If the string

represents more than one line, with several empty lines at the

end, the line terminators for all the blank lines will be removed:

>>> lines = ("line 1 \r\n"

... "\r\n"

... "\r\n")

>>> lines.rstrip("\n\r")

'line 1 '

Since this is typically only desired when reading text one line at a time, using

S.rstrip()

this way works well.

For older versions of Python, there are two partial substitutes:

If you want to remove all trailing whitespace, use the

rstrip()

method of string

objects. This removes all trailing whitespace, not just a single newline.

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