In order to make Python look more familiar, I've tried to assign an operator symbol to a variable's name,just for educational purposes: import operator equals = operator.eq
This seems to work fine for equals(a,b) but not for a equals b
Is there a way to express that a equals b instead of a == b
No, Python (and most mainstream languages) does not allow this kind of customization. In Python the restriction is quite intentional — an expression such as a equals b would look ungrammatical to any reader familiar with Python.
Not necessarily, but another SO answer shows how you can use this simple trick to "create" new operators. However, they only work if you surround the operator by | | or by << >>:
equals = Infix(lambda x, y: x == y):
print 2 |equals| 2 # True
The best you can do is
def equals(a,b):
return a == b
equals(1,5)
>> False
or
class my:
value = 0
def equals(self, b):
return self.value == b
a = my()
a.equals(3)
>>False
But you should use the built-in operator for readability. This way, a reader can distinguish, at once, an operator, a function, a symbol (a variable), a member function, etc...
Related
x = f1(x)
x = f2(x, x)
How do I write this in a single line? Obviously I don't want to write x = f2(f1(x), f1(x)) since it performs the same operation twice, but do I really have to do a two-liner here?
You should probably just keep it as two lines, it is perfectly clear that way. But if you must you can use an assignment expression:
>>> def f1(a): return a + 42
...
>>> def f2(b, c): return b + c
...
>>> f2(x:=f1(1), x)
86
>>>
But again, don't try to cram your code into one line. Rarely is a code improved by trying to make a "one-liner". Write clear, readable, and maintainable code. Don't try to write the shortest code possible. That is maybe fun if you are playing code-golf, but it isn't what you should do if you are trying to write software that is actually going to be used.
This is horrendous, and 2 clear lines is better than 1 obfuscated line, but...
x = f2(*itertools.repeat(f1(x), 2))
Example of use:
import itertools
def f1(x):
return 2*x
def f2(x1, x2):
return x1+x2
x = 1
x = f2(*itertools.repeat(f1(x), 2))
print(x)
Prints 4.
This really doesn't seem like a good place to condense things down to one line, but if you must, here's the way I would go about it.
Let's take the function f2. Normally, you'd pass in parameters like this:
x = f2("foo", "bar")
But you can also use a tuple containing "foo" and "bar" and extract the values as arguments for your function using this syntax:
t = ("foo", "bar")
x = f2(*t)
So if you construct a tuple with two of the same element, you can use that syntax to pass the same value to both arguments:
t = (f1(x),) * 2
x = f2(*t)
Now just eliminate the temporary variable t to make it a one-liner:
x = f2(*(f1(x),) * 2)
Obviously this isn't very intuitive or readable though, so I'd recommend against it.
One other option you have if you're using Python 3.8 or higher is to use the "walrus operator", which assigns a value and acts as an expression that evaluates to that value. For example, the below expression is equal to 5, but also sets x to 2 in the process of its evaluation:
(x := 2) + 3
Here's your solution for a one-liner using the walrus operator:
x = f2(x := f1(x), x)
Basically, x is set to f1(x), then reused for the second parameter of f2. This one might be a little more readable but it still isn't perfect.
Is it possible to insert an operation (e.g *, +) between two variables at runtime?
My solution without doing this is multiple if, elif statements, but I don't think that's the most efficient way to do it.
EDIT: What I meant is I get two integers, and I want to apply an operation on one of them with the other, e.g x * y, but I want to change * to another operator (maybe they're called functions? Not sure) e.g -, +,^ based on input.
Does that make sense? Basically think of it as a calculator.
I'm not sure if this is what you're looking for but the operator module has a lot of operations, e.g. add and mul (multiply):
import operator
var_1 = 2
var_2 = 3
print(operator.add(var_1, var_2))
print(operator.mul(var_1, var_2))
will print
5
6
#AaronHall's is the answer you're looking for, but for completeness, I'd mention you can also use eval.
var_1 = 2
var_2 = 3
op = '+'
print eval('%s%s%s' % (var_1, op, var_2))
However, eval is evil, so either don't use it, or use with caution.
To answer the follow-on question, for example, you can subclass int and then implement __xor__ to exponentiate instead of apply a bitwise or, which is called by the ^ operator:
import operator
class MyInt(int):
def __xor__(self, other):
return operator.pow(self, other)
and then:
>>> i = MyInt(2)
>>> i
2
>>> type(i)
<class '__main__.MyInt'>
>>> j = 3
>>> i^j
8
If your list of operators is small enough you can probably use lambdas.
def operator_factory(op):
if op == '+':
return lambda x,y: x + y
elif op == '-':
return lambda x,y: x - y
elif op == '*':
return lambda x,y: x * y
elif op == '/':
return lambda x,y: x / y
elif op == '^':
return lambda x,y: x ^ y
Your if statements can depend on user input. Then you just use this like so:
>>> f = operator_factory('+')
>>> f(2,3)
5
Is there a way to use variable like in operator overloading.
e.g.
a += 1
Instead of a = a + 1
in
a = max(a, some_other_variable)
The max() function is just an example.
NOTE:
My intent here is not to use the variable 'a' again, if possible. These two examples are different and not related to each other.
e.g.
a = some_function(a, b)
Here, the values returned from some_function() is assigned back to variable 'a' again.
Unless variable 'a' is a class variable I cannot access variable inside function some_function(), although if there is a way so that I can use it only once?
You cannot supplement Python's set of operators and statements directly in the Python code. However, you can write a wrapper that uses Python's language services to write a Pythonesque DSL which includes the operators you want.
I feel like you want something along these lines ...
>>> class Foo(object):
... def __iadd__(self, other):
... return max(self.num, other)
... def __init__(self, num):
... self.num = num
...
>>> a = Foo(5)
>>> a += 4
>>> print a
5
>>> a = Foo(4)
>>> a += 6
>>> a
6
But please note that I would consider this use of __iadd__ to be very impolite. Having __iadd__ return something other than self is generally inconsiderate if the type is mutable.
Instead of overloading an operator in a like the other answer, you could create a partial-like object for the second part. (I used the left shift operator for "coolness")
class partial(functools.partial):
def __rlshift__(self, val):
return self(val)
and use like this:
>>> a = 10
>>> a <<= partial(max, 20)
>>> a
20
So you don't need to mess with your variable types to execute the operation. Also you will not need to declare a new class for every function.
PS: Beware that the actual execution is max(20, a).
If I write the following in python, I get a syntax error, why so?
a = 1
b = (a+=1)
I am using python version 2.7
what I get when I run it, the following:
>>> a = 1
>>> b = (a +=1)
File "<stdin>", line 1
b = (a +=1)
^
SyntaxError: invalid syntax
>>>
Unlike in some other languages, assignment (including augmented assignment, like +=) in Python is not an expression. This also affects things like this:
(a=1) > 2
which is legal in C, and several other languages.
The reason generally given for this is because it helps to prevent a class of bugs like this:
if a = 1: # instead of ==
pass
else:
pass
since assignment isn't an expression, this is a SyntaxError in Python. In the equivalent C code, it is a subtle bug where the variable will be modified rather than checked, the check will always be true (in C, like in Python, a non-zero integer is always truthy), and the else block can never fire.
You can still do chained assignment in Python, so this works:
>>> a = 1
>>> a = b = a+1
>>> a
2
>>> b
2
a +=1 is a statement in Python and you can't assign a statement to a variable. Though it is a valid syntax in languages like C, PHP, etc but not Python.
b = (a+=1)
An equivalent version will be:
>>> a = 1
>>> a += 1
>>> b = a
As #Ashwini stated, a+=1 is an assigment, not a value. You can't assign it to b, or any variable. What you probably want is:
b = a+1
All the answers provided here are good, I just want to add that you can achieve what you want in a one-line expression, but written in a different manner:
b, a = a+1, a+1
Here you're doing almost the same thing: incrementing a by 1, and assigning the value of a+1 to b - I'm telling 'almost' because here we have two summations instead of one.
How do I use pre-increment/decrement operators (++, --), just like in C++?
Why does ++count run, but not change the value of the variable?
++ is not an operator. It is two + operators. The + operator is the identity operator, which does nothing. (Clarification: the + and - unary operators only work on numbers, but I presume that you wouldn't expect a hypothetical ++ operator to work on strings.)
++count
Parses as
+(+count)
Which translates to
count
You have to use the slightly longer += operator to do what you want to do:
count += 1
I suspect the ++ and -- operators were left out for consistency and simplicity. I don't know the exact argument Guido van Rossum gave for the decision, but I can imagine a few arguments:
Simpler parsing. Technically, parsing ++count is ambiguous, as it could be +, +, count (two unary + operators) just as easily as it could be ++, count (one unary ++ operator). It's not a significant syntactic ambiguity, but it does exist.
Simpler language. ++ is nothing more than a synonym for += 1. It was a shorthand invented because C compilers were stupid and didn't know how to optimize a += 1 into the inc instruction most computers have. In this day of optimizing compilers and bytecode interpreted languages, adding operators to a language to allow programmers to optimize their code is usually frowned upon, especially in a language like Python that is designed to be consistent and readable.
Confusing side-effects. One common newbie error in languages with ++ operators is mixing up the differences (both in precedence and in return value) between the pre- and post-increment/decrement operators, and Python likes to eliminate language "gotcha"-s. The precedence issues of pre-/post-increment in C are pretty hairy, and incredibly easy to mess up.
Python does not have pre and post increment operators.
In Python, integers are immutable. That is you can't change them. This is because the integer objects can be used under several names. Try this:
>>> b = 5
>>> a = 5
>>> id(a)
162334512
>>> id(b)
162334512
>>> a is b
True
a and b above are actually the same object. If you incremented a, you would also increment b. That's not what you want. So you have to reassign. Like this:
b = b + 1
Many C programmers who used python wanted an increment operator, but that operator would look like it incremented the object, while it actually reassigns it. Therefore the -= and += operators where added, to be shorter than the b = b + 1, while being clearer and more flexible than b++, so most people will increment with:
b += 1
Which will reassign b to b+1. That is not an increment operator, because it does not increment b, it reassigns it.
In short: Python behaves differently here, because it is not C, and is not a low level wrapper around machine code, but a high-level dynamic language, where increments don't make sense, and also are not as necessary as in C, where you use them every time you have a loop, for example.
While the others answers are correct in so far as they show what a mere + usually does (namely, leave the number as it is, if it is one), they are incomplete in so far as they don't explain what happens.
To be exact, +x evaluates to x.__pos__() and ++x to x.__pos__().__pos__().
I could imagine a VERY weird class structure (Children, don't do this at home!) like this:
class ValueKeeper(object):
def __init__(self, value): self.value = value
def __str__(self): return str(self.value)
class A(ValueKeeper):
def __pos__(self):
print 'called A.__pos__'
return B(self.value - 3)
class B(ValueKeeper):
def __pos__(self):
print 'called B.__pos__'
return A(self.value + 19)
x = A(430)
print x, type(x)
print +x, type(+x)
print ++x, type(++x)
print +++x, type(+++x)
TL;DR
Python does not have unary increment/decrement operators (--/++). Instead, to increment a value, use
a += 1
More detail and gotchas
But be careful here. If you're coming from C, even this is different in python. Python doesn't have "variables" in the sense that C does, instead python uses names and objects, and in python ints are immutable.
so lets say you do
a = 1
What this means in python is: create an object of type int having value 1 and bind the name a to it. The object is an instance of int having value 1, and the name a refers to it. The name a and the object to which it refers are distinct.
Now lets say you do
a += 1
Since ints are immutable, what happens here is as follows:
look up the object that a refers to (it is an int with id 0x559239eeb380)
look up the value of object 0x559239eeb380 (it is 1)
add 1 to that value (1 + 1 = 2)
create a new int object with value 2 (it has object id 0x559239eeb3a0)
rebind the name a to this new object
Now a refers to object 0x559239eeb3a0 and the original object (0x559239eeb380) is no longer refered to by the name a. If there aren't any other names refering to the original object it will be garbage collected later.
Give it a try yourself:
a = 1
print(hex(id(a)))
a += 1
print(hex(id(a)))
In python 3.8+ you can do :
(a:=a+1) #same as ++a (increment, then return new value)
(a:=a+1)-1 #same as a++ (return the incremented value -1) (useless)
You can do a lot of thinks with this.
>>> a = 0
>>> while (a:=a+1) < 5:
print(a)
1
2
3
4
Or if you want write somthing with more sophisticated syntaxe (the goal is not optimization):
>>> del a
>>> while (a := (a if 'a' in locals() else 0) + 1) < 5:
print(a)
1
2
3
4
It will return 0 even if 'a' doesn't exist without errors, and then will set it to 1
Python does not have these operators, but if you really need them you can write a function having the same functionality.
def PreIncrement(name, local={}):
#Equivalent to ++name
if name in local:
local[name]+=1
return local[name]
globals()[name]+=1
return globals()[name]
def PostIncrement(name, local={}):
#Equivalent to name++
if name in local:
local[name]+=1
return local[name]-1
globals()[name]+=1
return globals()[name]-1
Usage:
x = 1
y = PreIncrement('x') #y and x are both 2
a = 1
b = PostIncrement('a') #b is 1 and a is 2
Inside a function you have to add locals() as a second argument if you want to change local variable, otherwise it will try to change global.
x = 1
def test():
x = 10
y = PreIncrement('x') #y will be 2, local x will be still 10 and global x will be changed to 2
z = PreIncrement('x', locals()) #z will be 11, local x will be 11 and global x will be unaltered
test()
Also with these functions you can do:
x = 1
print(PreIncrement('x')) #print(x+=1) is illegal!
But in my opinion following approach is much clearer:
x = 1
x+=1
print(x)
Decrement operators:
def PreDecrement(name, local={}):
#Equivalent to --name
if name in local:
local[name]-=1
return local[name]
globals()[name]-=1
return globals()[name]
def PostDecrement(name, local={}):
#Equivalent to name--
if name in local:
local[name]-=1
return local[name]+1
globals()[name]-=1
return globals()[name]+1
I used these functions in my module translating javascript to python.
In Python, a distinction between expressions and statements is rigidly
enforced, in contrast to languages such as Common Lisp, Scheme, or
Ruby.
Wikipedia
So by introducing such operators, you would break the expression/statement split.
For the same reason you can't write
if x = 0:
y = 1
as you can in some other languages where such distinction is not preserved.
Yeah, I missed ++ and -- functionality as well. A few million lines of c code engrained that kind of thinking in my old head, and rather than fight it... Here's a class I cobbled up that implements:
pre- and post-increment, pre- and post-decrement, addition,
subtraction, multiplication, division, results assignable
as integer, printable, settable.
Here 'tis:
class counter(object):
def __init__(self,v=0):
self.set(v)
def preinc(self):
self.v += 1
return self.v
def predec(self):
self.v -= 1
return self.v
def postinc(self):
self.v += 1
return self.v - 1
def postdec(self):
self.v -= 1
return self.v + 1
def __add__(self,addend):
return self.v + addend
def __sub__(self,subtrahend):
return self.v - subtrahend
def __mul__(self,multiplier):
return self.v * multiplier
def __div__(self,divisor):
return self.v / divisor
def __getitem__(self):
return self.v
def __str__(self):
return str(self.v)
def set(self,v):
if type(v) != int:
v = 0
self.v = v
You might use it like this:
c = counter() # defaults to zero
for listItem in myList: # imaginary task
doSomething(c.postinc(),listItem) # passes c, but becomes c+1
...already having c, you could do this...
c.set(11)
while c.predec() > 0:
print c
....or just...
d = counter(11)
while d.predec() > 0:
print d
...and for (re-)assignment into integer...
c = counter(100)
d = c + 223 # assignment as integer
c = c + 223 # re-assignment as integer
print type(c),c # <type 'int'> 323
...while this will maintain c as type counter:
c = counter(100)
c.set(c + 223)
print type(c),c # <class '__main__.counter'> 323
EDIT:
And then there's this bit of unexpected (and thoroughly unwanted) behavior,
c = counter(42)
s = '%s: %d' % ('Expecting 42',c) # but getting non-numeric exception
print s
...because inside that tuple, getitem() isn't what used, instead a reference to the object is passed to the formatting function. Sigh. So:
c = counter(42)
s = '%s: %d' % ('Expecting 42',c.v) # and getting 42.
print s
...or, more verbosely, and explicitly what we actually wanted to happen, although counter-indicated in actual form by the verbosity (use c.v instead)...
c = counter(42)
s = '%s: %d' % ('Expecting 42',c.__getitem__()) # and getting 42.
print s
There are no post/pre increment/decrement operators in python like in languages like C.
We can see ++ or -- as multiple signs getting multiplied, like we do in maths (-1) * (-1) = (+1).
E.g.
---count
Parses as
-(-(-count)))
Which translates to
-(+count)
Because, multiplication of - sign with - sign is +
And finally,
-count
A straight forward workaround
c = 0
c = (lambda c_plusplus: plusplus+1)(c)
print(c)
1
No more typing
c = c + 1
Also, you could just write
c++
and finish all your code and then do search/replace for "c++", replace with "c=c+1". Just make sure regular expression search is off.
Extending Henry's answer, I experimentally implemented a syntax sugar library realizing a++: hdytto.
The usage is simple. After installing from PyPI, place sitecustomize.py:
from hdytto import register_hdytto
register_hdytto()
in your project directory. Then, make main.py:
# coding: hdytto
a = 5
print(a++)
print(++a)
b = 10 - --a
print(b--)
and run it by PYTHONPATH=. python main.py. The output will be
5
7
4
hdytto replaces a++ as ((a:=a+1)-1) when decoding the script file, so it works.