Swapping 1 with 0 and 0 with 1 in a Pythonic way
Question:
In some part of my Python program I have a val variable that can be 1 or 0. If it’s 1 I must change to 0, if it’s 0 I must change to 1.
How do you do it in a Pythonic way?
if val == 1:
val = 0
elif val == 0:
val = 1
it’s too long!
I did:
swap = {0: 1, 1:0}
So I can use it:
swap[val]
Other ideas?
Answers:
This isn’t pythonic, but it is language neutral. Often val = 1 - val is simplest.
Since True == 1 and False == 0 in python,
you could just use var = not var
It will just swap it.
If you want to be short:
f = lambda val: 0 if val else 1
Then:
>>> f(0)
1
>>> f(1)
0
In your case I recommend the ternary:
val = 0 if val else 1
If you had 2 variables to swap you could say:
(a, b) = (b, a)
Just another possibility:
i = (1, 0)[i]
This works well as long as i is positive, as dbr pointed out in the comments it doesn’t work for i < 0.
Are you sure you don’t want to use False and True? It sounds almost like it.
The shortest approach is using the bitwise operator XOR.
If you want val to be reassigned:
val ^= 1
If you do not want val to be reassigned:
val ^ 1
Function with mutable argument. Calling the swaper() will return different value every time.
def swaper(x=[1]):
x[0] ^= 1
return x[0]
Just another way:
val = ~val + 2
(0,1)[not val]
flips the val from 0 to 1 and vice versa.
To expand upon the answer by "YOU", use:
int(not(val))
Examples:
>>> val = 0
>>> int(not(val))
1
>>> val = 1
>>> int(not(val))
0
Note that this answer is only meant to be descriptive, not prescriptive.
Your way works well!
What about:
val = abs(val - 1)
short and simple!
The most pythonic way would probably be
int(not val)
But a shorter way would be
-~-val
After seeing all these simpler answers i thought of adding an abstract one , it’s Pythonic though :
val = set(range(0,2)).symmetric_difference(set(range(0 + val, val + 1))).pop()
All we do is return the difference of 2 sets namely [0, 1] and [val] where val is either 0 or 1.
we use symmetric_difference() to create the set [0, 1] – [val] and pop() to assign that value to variable val.
use np.where
ex.
np.where(np.array(val)==0,1,0)
this gives 1 where val is 0 and gives 0 where val is anything else, in your case 1
EDIT: val has to be array
I have swapped 0s and 1s in a list.
Here’s my list:
list1 = [1,0,0,1,0]
list1 = [i^1 for i in list1]
#xor each element is the list
print(list1)
So the outcome is: [0,1,1,0,1]
Here’s a simple way:
val = val + 1 - val * 2
For Example:
If val is 0
0+1-0*2=1
If val is 1
1+1-1*2=0
Another option:
val = (not val) * 1
If you are using an array and it is in numpy you can use
>> y = np.array([1,0,1,1])
>> y^1
[0,1,0,0]
In some part of my Python program I have a val variable that can be 1 or 0. If it’s 1 I must change to 0, if it’s 0 I must change to 1.
How do you do it in a Pythonic way?
if val == 1:
val = 0
elif val == 0:
val = 1
it’s too long!
I did:
swap = {0: 1, 1:0}
So I can use it:
swap[val]
Other ideas?
This isn’t pythonic, but it is language neutral. Often val = 1 - val is simplest.
Since True == 1 and False == 0 in python,
you could just use var = not var
It will just swap it.
If you want to be short:
f = lambda val: 0 if val else 1
Then:
>>> f(0)
1
>>> f(1)
0
In your case I recommend the ternary:
val = 0 if val else 1
If you had 2 variables to swap you could say:
(a, b) = (b, a)
Just another possibility:
i = (1, 0)[i]
This works well as long as i is positive, as dbr pointed out in the comments it doesn’t work for i < 0.
Are you sure you don’t want to use False and True? It sounds almost like it.
The shortest approach is using the bitwise operator XOR.
If you want val to be reassigned:
val ^= 1
If you do not want val to be reassigned:
val ^ 1
Function with mutable argument. Calling the swaper() will return different value every time.
def swaper(x=[1]):
x[0] ^= 1
return x[0]
Just another way:
val = ~val + 2
(0,1)[not val]
flips the val from 0 to 1 and vice versa.
To expand upon the answer by "YOU", use:
int(not(val))
Examples:
>>> val = 0
>>> int(not(val))
1
>>> val = 1
>>> int(not(val))
0
Note that this answer is only meant to be descriptive, not prescriptive.
Your way works well!
What about:
val = abs(val - 1)
short and simple!
The most pythonic way would probably be
int(not val)
But a shorter way would be
-~-val
After seeing all these simpler answers i thought of adding an abstract one , it’s Pythonic though :
val = set(range(0,2)).symmetric_difference(set(range(0 + val, val + 1))).pop()
All we do is return the difference of 2 sets namely [0, 1] and [val] where val is either 0 or 1.
we use symmetric_difference() to create the set [0, 1] – [val] and pop() to assign that value to variable val.
use np.where
ex.
np.where(np.array(val)==0,1,0)
this gives 1 where val is 0 and gives 0 where val is anything else, in your case 1
EDIT: val has to be array
I have swapped 0s and 1s in a list.
Here’s my list:
list1 = [1,0,0,1,0]
list1 = [i^1 for i in list1]
#xor each element is the list
print(list1)
So the outcome is: [0,1,1,0,1]
Here’s a simple way:
val = val + 1 - val * 2
For Example:
If val is 0
0+1-0*2=1
If val is 1
1+1-1*2=0
Another option:
val = (not val) * 1
If you are using an array and it is in numpy you can use
>> y = np.array([1,0,1,1])
>> y^1
[0,1,0,0]