working of exponential operator ** in Python
I am running a python script where I am computing following:
t - 2 ** (j - 1) * l
Where t = 302536, j = 6, l = 0.
This returns me 302536 (t), I am not able to understand how. As per me the result should have been 302535 (t - 1).
2 ** (j - 1) * l results in 0 which according to me should have resulted in 1 as (j - 1) * l results in 0.
How is this being computed?
python pow
edited Nov 21 '18 at 12:29
Mad Physicist
34.6k156896
asked Nov 21 '18 at 11:54
apoorv parmarapoorv parmar
317
add a comment |
I am running a python script where I am computing following:
t - 2 ** (j - 1) * l
Where t = 302536, j = 6, l = 0.
This returns me 302536 (t), I am not able to understand how. As per me the result should have been 302535 (t - 1).
2 ** (j - 1) * l results in 0 which according to me should have resulted in 1 as (j - 1) * l results in 0.
How is this being computed?
python pow
edited Nov 21 '18 at 12:29
Mad Physicist
34.6k156896
asked Nov 21 '18 at 11:54
apoorv parmarapoorv parmar
317
1
Why pandas? Can you explain it?
– jezrael
Nov 21 '18 at 11:56
this is just a line of code. All other computation in the script require pandas. May be I should have written python only
– apoorv parmar
Nov 21 '18 at 11:59
1
I think you can checkoperator-precedence- first is evaluate(), then**and last*
– jezrael
Nov 21 '18 at 12:03
add a comment |
I am running a python script where I am computing following:
t - 2 ** (j - 1) * l
Where t = 302536, j = 6, l = 0.
This returns me 302536 (t), I am not able to understand how. As per me the result should have been 302535 (t - 1).
2 ** (j - 1) * l results in 0 which according to me should have resulted in 1 as (j - 1) * l results in 0.
How is this being computed?
python pow
edited Nov 21 '18 at 12:29
Mad Physicist
34.6k156896
asked Nov 21 '18 at 11:54
apoorv parmarapoorv parmar
317
I am running a python script where I am computing following:
t - 2 ** (j - 1) * l
Where t = 302536, j = 6, l = 0.
This returns me 302536 (t), I am not able to understand how. As per me the result should have been 302535 (t - 1).
2 ** (j - 1) * l results in 0 which according to me should have resulted in 1 as (j - 1) * l results in 0.
How is this being computed?
python pow
python pow
edited Nov 21 '18 at 12:29
Mad Physicist
34.6k156896
asked Nov 21 '18 at 11:54
apoorv parmarapoorv parmar
317
edited Nov 21 '18 at 12:29
Mad Physicist
34.6k156896
asked Nov 21 '18 at 11:54
apoorv parmarapoorv parmar
317
edited Nov 21 '18 at 12:29
Mad Physicist
34.6k156896
edited Nov 21 '18 at 12:29
Mad Physicist
34.6k156896
edited Nov 21 '18 at 12:29
Mad Physicist
34.6k156896
34.6k156896
asked Nov 21 '18 at 11:54
apoorv parmarapoorv parmar
317
asked Nov 21 '18 at 11:54
apoorv parmarapoorv parmar
317
asked Nov 21 '18 at 11:54
apoorv parmarapoorv parmar
317
317
1
Why pandas? Can you explain it?
– jezrael
Nov 21 '18 at 11:56
this is just a line of code. All other computation in the script require pandas. May be I should have written python only
– apoorv parmar
Nov 21 '18 at 11:59
1
I think you can checkoperator-precedence- first is evaluate(), then**and last*
– jezrael
Nov 21 '18 at 12:03
add a comment |
1
Why pandas? Can you explain it?
– jezrael
Nov 21 '18 at 11:56
this is just a line of code. All other computation in the script require pandas. May be I should have written python only
– apoorv parmar
Nov 21 '18 at 11:59
1
I think you can checkoperator-precedence- first is evaluate(), then**and last*
– jezrael
Nov 21 '18 at 12:03
1
1
Why pandas? Can you explain it?
– jezrael
Nov 21 '18 at 11:56
Why pandas? Can you explain it?
– jezrael
Nov 21 '18 at 11:56
this is just a line of code. All other computation in the script require pandas. May be I should have written python only
– apoorv parmar
Nov 21 '18 at 11:59
this is just a line of code. All other computation in the script require pandas. May be I should have written python only
– apoorv parmar
Nov 21 '18 at 11:59
1
1
I think you can check
operator-precedence - first is evaluate (), then ** and last *– jezrael
Nov 21 '18 at 12:03
I think you can check
operator-precedence - first is evaluate (), then ** and last *– jezrael
Nov 21 '18 at 12:03
add a comment |
1 Answer
1
active
oldest
votes
The only thing that binds tighter than power is parentheses. Python (and every other language that natively supports a power operator that I've seen) follows arithmetic convention on this one, so you don't need to memorize different sets of conflicting rules. You operation can be explicitly rewritten as follows:
t - ((2 ** (j - 1)) * l)
As you pointed out, setting l = 0 discards much of the computation. It's just that it discards everything but t itself.
You could make such things explicit by using the function form of the power operator. Any of the following imports would work for the example below:
from math import pow
from operator import pow
from operator import __pow__ as pow
from numpy import pow
It seems like you wanted/expected
t - pow(2, (j - 1) * l)
But instead got
t - pow(2, j - 1) * l
answered Nov 21 '18 at 12:07
Mad PhysicistMad Physicist
34.6k156896
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
The only thing that binds tighter than power is parentheses. Python (and every other language that natively supports a power operator that I've seen) follows arithmetic convention on this one, so you don't need to memorize different sets of conflicting rules. You operation can be explicitly rewritten as follows:
t - ((2 ** (j - 1)) * l)
As you pointed out, setting l = 0 discards much of the computation. It's just that it discards everything but t itself.
You could make such things explicit by using the function form of the power operator. Any of the following imports would work for the example below:
from math import pow
from operator import pow
from operator import __pow__ as pow
from numpy import pow
It seems like you wanted/expected
t - pow(2, (j - 1) * l)
But instead got
t - pow(2, j - 1) * l
answered Nov 21 '18 at 12:07
Mad PhysicistMad Physicist
34.6k156896
add a comment |
The only thing that binds tighter than power is parentheses. Python (and every other language that natively supports a power operator that I've seen) follows arithmetic convention on this one, so you don't need to memorize different sets of conflicting rules. You operation can be explicitly rewritten as follows:
t - ((2 ** (j - 1)) * l)
As you pointed out, setting l = 0 discards much of the computation. It's just that it discards everything but t itself.
You could make such things explicit by using the function form of the power operator. Any of the following imports would work for the example below:
from math import pow
from operator import pow
from operator import __pow__ as pow
from numpy import pow
It seems like you wanted/expected
t - pow(2, (j - 1) * l)
But instead got
t - pow(2, j - 1) * l
answered Nov 21 '18 at 12:07
Mad PhysicistMad Physicist
34.6k156896
add a comment |
The only thing that binds tighter than power is parentheses. Python (and every other language that natively supports a power operator that I've seen) follows arithmetic convention on this one, so you don't need to memorize different sets of conflicting rules. You operation can be explicitly rewritten as follows:
t - ((2 ** (j - 1)) * l)
As you pointed out, setting l = 0 discards much of the computation. It's just that it discards everything but t itself.
You could make such things explicit by using the function form of the power operator. Any of the following imports would work for the example below:
from math import pow
from operator import pow
from operator import __pow__ as pow
from numpy import pow
It seems like you wanted/expected
t - pow(2, (j - 1) * l)
But instead got
t - pow(2, j - 1) * l
answered Nov 21 '18 at 12:07
Mad PhysicistMad Physicist
34.6k156896
The only thing that binds tighter than power is parentheses. Python (and every other language that natively supports a power operator that I've seen) follows arithmetic convention on this one, so you don't need to memorize different sets of conflicting rules. You operation can be explicitly rewritten as follows:
t - ((2 ** (j - 1)) * l)
As you pointed out, setting l = 0 discards much of the computation. It's just that it discards everything but t itself.
You could make such things explicit by using the function form of the power operator. Any of the following imports would work for the example below:
from math import pow
from operator import pow
from operator import __pow__ as pow
from numpy import pow
It seems like you wanted/expected
t - pow(2, (j - 1) * l)
But instead got
t - pow(2, j - 1) * l
answered Nov 21 '18 at 12:07
Mad PhysicistMad Physicist
34.6k156896
edited Nov 21 '18 at 20:45
answered Nov 21 '18 at 12:07
Mad PhysicistMad Physicist
34.6k156896
answered Nov 21 '18 at 12:07
Mad PhysicistMad Physicist
34.6k156896
answered Nov 21 '18 at 12:07
Mad PhysicistMad Physicist
34.6k156896
34.6k156896
add a comment |
add a comment |
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1
Why pandas? Can you explain it?
– jezrael
Nov 21 '18 at 11:56
this is just a line of code. All other computation in the script require pandas. May be I should have written python only
– apoorv parmar
Nov 21 '18 at 11:59
1
I think you can check
operator-precedence- first is evaluate(), then**and last*– jezrael
Nov 21 '18 at 12:03