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How to estimate how big one number is from another?
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I'm trying to find an efficient solution to this problem. I receive two numbers from an API call (we can call them n1, n2). Suppose n2 is bigger than n1. I want to know how much bigger n2 is. The difference between the two is not enough because I don't know how to evaluate the result of the subtraction. I don't see any other solution but to define a tolerance range that fits my domain of application and check if their difference falls within that range. Any idea?
python numbers
asked Mar 28 at 11:08
MartinMartin
1
add a comment
|
I'm trying to find an efficient solution to this problem. I receive two numbers from an API call (we can call them n1, n2). Suppose n2 is bigger than n1. I want to know how much bigger n2 is. The difference between the two is not enough because I don't know how to evaluate the result of the subtraction. I don't see any other solution but to define a tolerance range that fits my domain of application and check if their difference falls within that range. Any idea?
python numbers
asked Mar 28 at 11:08
MartinMartin
1
1
Your question is specific to your requirements, which none of us know. Any answers you get are going to be pure guesswork as to your actual need. stackoverflow.com/help/how-to-ask
– David Buck
Mar 28 at 11:14
add a comment
|
I'm trying to find an efficient solution to this problem. I receive two numbers from an API call (we can call them n1, n2). Suppose n2 is bigger than n1. I want to know how much bigger n2 is. The difference between the two is not enough because I don't know how to evaluate the result of the subtraction. I don't see any other solution but to define a tolerance range that fits my domain of application and check if their difference falls within that range. Any idea?
python numbers
asked Mar 28 at 11:08
MartinMartin
1
I'm trying to find an efficient solution to this problem. I receive two numbers from an API call (we can call them n1, n2). Suppose n2 is bigger than n1. I want to know how much bigger n2 is. The difference between the two is not enough because I don't know how to evaluate the result of the subtraction. I don't see any other solution but to define a tolerance range that fits my domain of application and check if their difference falls within that range. Any idea?
python numbers
python numbers
asked Mar 28 at 11:08
MartinMartin
1
asked Mar 28 at 11:08
MartinMartin
1
asked Mar 28 at 11:08
MartinMartin
1
asked Mar 28 at 11:08
MartinMartin
1
asked Mar 28 at 11:08
MartinMartin
1
1
1
Your question is specific to your requirements, which none of us know. Any answers you get are going to be pure guesswork as to your actual need. stackoverflow.com/help/how-to-ask
– David Buck
Mar 28 at 11:14
add a comment
|
1
Your question is specific to your requirements, which none of us know. Any answers you get are going to be pure guesswork as to your actual need. stackoverflow.com/help/how-to-ask
– David Buck
Mar 28 at 11:14
1
1
Your question is specific to your requirements, which none of us know. Any answers you get are going to be pure guesswork as to your actual need. stackoverflow.com/help/how-to-ask
– David Buck
Mar 28 at 11:14
Your question is specific to your requirements, which none of us know. Any answers you get are going to be pure guesswork as to your actual need. stackoverflow.com/help/how-to-ask
– David Buck
Mar 28 at 11:14
add a comment
|
2 Answers
2
active
oldest
votes
Why don't you just take the division?
Suppose n1 is 1 and n2 is 10.
If you divide n2 by n1 you will see that n1 is 10 times bigger than n2.
answered Mar 28 at 11:13
SyriusSyrius
6695 silver badges17 bronze badges
add a comment
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It's a common problem when using floating-point, that results can be almost equal and you want to treat them as if they were equal.
You need a formula to measure the difference between the numbers relative to the size of the numbers, and it's remarkably tricky to find. One problem is avoiding division by zero. As always, first check the libraries: in this case, the math module includes math.isclose(a, b, *, rel_tol=1e-09, abs_tol=0.0)
So let the mathematics experts handle it for you:
import math
if math.isclose(a,b):
# they agree to around nine decimal places
answered Mar 28 at 12:37
nigel222nigel222
2,6331 gold badge7 silver badges13 bronze badges
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Why don't you just take the division?
Suppose n1 is 1 and n2 is 10.
If you divide n2 by n1 you will see that n1 is 10 times bigger than n2.
answered Mar 28 at 11:13
SyriusSyrius
6695 silver badges17 bronze badges
add a comment
|
Why don't you just take the division?
Suppose n1 is 1 and n2 is 10.
If you divide n2 by n1 you will see that n1 is 10 times bigger than n2.
answered Mar 28 at 11:13
SyriusSyrius
6695 silver badges17 bronze badges
add a comment
|
Why don't you just take the division?
Suppose n1 is 1 and n2 is 10.
If you divide n2 by n1 you will see that n1 is 10 times bigger than n2.
answered Mar 28 at 11:13
SyriusSyrius
6695 silver badges17 bronze badges
Why don't you just take the division?
Suppose n1 is 1 and n2 is 10.
If you divide n2 by n1 you will see that n1 is 10 times bigger than n2.
answered Mar 28 at 11:13
SyriusSyrius
6695 silver badges17 bronze badges
answered Mar 28 at 11:13
SyriusSyrius
6695 silver badges17 bronze badges
answered Mar 28 at 11:13
SyriusSyrius
6695 silver badges17 bronze badges
answered Mar 28 at 11:13
SyriusSyrius
6695 silver badges17 bronze badges
6695 silver badges17 bronze badges
add a comment
|
add a comment
|
It's a common problem when using floating-point, that results can be almost equal and you want to treat them as if they were equal.
You need a formula to measure the difference between the numbers relative to the size of the numbers, and it's remarkably tricky to find. One problem is avoiding division by zero. As always, first check the libraries: in this case, the math module includes math.isclose(a, b, *, rel_tol=1e-09, abs_tol=0.0)
So let the mathematics experts handle it for you:
import math
if math.isclose(a,b):
# they agree to around nine decimal places
answered Mar 28 at 12:37
nigel222nigel222
2,6331 gold badge7 silver badges13 bronze badges
add a comment
|
It's a common problem when using floating-point, that results can be almost equal and you want to treat them as if they were equal.
You need a formula to measure the difference between the numbers relative to the size of the numbers, and it's remarkably tricky to find. One problem is avoiding division by zero. As always, first check the libraries: in this case, the math module includes math.isclose(a, b, *, rel_tol=1e-09, abs_tol=0.0)
So let the mathematics experts handle it for you:
import math
if math.isclose(a,b):
# they agree to around nine decimal places
answered Mar 28 at 12:37
nigel222nigel222
2,6331 gold badge7 silver badges13 bronze badges
add a comment
|
It's a common problem when using floating-point, that results can be almost equal and you want to treat them as if they were equal.
You need a formula to measure the difference between the numbers relative to the size of the numbers, and it's remarkably tricky to find. One problem is avoiding division by zero. As always, first check the libraries: in this case, the math module includes math.isclose(a, b, *, rel_tol=1e-09, abs_tol=0.0)
So let the mathematics experts handle it for you:
import math
if math.isclose(a,b):
# they agree to around nine decimal places
answered Mar 28 at 12:37
nigel222nigel222
2,6331 gold badge7 silver badges13 bronze badges
It's a common problem when using floating-point, that results can be almost equal and you want to treat them as if they were equal.
You need a formula to measure the difference between the numbers relative to the size of the numbers, and it's remarkably tricky to find. One problem is avoiding division by zero. As always, first check the libraries: in this case, the math module includes math.isclose(a, b, *, rel_tol=1e-09, abs_tol=0.0)
So let the mathematics experts handle it for you:
import math
if math.isclose(a,b):
# they agree to around nine decimal places
answered Mar 28 at 12:37
nigel222nigel222
2,6331 gold badge7 silver badges13 bronze badges
answered Mar 28 at 12:37
nigel222nigel222
2,6331 gold badge7 silver badges13 bronze badges
answered Mar 28 at 12:37
nigel222nigel222
2,6331 gold badge7 silver badges13 bronze badges
answered Mar 28 at 12:37
nigel222nigel222
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Your question is specific to your requirements, which none of us know. Any answers you get are going to be pure guesswork as to your actual need. stackoverflow.com/help/how-to-ask
– David Buck
Mar 28 at 11:14