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Astropy cannot parse inches
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
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I am trying to convert inches to mm with astropy.
In input I have unit as string ("inch","mm"). I create for it example function:
def astro_conv(self, amount: float, fromm: str, to: str) -> float:
u_from = u.Unit(fromm)
u_to = u.Unit(to)
return u_from.to(u_to, amount)
I got message:ValueError'inch' did not parse as unit: At col 0, inch is not a valid unit. I have check documentation and inch should be available: http://docs.astropy.org/en/stable/units/
What am I doing wrong?
python units-of-measurement astropy
asked Mar 8 at 13:30
Kuba WentaKuba Wenta
191115
add a comment |
I am trying to convert inches to mm with astropy.
In input I have unit as string ("inch","mm"). I create for it example function:
def astro_conv(self, amount: float, fromm: str, to: str) -> float:
u_from = u.Unit(fromm)
u_to = u.Unit(to)
return u_from.to(u_to, amount)
I got message:ValueError'inch' did not parse as unit: At col 0, inch is not a valid unit. I have check documentation and inch should be available: http://docs.astropy.org/en/stable/units/
What am I doing wrong?
python units-of-measurement astropy
asked Mar 8 at 13:30
Kuba WentaKuba Wenta
191115
2
From the docs ""This package defines colloquially used Imperial units. They are available in the astropy.units.imperial namespace, but not in the top-level astropy.units namespace, To include them:import astropy.units as u, u.imperial.enable()
– BHC
Mar 8 at 13:40
Yes, You are right! I was blind :)
– Kuba Wenta
Mar 8 at 13:43
add a comment |
I am trying to convert inches to mm with astropy.
In input I have unit as string ("inch","mm"). I create for it example function:
def astro_conv(self, amount: float, fromm: str, to: str) -> float:
u_from = u.Unit(fromm)
u_to = u.Unit(to)
return u_from.to(u_to, amount)
I got message:ValueError'inch' did not parse as unit: At col 0, inch is not a valid unit. I have check documentation and inch should be available: http://docs.astropy.org/en/stable/units/
What am I doing wrong?
python units-of-measurement astropy
asked Mar 8 at 13:30
Kuba WentaKuba Wenta
191115
I am trying to convert inches to mm with astropy.
In input I have unit as string ("inch","mm"). I create for it example function:
def astro_conv(self, amount: float, fromm: str, to: str) -> float:
u_from = u.Unit(fromm)
u_to = u.Unit(to)
return u_from.to(u_to, amount)
I got message:ValueError'inch' did not parse as unit: At col 0, inch is not a valid unit. I have check documentation and inch should be available: http://docs.astropy.org/en/stable/units/
What am I doing wrong?
python units-of-measurement astropy
python units-of-measurement astropy
asked Mar 8 at 13:30
Kuba WentaKuba Wenta
191115
asked Mar 8 at 13:30
Kuba WentaKuba Wenta
191115
asked Mar 8 at 13:30
Kuba WentaKuba Wenta
191115
asked Mar 8 at 13:30
Kuba WentaKuba Wenta
191115
asked Mar 8 at 13:30
Kuba WentaKuba Wenta
191115
191115
2
From the docs ""This package defines colloquially used Imperial units. They are available in the astropy.units.imperial namespace, but not in the top-level astropy.units namespace, To include them:import astropy.units as u, u.imperial.enable()
– BHC
Mar 8 at 13:40
Yes, You are right! I was blind :)
– Kuba Wenta
Mar 8 at 13:43
add a comment |
2
From the docs ""This package defines colloquially used Imperial units. They are available in the astropy.units.imperial namespace, but not in the top-level astropy.units namespace, To include them:import astropy.units as u, u.imperial.enable()
– BHC
Mar 8 at 13:40
Yes, You are right! I was blind :)
– Kuba Wenta
Mar 8 at 13:43
2
2
From the docs ""This package defines colloquially used Imperial units. They are available in the astropy.units.imperial namespace, but not in the top-level astropy.units namespace, To include them:
import astropy.units as u, u.imperial.enable() – BHC
Mar 8 at 13:40
From the docs ""This package defines colloquially used Imperial units. They are available in the astropy.units.imperial namespace, but not in the top-level astropy.units namespace, To include them:
import astropy.units as u, u.imperial.enable() – BHC
Mar 8 at 13:40
Yes, You are right! I was blind :)
– Kuba Wenta
Mar 8 at 13:43
Yes, You are right! I was blind :)
– Kuba Wenta
Mar 8 at 13:43
add a comment |
1 Answer
1
active
oldest
votes
Per the docs, Astropy does not include imperial units defined by default. I think this is in part to reduce the default units namespace and the overhead involved in creating and searching it, and imperial units get sacrificed in this case since they are less used in astronomy for the most part:
This package defines colloquially used Imperial units. They are available in the astropy.units.imperial namespace, but not in the top-level astropy.units namespace, e.g.:
>>> import astropy.units as u
>>> mph = u.imperial.mile / u.hour
>>> mph
Unit("mi / h")
To include them in compose and the results of
find_equivalent_units, do:
>>> import astropy.units as u
>>> u.imperial.enable()
answered Mar 11 at 17:21
IguananautIguananaut
8,99513041
add a comment |
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1 Answer
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1 Answer
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votes
Per the docs, Astropy does not include imperial units defined by default. I think this is in part to reduce the default units namespace and the overhead involved in creating and searching it, and imperial units get sacrificed in this case since they are less used in astronomy for the most part:
This package defines colloquially used Imperial units. They are available in the astropy.units.imperial namespace, but not in the top-level astropy.units namespace, e.g.:
>>> import astropy.units as u
>>> mph = u.imperial.mile / u.hour
>>> mph
Unit("mi / h")
To include them in compose and the results of
find_equivalent_units, do:
>>> import astropy.units as u
>>> u.imperial.enable()
answered Mar 11 at 17:21
IguananautIguananaut
8,99513041
add a comment |
Per the docs, Astropy does not include imperial units defined by default. I think this is in part to reduce the default units namespace and the overhead involved in creating and searching it, and imperial units get sacrificed in this case since they are less used in astronomy for the most part:
This package defines colloquially used Imperial units. They are available in the astropy.units.imperial namespace, but not in the top-level astropy.units namespace, e.g.:
>>> import astropy.units as u
>>> mph = u.imperial.mile / u.hour
>>> mph
Unit("mi / h")
To include them in compose and the results of
find_equivalent_units, do:
>>> import astropy.units as u
>>> u.imperial.enable()
answered Mar 11 at 17:21
IguananautIguananaut
8,99513041
add a comment |
Per the docs, Astropy does not include imperial units defined by default. I think this is in part to reduce the default units namespace and the overhead involved in creating and searching it, and imperial units get sacrificed in this case since they are less used in astronomy for the most part:
This package defines colloquially used Imperial units. They are available in the astropy.units.imperial namespace, but not in the top-level astropy.units namespace, e.g.:
>>> import astropy.units as u
>>> mph = u.imperial.mile / u.hour
>>> mph
Unit("mi / h")
To include them in compose and the results of
find_equivalent_units, do:
>>> import astropy.units as u
>>> u.imperial.enable()
answered Mar 11 at 17:21
IguananautIguananaut
8,99513041
Per the docs, Astropy does not include imperial units defined by default. I think this is in part to reduce the default units namespace and the overhead involved in creating and searching it, and imperial units get sacrificed in this case since they are less used in astronomy for the most part:
This package defines colloquially used Imperial units. They are available in the astropy.units.imperial namespace, but not in the top-level astropy.units namespace, e.g.:
>>> import astropy.units as u
>>> mph = u.imperial.mile / u.hour
>>> mph
Unit("mi / h")
To include them in compose and the results of
find_equivalent_units, do:
>>> import astropy.units as u
>>> u.imperial.enable()
answered Mar 11 at 17:21
IguananautIguananaut
8,99513041
answered Mar 11 at 17:21
IguananautIguananaut
8,99513041
answered Mar 11 at 17:21
IguananautIguananaut
8,99513041
answered Mar 11 at 17:21
IguananautIguananaut
8,99513041
8,99513041
add a comment |
add a comment |
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From the docs ""This package defines colloquially used Imperial units. They are available in the astropy.units.imperial namespace, but not in the top-level astropy.units namespace, To include them:
import astropy.units as u, u.imperial.enable()– BHC
Mar 8 at 13:40
Yes, You are right! I was blind :)
– Kuba Wenta
Mar 8 at 13:43