| lambda functions in IDL-python bridge [message #91857] |
Fri, 04 September 2015 06:55  |
Helder Marchetto
Messages: 520
Registered: November 2011
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Senior Member |
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Hi,
I've got the next problem, using a lambda function in the python sorted function.
Consider this easy python example:
d0 = {'sorted': 1, 'unsorted': 0}
d1 = {'sorted': 5, 'unsorted': 1}
d2 = {'sorted': 3, 'unsorted': 2}
d3 = {'sorted': 7, 'unsorted': 3}
d4 = {'sorted': 4, 'unsorted': 4}
l = [d0,d1,d2,d3,d4]
sorted(l, key = lambda x:x['sorted'])
The result would be:
[{'sorted': 1, 'unsorted': 0},
{'sorted': 3, 'unsorted': 2},
{'sorted': 4, 'unsorted': 4},
{'sorted': 5, 'unsorted': 1},
{'sorted': 7, 'unsorted': 3}]
Now if I would want to do this in IDL, things would get difficult. Suppose I get a dictionary (or a simple object with a property...) as a result of an operation and I want to sort this result according to a specific property... how do I do that in IDL?
The synthax:
matches = Python.sorted(matches, key = lambda(x: x.distance))
will give me a compilation error.
Thanks,
Helder
PS: I found a different solution to this problem using IDL. I import all properties in an idl variable, sort the variable and apply this to the list. A python solution would be though nice, just to know how to deal with lambda functions.
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| Re: lambda functions in IDL-python bridge [message #91871 is a reply to message #91857] |
Tue, 08 September 2015 08:34  |
chris_torrence@NOSPAM
Messages: 528
Registered: March 2007
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Senior Member |
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On Friday, September 4, 2015 at 7:55:30 AM UTC-6, Helder wrote:
> Hi,
> I've got the next problem, using a lambda function in the python sorted function.
>
> Consider this easy python example:
>
> d0 = {'sorted': 1, 'unsorted': 0}
> d1 = {'sorted': 5, 'unsorted': 1}
> d2 = {'sorted': 3, 'unsorted': 2}
> d3 = {'sorted': 7, 'unsorted': 3}
> d4 = {'sorted': 4, 'unsorted': 4}
> l = [d0,d1,d2,d3,d4]
> sorted(l, key = lambda x:x['sorted'])
>
> The result would be:
> [{'sorted': 1, 'unsorted': 0},
> {'sorted': 3, 'unsorted': 2},
> {'sorted': 4, 'unsorted': 4},
> {'sorted': 5, 'unsorted': 1},
> {'sorted': 7, 'unsorted': 3}]
>
> Now if I would want to do this in IDL, things would get difficult. Suppose I get a dictionary (or a simple object with a property...) as a result of an operation and I want to sort this result according to a specific property... how do I do that in IDL?
>
> The synthax:
> matches = Python.sorted(matches, key = lambda(x: x.distance))
>
> will give me a compilation error.
>
> Thanks,
> Helder
>
> PS: I found a different solution to this problem using IDL. I import all properties in an idl variable, sort the variable and apply this to the list. A python solution would be though nice, just to know how to deal with lambda functions.
Hi Helder,
Unfortunately, IDL can't translate its Lambda functions into Python lambda functions. If you wanted to have Python do this, you could transfer the variable over to Python, and then execute a Python command. Like this:
IDL> d0 = hash('sorted', 1, 'unsorted', -1)
IDL> d1 = hash('sorted', 5, 'unsorted', 1)
IDL> d2 = hash('sorted', 3, 'unsorted', 2)
IDL> d3 = hash('sorted', 7, 'unsorted', 3)
IDL> d4 = hash('sorted', 4, 'unsorted', 4)
IDL> lst = list(d0,d1,d2,d3,d4)
IDL> Python.l = lst
IDL> >>>result = sorted(l, key = lambda x:x['sorted'])
IDL> result = Python.result
IDL> print, result, /implied
Now, if you wanted to do this purely in IDL, you could use List::Sort with a Lambda compare function, and the new .Compare method, which exists on IDL variables:
d0 = {sorted: 1, unsorted: 0}
d1 = {sorted: 5, unsorted: 1}
d2 = {sorted: 3, unsorted: 2}
d3 = {sorted: 7, unsorted: 3}
d4 = {sorted: 4, unsorted: 4}
lst = List(d0,d1,d2,d3,d4)
result = lst.Sort(COMPARE=Lambda(a,b:(a.sorted).Compare(b.sorted)))
print, result, /implied
Here's an example of how the .Compare method works:
IDL> print, (1).compare(2)
-1
IDL> print, (1).compare(1)
0
IDL> print, (1).compare(0)
1
Cheers,
Chris
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| Re: lambda functions in IDL-python bridge [message #91891 is a reply to message #91871] |
Wed, 09 September 2015 00:27 |
Helder Marchetto
Messages: 520
Registered: November 2011
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Senior Member |
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On Tuesday, September 8, 2015 at 5:34:11 PM UTC+2, Chris Torrence wrote:
> On Friday, September 4, 2015 at 7:55:30 AM UTC-6, Helder wrote:
>> Hi,
>> I've got the next problem, using a lambda function in the python sorted function.
>>
>> Consider this easy python example:
>>
>> d0 = {'sorted': 1, 'unsorted': 0}
>> d1 = {'sorted': 5, 'unsorted': 1}
>> d2 = {'sorted': 3, 'unsorted': 2}
>> d3 = {'sorted': 7, 'unsorted': 3}
>> d4 = {'sorted': 4, 'unsorted': 4}
>> l = [d0,d1,d2,d3,d4]
>> sorted(l, key = lambda x:x['sorted'])
>>
>> The result would be:
>> [{'sorted': 1, 'unsorted': 0},
>> {'sorted': 3, 'unsorted': 2},
>> {'sorted': 4, 'unsorted': 4},
>> {'sorted': 5, 'unsorted': 1},
>> {'sorted': 7, 'unsorted': 3}]
>>
>> Now if I would want to do this in IDL, things would get difficult. Suppose I get a dictionary (or a simple object with a property...) as a result of an operation and I want to sort this result according to a specific property... how do I do that in IDL?
>>
>> The synthax:
>> matches = Python.sorted(matches, key = lambda(x: x.distance))
>>
>> will give me a compilation error.
>>
>> Thanks,
>> Helder
>>
>> PS: I found a different solution to this problem using IDL. I import all properties in an idl variable, sort the variable and apply this to the list. A python solution would be though nice, just to know how to deal with lambda functions.
>
> Hi Helder,
>
> Unfortunately, IDL can't translate its Lambda functions into Python lambda functions. If you wanted to have Python do this, you could transfer the variable over to Python, and then execute a Python command. Like this:
>
> IDL> d0 = hash('sorted', 1, 'unsorted', -1)
> IDL> d1 = hash('sorted', 5, 'unsorted', 1)
> IDL> d2 = hash('sorted', 3, 'unsorted', 2)
> IDL> d3 = hash('sorted', 7, 'unsorted', 3)
> IDL> d4 = hash('sorted', 4, 'unsorted', 4)
> IDL> lst = list(d0,d1,d2,d3,d4)
> IDL> Python.l = lst
> IDL> >>>result = sorted(l, key = lambda x:x['sorted'])
> IDL> result = Python.result
> IDL> print, result, /implied
>
> Now, if you wanted to do this purely in IDL, you could use List::Sort with a Lambda compare function, and the new .Compare method, which exists on IDL variables:
>
> d0 = {sorted: 1, unsorted: 0}
> d1 = {sorted: 5, unsorted: 1}
> d2 = {sorted: 3, unsorted: 2}
> d3 = {sorted: 7, unsorted: 3}
> d4 = {sorted: 4, unsorted: 4}
> lst = List(d0,d1,d2,d3,d4)
> result = lst.Sort(COMPARE=Lambda(a,b:(a.sorted).Compare(b.sorted)))
> print, result, /implied
>
> Here's an example of how the .Compare method works:
> IDL> print, (1).compare(2)
> -1
> IDL> print, (1).compare(1)
> 0
> IDL> print, (1).compare(0)
> 1
>
> Cheers,
> Chris
Hi Chris,
thanks for the info. I'm fighting to understand how to interface python and IDL. Your examples are very nice and hope that more of such will come (that's a suggestion...).
Your IDL solution looks very exotic (=cool).
Cheers,
Helder
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