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RE: st: best practice for dates and times
From
Nick Cox <[email protected]>
To
"'[email protected]'" <[email protected]>
Subject
RE: st: best practice for dates and times
Date
Mon, 21 Feb 2011 20:11:17 +0000
I am afraid that for me your verbalisation of what you think is happening casts no light on the most recent specific examples. I am wary of claiming "proof" of internal operation when all I see are inputs and outputs.
Nick
[email protected]
Liao, Junlin
Nick,
What you observed is explained by another mechanism and this mechanism further proves that -replace- takes consideration of inputs. -replace- will use double if a variable with double type is involved in calculation, and it will use float if float variable is involved. It goes float if both double and float are involved. It goes to double again if float cancels each other out. Same is true with double. It does not make sense to go float when both double and float are involved (it does not matter if you set type to double).
Junlin
-----Original Message-----
From: [email protected] [mailto:[email protected]] On Behalf Of Nick Cox
Sent: Monday, February 21, 2011 1:17 PM
To: '[email protected]'
Subject: RE: st: best practice for dates and times
Good question. There appears to be some switch in the code and I don't know what it is.
As said, I'd be surprised if the switch was -replace- detecting the use of -clock()- but I might well be wrong.
Here is another experiment. If you also go
gen byte x2 = 1
replace x2 = x
-x2- is also replaced as double. In what sense does the second -replace- also know that its argument came from a -clock()- calculation, as you seem to be supposing? I doubt that it does, but any amount of guesswork can only be settled by looking at the source code.
Nick
[email protected]
Liao, Junlin
Maarten,
How do you explain the following? I found out that it does not matter if you set type to float or double. -replace- always promote decimal numbers to float for most of calculations. However, for -clock()- results, it chooses double. If you and Nick are correct, then the results should be consistently of one type or the type according to setting.
. clear
. set type float
. set obs 1
obs was 0, now 1
. gen byte x=1
. replace x=clock("1may2001 1:20","DMY hm") x was byte now double
(1 real change made)
. gen byte y=1
. replace y=2.123456^3
y was byte now float
(1 real change made)
. des
Contains data
obs: 1
vars: 2
size: 16 (99.9% of memory free)
--------------------------------------------------------------------------------------------------------------------------------------------------------------
storage display value
variable name type format label variable label
--------------------------------------------------------------------------------------------------------------------------------------------------------------
x double %8.0g
y float %8.0g
--------------------------------------------------------------------------------------------------------------------------------------------------------------
Sorted by:
Note: dataset has changed since last saved
Junlin
-----Original Message-----
From: [email protected] [mailto:[email protected]] On Behalf Of Maarten buis
Sent: Monday, February 21, 2011 12:45 PM
To: [email protected]
Subject: RE: st: best practice for dates and times
--- Liao, Junlin
> Please note in previous code I had intentionally set type to float.
> -replace- still choose double for results of -clock-. Suppose I do
> another calculation, -replace- will choose float. Therefore, -replace-
> must know about -clock-.
--- On Mon, 21/2/11, Nick Cox wrote:
> As I understand it, -replace- looks at values, not what produced them.
> If it sees a particular number that should be stored as -double-, it
> is not because it comes tagged with a history "produced by -clock()-;
> act accordingly".
Nick seems to be right, -replace- will increase the storage type regarless of what function produced it, just not float, which I think is a sensible default given what was said before.
set obs 1
gen byte foo = 1
replace foo = 10000
desc
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