| Re: Five days mean values [message #45752] |
Fri, 07 October 2005 05:10 |
btt
Messages: 345
Registered: December 2000
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Senior Member |
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Julio wrote:
> Thanks Mike, now I undestand how rebin works...
>
> I'll try this, but I'm not sure rebin will help me... The fact is my
> temperature array size is not always multiple of five, due to the
> number of days in a month (28, 30 or 31).
>
> Kind Regards,
> J�lio
>
Hello,
I'm not sure what you want to do with the 'extra' days each month. But
if what you want is to bin the dates into 5-day consecutive groups then
I suggest you...
(1) Convert your dates from strings to Julian day values. You need to
used some string processing before you get to using JULDAY().
(2) Sort your dates into 5 day bins using HISTOGRAM().
H = HISTOGRAM(myJulianDayValues, START = JULDAY(8,1,2001), BIN = 5, $
REVERSE = r, LOCATION = binStartDates)
(3) Use the REVERSE_INDICES to extract the records for each 5 day group.
Hope that helps.
Cheers,
Ben
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| Re: Five days mean values [message #45757 is a reply to message #45752] |
Thu, 06 October 2005 11:04  |
Michael Wallace
Messages: 409
Registered: December 2003
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Senior Member |
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> I'll try this, but I'm not sure rebin will help me... The fact is my
> temperature array size is not always multiple of five, due to the
> number of days in a month (28, 30 or 31).
With or without rebin, what do you with the days at the end of a month.
You said previously that you were calculating 5-day averages. What do
you want to do with the left over days?
-Mike
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| Re: Five days mean values [message #45758 is a reply to message #45757] |
Thu, 06 October 2005 10:54 |
Julio[1]
Messages: 52
Registered: May 2005
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Member |
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Thanks Mike, now I undestand how rebin works...
I'll try this, but I'm not sure rebin will help me... The fact is my
temperature array size is not always multiple of five, due to the
number of days in a month (28, 30 or 31).
Kind Regards,
Júlio
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| Re: Five days mean values [message #45760 is a reply to message #45758] |
Thu, 06 October 2005 07:23 |
Michael Wallace
Messages: 409
Registered: December 2003
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Senior Member |
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I wouldn't have thought of using rebin, but rebin will do the job for
you as Ken describes. Rebin is used to resize arrays. When you
compress an array, rebin take averages of the data points in the array.
You can use rebin to compress your temperature array and when it does
so, averages will automatically be taken.
In order for the results to be accurate, you can neither have any
missing days nor can your temperature array size not be a multiple of
five. The array has to be a multiple of five because the compression
and averaging is occurring over every five data points.
; temp is your temperature array
avgtemp = rebin(temp, n_elements(temp) / 5)
The n_elements(temp) / 5 tells us how many five element groups there are
in your temperature array. The first five elements will be averaged and
stored in avgtemp[0]. The second five elements will be averaged and
stored in avgtemp[1], etc.
-Mike
Julio wrote:
> Hi Kenneth,
>
> I don't see how rebin can help me... An example, considering the
> meteorological data below, I want to find the temp(max) mean values
> from 01/08/2001 to 05/08/2001 and from 06/08/2001 to 10/08/2001...
>
> So I have Temp (max) = 27.2 and 27.4
>
> The problem is I have a very large amount of data. I can't see a way to
> find the mean value every five days.
>
> Any comments welcome!
>
> Regards,
> J�lio
>
> Kenneth Bowman escreveu:
>
>
>> In article <1128546470.965721.312300@g49g2000cwa.googlegroups.com>,
>> "Julio" <julio@cpa.unicamp.br> wrote:
>>
>>
>>> Hello people,
>>>
>>> I have meteorological data from several years, like these:
>>>
>>> Date Temp (max) Temp (min)
>>> 01/08/2001 27.2 12.7
>>> 02/08/2001 27.8 12.4
>>> 03/08/2001 26.8 16.3
>>> 04/08/2001 26.6 12
>>> 05/08/2001 27.4 11
>>> 06/08/2001 27.6 16.1
>>> 07/08/2001 27.6 11.2
>>> 08/08/2001 28.4 13.4
>>> 09/08/2001 27.2 10.9
>>> 10/08/2001 26 9.7
>>>
>>> I want to calculate mean value from 01-05 days, 05-10 days, 10-15 days
>>> and so on.
>>> In other words, I must get the mean value at each five days.
>>> I'm trying to make some code to get it automatically. I think this kind
>>> of work is trivial for meteorological data users. Does anybody have
>>> some idea?
>>>
>>> Regards,
>>> J�lio
>>
>> If the data are in an array T, this is a very quick way to compute averages
>>
>> IDL> T = findgen(15)
>> IDL> print, T
>> 0.00000 1.00000 2.00000 3.00000 4.00000 5.00000
>> 6.00000
>> 7.00000 8.00000 9.00000 10.0000 11.0000 12.0000
>> 13.0000
>> 14.0000
>> IDL> print, rebin(T, 3)
>> 2.00000 7.00000 12.0000
>>
>> but make sure that the number of days is a multiple of 5 and that you have no
>> missing data. Also, watch out for leap days.
>>
>> Ken Bowman
>
>
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| Re: Five days mean values [message #45761 is a reply to message #45760] |
Thu, 06 October 2005 05:56 |
Julio[1]
Messages: 52
Registered: May 2005
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Member |
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Hi Kenneth,
I don't see how rebin can help me... An example, considering the
meteorological data below, I want to find the temp(max) mean values
from 01/08/2001 to 05/08/2001 and from 06/08/2001 to 10/08/2001...
So I have Temp (max) = 27.2 and 27.4
The problem is I have a very large amount of data. I can't see a way to
find the mean value every five days.
Any comments welcome!
Regards,
Júlio
Kenneth Bowman escreveu:
> In article <1128546470.965721.312300@g49g2000cwa.googlegroups.com>,
> "Julio" <julio@cpa.unicamp.br> wrote:
>
>> Hello people,
>>
>> I have meteorological data from several years, like these:
>>
>> Date Temp (max) Temp (min)
>> 01/08/2001 27.2 12.7
>> 02/08/2001 27.8 12.4
>> 03/08/2001 26.8 16.3
>> 04/08/2001 26.6 12
>> 05/08/2001 27.4 11
>> 06/08/2001 27.6 16.1
>> 07/08/2001 27.6 11.2
>> 08/08/2001 28.4 13.4
>> 09/08/2001 27.2 10.9
>> 10/08/2001 26 9.7
>>
>> I want to calculate mean value from 01-05 days, 05-10 days, 10-15 days
>> and so on.
>> In other words, I must get the mean value at each five days.
>> I'm trying to make some code to get it automatically. I think this kind
>> of work is trivial for meteorological data users. Does anybody have
>> some idea?
>>
>> Regards,
>> Júlio
>
> If the data are in an array T, this is a very quick way to compute averages
>
> IDL> T = findgen(15)
> IDL> print, T
> 0.00000 1.00000 2.00000 3.00000 4.00000 5.00000
> 6.00000
> 7.00000 8.00000 9.00000 10.0000 11.0000 12.0000
> 13.0000
> 14.0000
> IDL> print, rebin(T, 3)
> 2.00000 7.00000 12.0000
>
> but make sure that the number of days is a multiple of 5 and that you have no
> missing data. Also, watch out for leap days.
>
> Ken Bowman
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| Re: Five days mean values [message #45764 is a reply to message #45761] |
Wed, 05 October 2005 14:57 |
K. Bowman
Messages: 330
Registered: May 2000
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Senior Member |
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In article <1128546470.965721.312300@g49g2000cwa.googlegroups.com>,
"Julio" <julio@cpa.unicamp.br> wrote:
> Hello people,
>
> I have meteorological data from several years, like these:
>
> Date Temp (max) Temp (min)
> 01/08/2001 27.2 12.7
> 02/08/2001 27.8 12.4
> 03/08/2001 26.8 16.3
> 04/08/2001 26.6 12
> 05/08/2001 27.4 11
> 06/08/2001 27.6 16.1
> 07/08/2001 27.6 11.2
> 08/08/2001 28.4 13.4
> 09/08/2001 27.2 10.9
> 10/08/2001 26 9.7
>
> I want to calculate mean value from 01-05 days, 05-10 days, 10-15 days
> and so on.
> In other words, I must get the mean value at each five days.
> I'm trying to make some code to get it automatically. I think this kind
> of work is trivial for meteorological data users. Does anybody have
> some idea?
>
> Regards,
> J�lio
If the data are in an array T, this is a very quick way to compute averages
IDL> T = findgen(15)
IDL> print, T
0.00000 1.00000 2.00000 3.00000 4.00000 5.00000
6.00000
7.00000 8.00000 9.00000 10.0000 11.0000 12.0000
13.0000
14.0000
IDL> print, rebin(T, 3)
2.00000 7.00000 12.0000
but make sure that the number of days is a multiple of 5 and that you have no
missing data. Also, watch out for leap days.
Ken Bowman
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