<ul>
<li>Date/timestamps are tricky to convert without further information; are the values you show days (unlikely, if 990555 is your date/time value), hours, seconds? See if you can figure out the system used (either by knowing which date your timestamp is supposed to represent, or looking at the step sizes for the data); this will generally give you an idea of what is being represented, and how to convert to an actual date.</li>
<li>Coordinate and long/lat are basically the same thing, only built up from decimals or hours/minutes (e.g. 0.5 representing 30 minutes). Conversion can be found <a href="http://transition.fcc.gov/mb/audio/bickel/DDDMMSS-decimal.html" rel="nofollow">http://transition.fcc.gov/mb/audio/bickel/DDDMMSS-decimal.html</a>, for instance. Mind that other coordinate systems are also possible, which will make the conversion different; again, this is important to figure out before you can convert your data.</li>
<li>Comparison options will depend on: 1) whether your data corresponds (do you have measurements on the same date/time, as well as the same location?), 2) what you are interested in.</li>
</ul>

CC BY-SA 3.0
2014-05-27T09:15:58.143Z
Thank you for help! My data is representing "hour since 1900-01-01 00:00:00"
Student
CC BY-SA 3.0
2014-05-27T17:42:06.543Z
In that case, for example: 990555/24 = 41273.125 days since 1-1-1900. You can calculate the actual date from that manually (keep in mind to account for leap years, which may be quite troublesome). Another way is to enter your "number of days since 1900" in excel and converting the cells containing these values from a number to a date/time format, which will do the conversion for you (in this case resulting in 30-12-2012 3:00).
Pytrik
CC BY-SA 3.0
2014-05-27T22:44:24.403Z
@Student: Here's a way to make sure your computations are correct, using Wolframm: "[1-1-1900 + 990555  hours](http://www.wolframalpha.com/input/?i=1-1-1900+%2B+990555++hours)" = Jan 1, 2013; this assumes Wolframm is correct, which I'm pretty sure they are, though you could always find another source too.
blunders
CC BY-SA 3.0
2014-05-28T15:46:30.023Z
Blunders, your answer is correct. But how you got it? Can you write a formula?
Student
CC BY-SA 3.0
2014-06-16T08:32:22.960Z
@Student To stay in R you can use this snippet `as.POSIXct(990555*3600, origin="1900-01-01 00:00:00")`. The standard to represent time in R and in most modern languages is POSIX and in this standard time is represented in seconds not hours, hence the multiplication by 3600.
plannapus
CC BY-SA 3.0
2014-05-26T21:29:14.983Z
2014-05-26T21:29:14.983Z
Pytrik
<p>I want to make precipitation data analysis for the Latvian territory.
I only have meteorological stations and reanalysis  data. 
My data from meteorological stations are in the format:</p>

<pre><code>year  month  day    time      station1 station2 station3
2013    1     1    03.00        0                 0.1
2013    1     1    06.00        0.1               0.2
2013    1     1    15.00        0       0.1         0
2013    1     1    18.00        0.2     0         0.3
2013    1     2     03.00        0                 0.4
                  ...  
</code></pre>

<p>I also have a file with coordinates for meteorological stations.
And data from reanalysis is in format </p>

<pre><code>&gt; head(data)
  longitude latitude   time value
1     -27.0      -27 990555     0
2     -25.5      -27 990555     0
3     -24.0      -27 990555     0
4     -22.5      -27 990555     0
5     -21.0      -27 990555     0
6     -19.5      -27 990555     0
</code></pre>

<p>Edited: My main questions are:
1) What can tell me if these data are similar, should I look their summary and compare, or it is better to use some tests. Could you advice methods how to compare different data.
2) If there are missing values, should I just get them out and then analyse data?<br>
3) Maybe someone can advice some tutorial with R code? Can someone tell me how to decode time values from format like 990555 (I need formula or function for R)?</p>

CC BY-SA 3.0
2014-05-27T12:50:57.477Z
Unclear what you're asking - we need to know what you actually want to *do* with this data; what you want to find out. voting to close, but perhaps you could add some more detail and see if it gets reopened?
Semidiurnal Simon
CC BY-SA 3.0
2014-05-29T08:52:02.487Z
Hi Student, thanks for the revisions and clarification. Now that I can see what you're asking, I think that you might be better off asking the first half on statistics.SE, and for the second half doing some searching (I think Stackoverflow would want to see some more research before helping)
Semidiurnal Simon
<meteorology><precipitation><data-formats>
How to perform precipitation data analysis?
Student
# How to perform precipitation data analysis?

(Asked by: username_0 on 2014-05-25)

I want to make precipitation data analysis for the Latvian territory.
I only have meteorological stations and reanalysis data.
My data from meteorological stations are in the format:

```
year  month  day    time      station1 station2 station3
2013    1     1    03.00        0                 0.1
2013    1     1    06.00        0.1               0.2
2013    1     1    15.00        0       0.1         0
2013    1     1    18.00        0.2     0         0.3
2013    1     2     03.00        0                 0.4
                  ...  

```

I also have a file with coordinates for meteorological stations.
And data from reanalysis is in format

```
> head(data)
  longitude latitude   time value
1     -27.0      -27 990555     0
2     -25.5      -27 990555     0
3     -24.0      -27 990555     0
4     -22.5      -27 990555     0
5     -21.0      -27 990555     0
6     -19.5      -27 990555     0

```

Edited: My main questions are:
1) What can tell me if these data are similar, should I look their summary and compare, or it is better to use some tests. Could you advice methods how to compare different data.
2) If there are missing values, should I just get them out and then analyse data?  
3) Maybe someone can advice some tutorial with R code? Can someone tell me how to decode time values from format like 990555 (I need formula or function for R)?

**Tags:** meteorology, precipitation, data-formats

> **Comments:**
> **username_1**: Unclear what you're asking - we need to know what you actually want to *do* with this data; what you want to find out. voting to close, but perhaps you could add some more detail and see if it gets reopened?  
> **username_1**: Hi username_0, thanks for the revisions and clarification. Now that I can see what you're asking, I think that you might be better off asking the first half on statistics.SE, and for the second half doing some searching (I think Stackoverflow would want to see some more research before helping)  

---

## 1 answers

### Answer by username_2 on 2014-05-26. Score: 3

* Date/timestamps are tricky to convert without further information; are the values you show days (unlikely, if 990555 is your date/time value), hours, seconds? See if you can figure out the system used (either by knowing which date your timestamp is supposed to represent, or looking at the step sizes for the data); this will generally give you an idea of what is being represented, and how to convert to an actual date.
* Coordinate and long/lat are basically the same thing, only built up from decimals or hours/minutes (e.g. 0.5 representing 30 minutes). Conversion can be found <http://transition.fcc.gov/mb/audio/bickel/DDDMMSS-decimal.html>, for instance. Mind that other coordinate systems are also possible, which will make the conversion different; again, this is important to figure out before you can convert your data.
* Comparison options will depend on: 1) whether your data corresponds (do you have measurements on the same date/time, as well as the same location?), 2) what you are interested in.

> **Comments:**
> **username_0**: Thank you for help! My data is representing "hour since 1900-01-01 00:00:00"  
> **username_2**: In that case, for example: 990555/24 = 41273.125 days since 1-1-1900. You can calculate the actual date from that manually (keep in mind to account for leap years, which may be quite troublesome). Another way is to enter your "number of days since 1900" in excel and converting the cells containing these values from a number to a date/time format, which will do the conversion for you (in this case resulting in 30-12-2012 3:00).  
> **username_3**: @username_0: Here's a way to make sure your computations are correct, using Wolframm: "[1-1-1900 + 990555  hours](http://www.wolframalpha.com/input/?i=1-1-1900+%2B+990555++hours)" = Jan 1, 2013; this assumes Wolframm is correct, which I'm pretty sure they are, though you could always find another source too.  
> **username_0**: username_3, your answer is correct. But how you got it? Can you write a formula?  
> **username_4**: @username_0 To stay in R you can use this snippet `as.POSIXct(990555*3600, origin="1900-01-01 00:00:00")`. The standard to represent time in R and in most modern languages is POSIX and in this standard time is represented in seconds not hours, hence the multiplication by 3600.  

---

