Re: [libreoffice-users] Re: Not receiving posts from the List

[Marion & Noel Lodge]

Hi,

Thanks for the suggestions.

I eventually discovered that the Gmail filter I had set up, (to add the
label 'LibreOffice'), had become corrupted and the label was blank.  Not
sure where the emails were going.  They didn't seem to be in Trash, but a
global search found them. Fixing the filter fixed the problem.

​Noel​
--
Marion & Noel Lodge
lodg...@gmail.com

On 9 May 2017 at 05:35, Kolbjørn Stuestøl  wrote:

> Den 08.05.2017 16:55, anne-ology skreiv:
>
>> possibly
>>
>>   your spam folder has been re-set where it collects even
>> those
>> not really spam;
>>
>>   your ISP is at fault - either not sending what it claims as
>> spam or losing e-mails in some glitch or  ???;
>>
>>   ??? - these 'bits & bytes' sometime seem to have a mind of
>> their own  ;-)
>>
> I had the same problem some times ago. My web host (domeneshop.no) had
> banned the IP addresses 5.9.148.85 and 2a01:4f8:190:3144::2. An e-mail to
> the host and they removed the ban.
>
> Kolbjørn
>
>
>
>>
>> From: Marion & Noel Lodge 
>> Date: Mon, May 8, 2017 at 4:51 AM
>> Subject: Not receiving posts from the List
>> To: users+ow...@global.libreoffice.org
>>
>>
>> Hi,
>>
>> The last messages I received from this list were on 19th April 2017.  I
>> checked on Nabble and there have been messages posted there since that
>> date, but they have not been forwarded to my email address.
>>
>> I thought that my subscription might have dropped out so I tried
>> re-subscribing, but received the message -
>>   You were unable to be subscribed to the list because you are already
>> subscribed.
>>
>> Any suggestions on how I might trouble shoot this problem?
>>
>> Thanks,
>>
>> Noel
>>
>>
>
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Re: [libreoffice-users] COUNTIF / DAYS / ?

[Hylton Conacher (ZR1HPC)]

Michael,

Thank-you for the time and response

On 08/05/2017 08:54, Michael D. Setzer II wrote:

Not sure if I fully understand what you are looking at,
but here is what I have attempted.
Assuming you column data is:
A is month
B is day of Month
C is the rain on day in 2016
D is the rain on day in 2017


Correct :)


I've added Column E and Column F to be real date value.
E2 =DATE(2016,MATCH(A2,$G$1:$G$12),B2)
F2 =DATE(2017,MATCH(A2,$G$1:$G$12),B2)


This turns out to be a problem as I am working with five(5) years of 
data and whilst my example did not shot it there are multiple values 
under each month, even if it is zero.


I don't think five columns of dates is a smart way to do this.
I am rather going to investigate making my date column a rolling year 
numeric month/day for 365 rows, with the columns being divided into years.


Example:

Date201620172018
01010   1   5
01021   0   0
01030   5   10
05011   0   0
09225   1   1
31120   1   0

I then reckon use of MATCH, DAYS, COUNTIF and SUM should deliver what I 
seek.


Your thoughts on how to get the rows in a column formatted as Date to 
show a sequential month and day number?


I have Bcc'd you as the message contain an attachment of my existing sheet.

Regards
Hylton

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Re: [libreoffice-users] COUNTIF / DAYS / ?

[Hylton Conacher (ZR1HPC)]

On 07/05/2017 23:48, Remy Gauthier wrote:

Hi,

If you have one row per day of the year, you can find out easily where 
the row corresponding to "TODAY()-1" is located by finding out the 
number of days between Jan 1st of that year and "yesterday". This would 
be something like (you may need to adjust based on the heading of your 
column(s)):


=DAYS(YEAR(TODAY()-1)-1;MONTH(TODAY()-1);DAY(TODAY()-1);YEAR(TODAY()-1)-1;1;1)

The formula above calculates the number of days between yesterday last 
year and Jan 1st of last year. Now, you need to find in which column to 
add. Since you seem to have a heading providing the year, you can use 
the MATCH() function to locate the column:


=MATCH(YEAR(TODAY()-1)-1;$a$1:$z$1;0)

This should give you the column where last year's data is located (I 
used a1 to z1, but any row reference will work).


To add the values, you now just need to use a combination of SUM() and 
OFFSET():


=SUM(OFFSET($A$2;0;MATCH(YEAR(TODAY()-1)-1;$a$1:$z$1;0)-1;DAYS(YEAR(TODAY()-1)-1;MONTH(TODAY()-1);DAY(TODAY()-1);YEAR(TODAY()-1)-1;1;1);1))

What that does:

OFFSET() allows you to create dynamic arrays for functions like SUM(). I 
used the anchor point $A$2 (I supposed your heading is in row 1), and 
told OFFSET to move from that reference 0 rows down and 
"MATCH(YEAR(TODAY()-1)-1;$a$1:$z$1;0)-1" columns to the right; then, 
from that new point, make an array that is 
"DAYS(YEAR(TODAY()-1)-1;MONTH(TODAY()-1);DAY(TODAY()-1);YEAR(TODAY()-1)-1;1;1)" 
rows high and 1 column wide. The resulting array is then used by SUM() 
to calculate the total rainfall.


This will probably need to be adjusted a bit to take into account the 
position of your column heading and to correctly find the row where to 
stop, but you should now have something to play with.


I hope this helps.

Rémy Gauthier.


Thank you for the in-depth reply! I will need to play with this  and the 
functions you mention to get it right but hopefully I will come right. I 
will post my final formula here, for anyone else wanting to do something 
similar.



Regards
Hylton

Le dimanche 07 mai 2017 à 21:55 +0200, Hylton Conacher (ZR1HPC) a écrit :

Hi,

I have a spreadsheet I use to draw graphs and extract information about
the rainfall in my area for the last two years.

I need to create a formula that will count how many days it rained
between the start of the year, both last year and this year, so I can
compare the amount of rainfall that was received during that time frame.

The example below is a sample dataset that shows how much rainfall was
received in 2016 or 2017, on each of the four days

Example:
Month Day   20162017
Jan  1  0   15
Feb  29 1   0   
May  6  0   1
Sept 22 15  5

  From this we can determine that 16mm of rain fell in 2016 over 2 days.
Likewise we can determine that 21mm of rain fell in 2017 but over 3 days.

Today is the 7th of May 2017 and I would like a formula to work out how
many days the rainfall received between(and incl) 2016/01/01 and the day
before TODAY last year. The answer for 2016 is 1 i.e. it only rained
once between the dates specified, however the answer for 2017 is 16 as
it rained twice between TODAY-1 and 2017/1/1. I had thought of using
COUNTIF or DAYS or =COUNTIF(C2:SUM(TODAY()-1,">0")), however the problem
is that I do not have a single date column, but three.

This formula will allow me to create a graph showing how much rainfall
had fallen last year compared to this year between the beginning of the
year(01/01) and the day prior to Today.

I had thought of converting the first two columns into a single and
having the text name of the month with each successive line being a new
date in that month until the month changed, however I was unable to get
it to work i.e. automatically change month after 31 days in January to
read February 01.
As you can imagine I have 367 rows of data per year, mostly with 0 as a
value, however there are odd days it does rain.

Any comments, and if you want the original spreadsheet, just yell where
to put it i.e. Nabble etc.

Regards
Hylton




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Re: [libreoffice-users] Frequency Distribution

[jorge]

Hi Brian:

I've searched information in the web but usually it is more complex
than my knowledge can evaluate. For example, in Wikipedia found
https://en.wikipedia.org/wiki/Beta_distribution, but it so deep for me,
but show some graph with different values of Beta and Alpha that could
help me.

I found some others like this, easier to understand the information:

https://stats.stackexchange.com/questions/47771/what-is-the-intuition-behind-beta-distribution

And found a program named "SciDavies" that could help me too.

But the major problem to me is my knowledge in statistics. It is not to
deep, although I already found some books that could help me too.

In general, when I made my question about it, I was trying to find a
way to make the process easier and fast to me.

Thank you very much because you try to help me, and clear me that
LibreOffice can't help with its formulas to solve my problem directly.

Regards,

-- 
Atentamente,

Jorge Rodríguez


El mar, 09-05-2017 a las 16:25 +0100, Brian Barker escribió:
> At 08:08 08/05/2017 -0600, Jorge Rodríguez wrote:
> >Hi Brian, well, thank you for your answer
> 
> My point, of course, was that it was sufficient - 
> indeed, better - to explain that you had a table 
> with consecutive integers in one column and real 
> values in the other, rather than listing 
> eighty-two values and asking readers to make that observation themselves.
> 
> At 14:10 07/05/2017 -0600, Jorge Rodríguez wrote:
> >If I have a number of data about some topic, how 
> >can I get the type of distribution using the 
> >LibreOffice's formula ? How can I know if it is 
> >normal, gamma, etc. ? There is a way ?
> 
> The simple answer is that your data will 
> certainly match none of those distributions - at least, not precisely.
> 
> Think of a straight line: this can be defined 
> exactly by two points. If you have three points, 
> they will not - except by an impossible 
> coincidence - lie on a straight line. But if you 
> believe that your data should lie on a straight 
> line, there are techniques that determine the 
> straight line which probably misses all of your 
> points but is in some sense the "best fit" to them.
> 
> Similarly, your analytic distributions will be 
> definable by a small number of points - certainly 
> a lot less than your forty-one. So the best you 
> can do is to attempt to fit various distributions 
> to your data and also to discover how good the 
> fit is in each case. That might help you to see 
> which distribution the data fits best. But a 
> better logic would be to decide from the nature 
> of your data - what it represents - which 
> distribution is likely, and then to fit the data 
> to such a distribution (if that is necessary).
> 
> In any case, Calc does not appear to have any 
> functionality that would help you here, but you 
> will certainly be able to build formulae to 
> achieve what you want in Calc once you have 
> unearthed the relevant mathematics. That will surely be out there on the web.
> 
> I trust this helps.
> 
> Brian Barker
> 
> 


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