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smart hendsome

2015-Feb-25 22:54 UTC

[R] Replace the value with 1 and 0

Hi everyone,
I have this kind of rainfall dataset:
?? Year Month Day Amount
1? 1950???? 1?? 1??? 0.0
2? 1950???? 1?? 2?? 35.5
3? 1950???? 1?? 3?? 17.8
4? 1950???? 1?? 4?? 24.5
5? 1950???? 1?? 5?? 12.3
6? 1950???? 1?? 6?? 11.5
7? 1950???? 1?? 7??? 5.7
8? 1950???? 1?? 8?? 13.2
9? 1950???? 1?? 9?? 11.3
10 1950???? 1? 10?? 14.7
11 1950???? 1? 11?? 11.9
12 1950???? 1? 12?? 17.5
13 1950???? 1? 13??? 8.1
14 1950???? 1? 14??? 0.4
15 1950???? 1? 15??? 0.0
16 1950???? 1? 16?? 19.5
17 1950???? 1? 17?? 10.7
18 1950???? 1? 18??? 0.5
19 1950???? 1? 19?? 12.7
20 1950???? 1? 20??? 6.3

I want to set as rain for Amount> 0 and not rain for Amount = 0.? I want to
replace the Amount>0 with 1 and Amount equal to zero with 0.? Then I want to
count how many rain in that particular month in that year. Anyone can help me?
This is what I want:
 Year Month Day Amount
1? 1950???? 1?? 1?? ? 0
2? 1950???? 1?? 2? ?? 1
3? 1950???? 1?? 3???? 1
4? 1950???? 1?? 4???? 1
5? 1950???? 1?? 5? ?? 1
6? 1950???? 1?? 6???? 1
7? 1950???? 1?? 7?? ? 1
8? 1950???? 1?? 8???? 1
9? 1950???? 1?? 9???? 1
10 1950???? 1? 10??? 1
11 1950???? 1? 11?? ? 1
12 1950???? 1? 12??? 1
13 1950???? 1? 13??? 1
14 1950???? 1? 14??? 1
15 1950???? 1? 15??? 1
16 1950???? 1? 16??? 1
17 1950???? 1? 17??? 1
18 1950???? 1? 18??? 1
19 1950???? 1? 19??? 1
20 1950???? 1? 20??? 1
Then become like this:
 
| 
 | Jan | Feb | Mar | Apr | May | Jun | Jul | Aug | Sep | Oct | Nov | Dec |
| 1950 | 17 | 6 | 23 | 20 | 19 | 9 | 17 | 23 | 18 | 20 | 20 | 17 |
| 1951 | 23 | 19 | 20 | 20 | 19 | 11 | 16 | 20 | 22 | 25 | 25 | 16 |
| 1952 | 15 | 21 | 30 | 24 | 23 | 20 | 16 | 19 | 20 | 19 | 21 | 15 |


Thanks.


	[[alternative HTML version deleted]]

Peter Alspach

2015-Feb-25 23:26 UTC

head link

[R] Replace the value with 1 and 0

Tena koe

Something like:

set.seed(153)
# Create some (unrealistic) rainfall data
yourData <- data.frame(year=rep(1950:1954, each=10), month=rep(rep(1:2,
each=5), 5), rain=sample(0:1, 50, replace=TRUE)*round(rnorm(50, 20, 2), 1))
tapply(yourData$rain>0, yourData[,c('year','month')], sum)

will give your final table, but it will need some tidying up.

HTH ....

Peter Alspach


-----Original Message-----
From: R-help [mailto:r-help-bounces at r-project.org] On Behalf Of smart
hendsome
Sent: Thursday, 26 February 2015 11:54 a.m.
To: r-help at r-project.org
Subject: [R] Replace the value with 1 and 0

Hi everyone,
I have this kind of rainfall dataset:
?? Year Month Day Amount
1? 1950???? 1?? 1??? 0.0
2? 1950???? 1?? 2?? 35.5
3? 1950???? 1?? 3?? 17.8
4? 1950???? 1?? 4?? 24.5
5? 1950???? 1?? 5?? 12.3
6? 1950???? 1?? 6?? 11.5
7? 1950???? 1?? 7??? 5.7
8? 1950???? 1?? 8?? 13.2
9? 1950???? 1?? 9?? 11.3
10 1950???? 1? 10?? 14.7
11 1950???? 1? 11?? 11.9
12 1950???? 1? 12?? 17.5
13 1950???? 1? 13??? 8.1
14 1950???? 1? 14??? 0.4
15 1950???? 1? 15??? 0.0
16 1950???? 1? 16?? 19.5
17 1950???? 1? 17?? 10.7
18 1950???? 1? 18??? 0.5
19 1950???? 1? 19?? 12.7
20 1950???? 1? 20??? 6.3

I want to set as rain for Amount> 0 and not rain for Amount = 0.? I want to
replace the Amount>0 with 1 and Amount equal to zero with 0.? Then I want to
count how many rain in that particular month in that year. Anyone can help me?
This is what I want:
 Year Month Day Amount
1? 1950???? 1?? 1?? ? 0
2? 1950???? 1?? 2? ?? 1
3? 1950???? 1?? 3???? 1
4? 1950???? 1?? 4???? 1
5? 1950???? 1?? 5? ?? 1
6? 1950???? 1?? 6???? 1
7? 1950???? 1?? 7?? ? 1
8? 1950???? 1?? 8???? 1
9? 1950???? 1?? 9???? 1
10 1950???? 1? 10??? 1
11 1950???? 1? 11?? ? 1
12 1950???? 1? 12??? 1
13 1950???? 1? 13??? 1
14 1950???? 1? 14??? 1
15 1950???? 1? 15??? 1
16 1950???? 1? 16??? 1
17 1950???? 1? 17??? 1
18 1950???? 1? 18??? 1
19 1950???? 1? 19??? 1
20 1950???? 1? 20??? 1
Then become like this:
 
| 
 | Jan | Feb | Mar | Apr | May | Jun | Jul | Aug | Sep | Oct | Nov | Dec |
| 1950 | 17 | 6 | 23 | 20 | 19 | 9 | 17 | 23 | 18 | 20 | 20 | 17 |
| 1951 | 23 | 19 | 20 | 20 | 19 | 11 | 16 | 20 | 22 | 25 | 25 | 16 |
| 1952 | 15 | 21 | 30 | 24 | 23 | 20 | 16 | 19 | 20 | 19 | 21 | 15 |


Thanks.


	[[alternative HTML version deleted]]

______________________________________________
R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

The contents of this e-mail are confidential and may be subject to legal
privilege.
 If you are not the intended recipient you must not use, disseminate, distribute
or
 reproduce all or any part of this e-mail or attachments.  If you have received
this
 e-mail in error, please notify the sender and delete all material pertaining to
this
 e-mail.  Any opinion or views expressed in this e-mail are those of the
individual
 sender and may not represent those of The New Zealand Institute for Plant and
 Food Research Limited.

JS Huang

2015-Feb-25 23:33 UTC

head link

[R] Replace the value with 1 and 0

Hi,

  Here is an implementation.  More data are added.  An extra column hasRain
is added instead of replacing column Amount.
> rain   Year Month Day Amount
1  1950     1   1    0.0
2  1950     1   2   35.5
3  1950     1   3   17.8
4  1950     1   4   24.5
5  1950     1   5   12.3
6  1950     1   6   11.5
7  1950     1   7    5.7
8  1950     1   8   13.2
9  1950     1   9   11.3
10 1950     1  10   14.7
11 1950     1  11   11.9
12 1950     1  12   17.5
13 1950     1  13    8.1
14 1950     1  14    0.4
15 1950     1  15    0.0
16 1950     1  16   19.5
17 1950     1  17   10.7
18 1950     1  18    0.5
19 1950     1  19   12.7
20 1950     1  20    6.3
21 1950     2   1    0.0
22 1950     2   2   35.5
23 1950     2   3   17.8
24 1950     2   4   24.5
25 1950     2   5   12.3
26 1950     2   6   11.5
27 1950     2   7    5.7
28 1950     2   8   13.2
29 1950     2   9   11.3
30 1950     2  10   14.7
31 1950     2  11   11.9
32 1950     2  12   17.5
33 1950     2  13    8.1
34 1950     2  14    0.4
35 1950     2  15    0.0
36 1950     2  16   19.5
37 1950     2  17   10.7
38 1950     2  18    0.0
39 1950     2  19    0.0
40 1950     2  20    0.0> rain$hasRain <- ifelse(rain$Amount>0,1,0)
> rain   Year Month Day Amount hasRain
1  1950     1   1    0.0       0
2  1950     1   2   35.5       1
3  1950     1   3   17.8       1
4  1950     1   4   24.5       1
5  1950     1   5   12.3       1
6  1950     1   6   11.5       1
7  1950     1   7    5.7       1
8  1950     1   8   13.2       1
9  1950     1   9   11.3       1
10 1950     1  10   14.7       1
11 1950     1  11   11.9       1
12 1950     1  12   17.5       1
13 1950     1  13    8.1       1
14 1950     1  14    0.4       1
15 1950     1  15    0.0       0
16 1950     1  16   19.5       1
17 1950     1  17   10.7       1
18 1950     1  18    0.5       1
19 1950     1  19   12.7       1
20 1950     1  20    6.3       1
21 1950     2   1    0.0       0
22 1950     2   2   35.5       1
23 1950     2   3   17.8       1
24 1950     2   4   24.5       1
25 1950     2   5   12.3       1
26 1950     2   6   11.5       1
27 1950     2   7    5.7       1
28 1950     2   8   13.2       1
29 1950     2   9   11.3       1
30 1950     2  10   14.7       1
31 1950     2  11   11.9       1
32 1950     2  12   17.5       1
33 1950     2  13    8.1       1
34 1950     2  14    0.4       1
35 1950     2  15    0.0       0
36 1950     2  16   19.5       1
37 1950     2  17   10.7       1
38 1950     2  18    0.0       0
39 1950     2  19    0.0       0
40 1950     2  20    0.0       0> tapply(rain$hasRain,list(rain$Year,rain$Month),sum)      1  2
1950 18 15> 


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View this message in context:
http://r.789695.n4.nabble.com/Replace-the-value-with-1-and-0-tp4703838p4703840.html
Sent from the R help mailing list archive at Nabble.com.

Clint Bowman

2015-Feb-25 23:40 UTC

head link

[R] Replace the value with 1 and 0

or:

with(yourData,table(year,month,yourData[["rain"]]>0))

Clint Bowman			INTERNET:	clint at ecy.wa.gov
Air Quality Modeler		INTERNET:	clint at math.utah.edu
Department of Ecology		VOICE:		(360) 407-6815
PO Box 47600			FAX:		(360) 407-7534
Olympia, WA 98504-7600

         USPS:           PO Box 47600, Olympia, WA 98504-7600
         Parcels:        300 Desmond Drive, Lacey, WA 98503-1274

On Thu, 26 Feb 2015, Peter Alspach wrote:
> Tena koe
>
> Something like:
>
> set.seed(153)
> # Create some (unrealistic) rainfall data
> yourData <- data.frame(year=rep(1950:1954, each=10), month=rep(rep(1:2,
each=5), 5), rain=sample(0:1, 50, replace=TRUE)*round(rnorm(50, 20, 2), 1))
> tapply(yourData$rain>0, yourData[,c('year','month')],
sum)
>
> will give your final table, but it will need some tidying up.
>
> HTH ....
>
> Peter Alspach
>
>
> -----Original Message-----
> From: R-help [mailto:r-help-bounces at r-project.org] On Behalf Of smart
hendsome
> Sent: Thursday, 26 February 2015 11:54 a.m.
> To: r-help at r-project.org
> Subject: [R] Replace the value with 1 and 0
>
> Hi everyone,
> I have this kind of rainfall dataset:
> ?? Year Month Day Amount
> 1? 1950???? 1?? 1??? 0.0
> 2? 1950???? 1?? 2?? 35.5
> 3? 1950???? 1?? 3?? 17.8
> 4? 1950???? 1?? 4?? 24.5
> 5? 1950???? 1?? 5?? 12.3
> 6? 1950???? 1?? 6?? 11.5
> 7? 1950???? 1?? 7??? 5.7
> 8? 1950???? 1?? 8?? 13.2
> 9? 1950???? 1?? 9?? 11.3
> 10 1950???? 1? 10?? 14.7
> 11 1950???? 1? 11?? 11.9
> 12 1950???? 1? 12?? 17.5
> 13 1950???? 1? 13??? 8.1
> 14 1950???? 1? 14??? 0.4
> 15 1950???? 1? 15??? 0.0
> 16 1950???? 1? 16?? 19.5
> 17 1950???? 1? 17?? 10.7
> 18 1950???? 1? 18??? 0.5
> 19 1950???? 1? 19?? 12.7
> 20 1950???? 1? 20??? 6.3
>
> I want to set as rain for Amount> 0 and not rain for Amount = 0.? I want
to replace the Amount>0 with 1 and Amount equal to zero with 0.? Then I want
to count how many rain in that particular month in that year. Anyone can help
me?
> This is what I want:
> Year Month Day Amount
> 1? 1950???? 1?? 1?? ? 0
> 2? 1950???? 1?? 2? ?? 1
> 3? 1950???? 1?? 3???? 1
> 4? 1950???? 1?? 4???? 1
> 5? 1950???? 1?? 5? ?? 1
> 6? 1950???? 1?? 6???? 1
> 7? 1950???? 1?? 7?? ? 1
> 8? 1950???? 1?? 8???? 1
> 9? 1950???? 1?? 9???? 1
> 10 1950???? 1? 10??? 1
> 11 1950???? 1? 11?? ? 1
> 12 1950???? 1? 12??? 1
> 13 1950???? 1? 13??? 1
> 14 1950???? 1? 14??? 1
> 15 1950???? 1? 15??? 1
> 16 1950???? 1? 16??? 1
> 17 1950???? 1? 17??? 1
> 18 1950???? 1? 18??? 1
> 19 1950???? 1? 19??? 1
> 20 1950???? 1? 20??? 1
> Then become like this:
>
> |
> | Jan | Feb | Mar | Apr | May | Jun | Jul | Aug | Sep | Oct | Nov | Dec |
> | 1950 | 17 | 6 | 23 | 20 | 19 | 9 | 17 | 23 | 18 | 20 | 20 | 17 |
> | 1951 | 23 | 19 | 20 | 20 | 19 | 11 | 16 | 20 | 22 | 25 | 25 | 16 |
> | 1952 | 15 | 21 | 30 | 24 | 23 | 20 | 16 | 19 | 20 | 19 | 21 | 15 |
>
>
> Thanks.
>
>
> 	[[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
> The contents of this e-mail are confidential and may be subject to legal
privilege.
> If you are not the intended recipient you must not use, disseminate,
distribute or
> reproduce all or any part of this e-mail or attachments.  If you have
received this
> e-mail in error, please notify the sender and delete all material
pertaining to this
> e-mail.  Any opinion or views expressed in this e-mail are those of the
individual
> sender and may not represent those of The New Zealand Institute for Plant
and
> Food Research Limited.
> ______________________________________________
> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>

Göran Broström

2015-Feb-26 08:59 UTC

head link

[R] Replace the value with 1 and 0

On 2015-02-26 00:33, JS Huang wrote:> Hi,
>
>    Here is an implementation.  More data are added.  An extra column
hasRain
> is added instead of replacing column Amount.
>
>> rain
>     Year Month Day Amount
> 1  1950     1   1    0.0
> 2  1950     1   2   35.5
> 3  1950     1   3   17.8
> 4  1950     1   4   24.5
> 5  1950     1   5   12.3
> 6  1950     1   6   11.5
> 7  1950     1   7    5.7
> 8  1950     1   8   13.2
> 9  1950     1   9   11.3
> 10 1950     1  10   14.7
> 11 1950     1  11   11.9
> 12 1950     1  12   17.5
> 13 1950     1  13    8.1
> 14 1950     1  14    0.4
> 15 1950     1  15    0.0
> 16 1950     1  16   19.5
> 17 1950     1  17   10.7
> 18 1950     1  18    0.5
> 19 1950     1  19   12.7
> 20 1950     1  20    6.3
> 21 1950     2   1    0.0
> 22 1950     2   2   35.5
> 23 1950     2   3   17.8
> 24 1950     2   4   24.5
> 25 1950     2   5   12.3
> 26 1950     2   6   11.5
> 27 1950     2   7    5.7
> 28 1950     2   8   13.2
> 29 1950     2   9   11.3
> 30 1950     2  10   14.7
> 31 1950     2  11   11.9
> 32 1950     2  12   17.5
> 33 1950     2  13    8.1
> 34 1950     2  14    0.4
> 35 1950     2  15    0.0
> 36 1950     2  16   19.5
> 37 1950     2  17   10.7
> 38 1950     2  18    0.0
> 39 1950     2  19    0.0
> 40 1950     2  20    0.0
>> rain$hasRain <- ifelse(rain$Amount>0,1,0)
No! Better is

rain$hasRain <- as.numeric(rain$Amount > 0)

See previous discussions about the use of 'ifelse'.

G?ran
>> rain
>     Year Month Day Amount hasRain
> 1  1950     1   1    0.0       0
> 2  1950     1   2   35.5       1
> 3  1950     1   3   17.8       1
> 4  1950     1   4   24.5       1
> 5  1950     1   5   12.3       1
> 6  1950     1   6   11.5       1
> 7  1950     1   7    5.7       1
> 8  1950     1   8   13.2       1
> 9  1950     1   9   11.3       1
> 10 1950     1  10   14.7       1
> 11 1950     1  11   11.9       1
> 12 1950     1  12   17.5       1
> 13 1950     1  13    8.1       1
> 14 1950     1  14    0.4       1
> 15 1950     1  15    0.0       0
> 16 1950     1  16   19.5       1
> 17 1950     1  17   10.7       1
> 18 1950     1  18    0.5       1
> 19 1950     1  19   12.7       1
> 20 1950     1  20    6.3       1
> 21 1950     2   1    0.0       0
> 22 1950     2   2   35.5       1
> 23 1950     2   3   17.8       1
> 24 1950     2   4   24.5       1
> 25 1950     2   5   12.3       1
> 26 1950     2   6   11.5       1
> 27 1950     2   7    5.7       1
> 28 1950     2   8   13.2       1
> 29 1950     2   9   11.3       1
> 30 1950     2  10   14.7       1
> 31 1950     2  11   11.9       1
> 32 1950     2  12   17.5       1
> 33 1950     2  13    8.1       1
> 34 1950     2  14    0.4       1
> 35 1950     2  15    0.0       0
> 36 1950     2  16   19.5       1
> 37 1950     2  17   10.7       1
> 38 1950     2  18    0.0       0
> 39 1950     2  19    0.0       0
> 40 1950     2  20    0.0       0
>> tapply(rain$hasRain,list(rain$Year,rain$Month),sum)
>        1  2
> 1950 18 15
>>
>
>
>
> --
> View this message in context:
http://r.789695.n4.nabble.com/Replace-the-value-with-1-and-0-tp4703838p4703840.html
> Sent from the R help mailing list archive at Nabble.com.
>
> ______________________________________________
> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>

R help - Feb 2015 - Replace the value with 1 and 0

[R] Replace the value with 1 and 0

[R] Replace the value with 1 and 0

[R] Replace the value with 1 and 0

[R] Replace the value with 1 and 0

[R] Replace the value with 1 and 0