Mixing
subset using percentile for gridded data
0
I have gridded data that has 24249 obs and 963 var for daily maximum temperatures (K). I am looking for a way in r to select all days with maximum temperatures higher than the 90th percentile.
> dim(DailyT)
[1] 24249 963
> DailyT[1:4,1:7]
x y 1988-05-01 1988-05-02 1988-05-03 1988-05-04 1988-05-05
1 34.000 33 291.7603 291.8044 291.6158 292.9659 293.7032
2 34.125 33 291.7240 291.7951 291.5439 292.9451 293.7017
3 34.250 33 291.6884 291.7866 291.4721 292.9250 293.7001
4 34.375 33 291.6521 291.7781 291.4010 292.9049 293.6986
I did this but did not work
df<- DailyT[DailyT[,3:963] <= quantile(DailyT[,3:963],.9, na.rm = T, type = 6) ]
r
asked Nov 22 '18 at 8:55
AliAli
337
add a comment |
0
I have gridded data that has 24249 obs and 963 var for daily maximum temperatures (K). I am looking for a way in r to select all days with maximum temperatures higher than the 90th percentile.
> dim(DailyT)
[1] 24249 963
> DailyT[1:4,1:7]
x y 1988-05-01 1988-05-02 1988-05-03 1988-05-04 1988-05-05
1 34.000 33 291.7603 291.8044 291.6158 292.9659 293.7032
2 34.125 33 291.7240 291.7951 291.5439 292.9451 293.7017
3 34.250 33 291.6884 291.7866 291.4721 292.9250 293.7001
4 34.375 33 291.6521 291.7781 291.4010 292.9049 293.6986
I did this but did not work
df<- DailyT[DailyT[,3:963] <= quantile(DailyT[,3:963],.9, na.rm = T, type = 6) ]
r
asked Nov 22 '18 at 8:55
AliAli
337
Maybe you find this helpful.
– A. Suliman
Nov 22 '18 at 9:07
add a comment |
0
0
0
I have gridded data that has 24249 obs and 963 var for daily maximum temperatures (K). I am looking for a way in r to select all days with maximum temperatures higher than the 90th percentile.
> dim(DailyT)
[1] 24249 963
> DailyT[1:4,1:7]
x y 1988-05-01 1988-05-02 1988-05-03 1988-05-04 1988-05-05
1 34.000 33 291.7603 291.8044 291.6158 292.9659 293.7032
2 34.125 33 291.7240 291.7951 291.5439 292.9451 293.7017
3 34.250 33 291.6884 291.7866 291.4721 292.9250 293.7001
4 34.375 33 291.6521 291.7781 291.4010 292.9049 293.6986
I did this but did not work
df<- DailyT[DailyT[,3:963] <= quantile(DailyT[,3:963],.9, na.rm = T, type = 6) ]
r
asked Nov 22 '18 at 8:55
AliAli
337
I have gridded data that has 24249 obs and 963 var for daily maximum temperatures (K). I am looking for a way in r to select all days with maximum temperatures higher than the 90th percentile.
> dim(DailyT)
[1] 24249 963
> DailyT[1:4,1:7]
x y 1988-05-01 1988-05-02 1988-05-03 1988-05-04 1988-05-05
1 34.000 33 291.7603 291.8044 291.6158 292.9659 293.7032
2 34.125 33 291.7240 291.7951 291.5439 292.9451 293.7017
3 34.250 33 291.6884 291.7866 291.4721 292.9250 293.7001
4 34.375 33 291.6521 291.7781 291.4010 292.9049 293.6986
I did this but did not work
df<- DailyT[DailyT[,3:963] <= quantile(DailyT[,3:963],.9, na.rm = T, type = 6) ]
r
r
asked Nov 22 '18 at 8:55
AliAli
337
asked Nov 22 '18 at 8:55
AliAli
337
asked Nov 22 '18 at 8:55
AliAli
337
asked Nov 22 '18 at 8:55
AliAli
337
asked Nov 22 '18 at 8:55
AliAli
337
337
Maybe you find this helpful.
– A. Suliman
Nov 22 '18 at 9:07
add a comment |
Maybe you find this helpful.
– A. Suliman
Nov 22 '18 at 9:07
Maybe you find this helpful.
– A. Suliman
Nov 22 '18 at 9:07
Maybe you find this helpful.
– A. Suliman
Nov 22 '18 at 9:07
add a comment |
1 Answer
1
active
oldest
votes
0
First, you need an id column to identify the rows later. Then, calculate the 90% quantile of all temperature values. At the end subset data witch any row cells exceeding q.
DailyT <- cbind(id=rownames(DailyT), DailyT) # to identify rows later
q <- quantile(as.matrix(DailyT[, -(1:3)]), .9, na.rm = T, type = 6) # 293.7003
DailyT.q <- DailyT[which(sapply(1:nrow(DailyT), function(x) any(DailyT[x, -(1:2)] >= q))), ]
Yields
> DailyT.q
id x y X1988.05.01 X1988.05.02 X1988.05.03 X1988.05.04 X1988.05.05
1 1 34.000 33 291.7603 291.8044 291.6158 292.9659 293.7032
2 2 34.125 33 291.7240 291.7951 291.5439 292.9451 293.7017
Edit:
To get the quantile rowwise use apply()
q90 <- apply(DailyT[, 4:8], MARGIN=1, quantile, .9,na.rm = T, type = 6)
> data.frame(DailyT, q90=q90)
id x y X1988.05.01 X1988.05.02 X1988.05.03 X1988.05.04 X1988.05.05 q90
1 1 34.000 33 291.7603 291.8044 291.6158 292.9659 293.7032 293.7032
2 2 34.125 33 291.7240 291.7951 291.5439 292.9451 293.7017 293.7017
3 3 34.250 33 291.6884 291.7866 291.4721 292.9250 293.7001 293.7001
4 4 34.375 33 291.6521 291.7781 291.4010 292.9049 293.6986 293.6986
Data
> dput(DailyT)
structure(list(x = c(34, 34.125, 34.25, 34.375), y = c(33L, 33L,
33L, 33L), X1988.05.01 = c(291.7603, 291.724, 291.6884, 291.6521
), X1988.05.02 = c(291.8044, 291.7951, 291.7866, 291.7781), X1988.05.03 = c(291.6158,
291.5439, 291.4721, 291.401), X1988.05.04 = c(292.9659, 292.9451,
292.925, 292.9049), X1988.05.05 = c(293.7032, 293.7017, 293.7001,
293.6986)), class = "data.frame", row.names = c(NA, -4L))
answered Nov 22 '18 at 9:21
jay.sfjay.sf
5,51631739
Thanks, I need to calculate the 90% quantile of each row not of all data.
– Ali
Nov 24 '18 at 11:09
Aha, please see my edit.
– jay.sf
Nov 24 '18 at 11:24
Worked.... Many thanks
– Ali
Nov 24 '18 at 13:32
Very good! - Please mark the question as answered when you're satisfied with the given answer and win +2 reputation. This stops people spending time on answering a question that has already been answered.
– jay.sf
Nov 24 '18 at 13:58
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53427098%2fsubset-using-percentile-for-gridded-data%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
0
First, you need an id column to identify the rows later. Then, calculate the 90% quantile of all temperature values. At the end subset data witch any row cells exceeding q.
DailyT <- cbind(id=rownames(DailyT), DailyT) # to identify rows later
q <- quantile(as.matrix(DailyT[, -(1:3)]), .9, na.rm = T, type = 6) # 293.7003
DailyT.q <- DailyT[which(sapply(1:nrow(DailyT), function(x) any(DailyT[x, -(1:2)] >= q))), ]
Yields
> DailyT.q
id x y X1988.05.01 X1988.05.02 X1988.05.03 X1988.05.04 X1988.05.05
1 1 34.000 33 291.7603 291.8044 291.6158 292.9659 293.7032
2 2 34.125 33 291.7240 291.7951 291.5439 292.9451 293.7017
Edit:
To get the quantile rowwise use apply()
q90 <- apply(DailyT[, 4:8], MARGIN=1, quantile, .9,na.rm = T, type = 6)
> data.frame(DailyT, q90=q90)
id x y X1988.05.01 X1988.05.02 X1988.05.03 X1988.05.04 X1988.05.05 q90
1 1 34.000 33 291.7603 291.8044 291.6158 292.9659 293.7032 293.7032
2 2 34.125 33 291.7240 291.7951 291.5439 292.9451 293.7017 293.7017
3 3 34.250 33 291.6884 291.7866 291.4721 292.9250 293.7001 293.7001
4 4 34.375 33 291.6521 291.7781 291.4010 292.9049 293.6986 293.6986
Data
> dput(DailyT)
structure(list(x = c(34, 34.125, 34.25, 34.375), y = c(33L, 33L,
33L, 33L), X1988.05.01 = c(291.7603, 291.724, 291.6884, 291.6521
), X1988.05.02 = c(291.8044, 291.7951, 291.7866, 291.7781), X1988.05.03 = c(291.6158,
291.5439, 291.4721, 291.401), X1988.05.04 = c(292.9659, 292.9451,
292.925, 292.9049), X1988.05.05 = c(293.7032, 293.7017, 293.7001,
293.6986)), class = "data.frame", row.names = c(NA, -4L))
answered Nov 22 '18 at 9:21
jay.sfjay.sf
5,51631739
Thanks, I need to calculate the 90% quantile of each row not of all data.
– Ali
Nov 24 '18 at 11:09
Aha, please see my edit.
– jay.sf
Nov 24 '18 at 11:24
Worked.... Many thanks
– Ali
Nov 24 '18 at 13:32
Very good! - Please mark the question as answered when you're satisfied with the given answer and win +2 reputation. This stops people spending time on answering a question that has already been answered.
– jay.sf
Nov 24 '18 at 13:58
add a comment |
0
First, you need an id column to identify the rows later. Then, calculate the 90% quantile of all temperature values. At the end subset data witch any row cells exceeding q.
DailyT <- cbind(id=rownames(DailyT), DailyT) # to identify rows later
q <- quantile(as.matrix(DailyT[, -(1:3)]), .9, na.rm = T, type = 6) # 293.7003
DailyT.q <- DailyT[which(sapply(1:nrow(DailyT), function(x) any(DailyT[x, -(1:2)] >= q))), ]
Yields
> DailyT.q
id x y X1988.05.01 X1988.05.02 X1988.05.03 X1988.05.04 X1988.05.05
1 1 34.000 33 291.7603 291.8044 291.6158 292.9659 293.7032
2 2 34.125 33 291.7240 291.7951 291.5439 292.9451 293.7017
Edit:
To get the quantile rowwise use apply()
q90 <- apply(DailyT[, 4:8], MARGIN=1, quantile, .9,na.rm = T, type = 6)
> data.frame(DailyT, q90=q90)
id x y X1988.05.01 X1988.05.02 X1988.05.03 X1988.05.04 X1988.05.05 q90
1 1 34.000 33 291.7603 291.8044 291.6158 292.9659 293.7032 293.7032
2 2 34.125 33 291.7240 291.7951 291.5439 292.9451 293.7017 293.7017
3 3 34.250 33 291.6884 291.7866 291.4721 292.9250 293.7001 293.7001
4 4 34.375 33 291.6521 291.7781 291.4010 292.9049 293.6986 293.6986
Data
> dput(DailyT)
structure(list(x = c(34, 34.125, 34.25, 34.375), y = c(33L, 33L,
33L, 33L), X1988.05.01 = c(291.7603, 291.724, 291.6884, 291.6521
), X1988.05.02 = c(291.8044, 291.7951, 291.7866, 291.7781), X1988.05.03 = c(291.6158,
291.5439, 291.4721, 291.401), X1988.05.04 = c(292.9659, 292.9451,
292.925, 292.9049), X1988.05.05 = c(293.7032, 293.7017, 293.7001,
293.6986)), class = "data.frame", row.names = c(NA, -4L))
answered Nov 22 '18 at 9:21
jay.sfjay.sf
5,51631739
Thanks, I need to calculate the 90% quantile of each row not of all data.
– Ali
Nov 24 '18 at 11:09
Aha, please see my edit.
– jay.sf
Nov 24 '18 at 11:24
Worked.... Many thanks
– Ali
Nov 24 '18 at 13:32
Very good! - Please mark the question as answered when you're satisfied with the given answer and win +2 reputation. This stops people spending time on answering a question that has already been answered.
– jay.sf
Nov 24 '18 at 13:58
add a comment |
0
0
0
First, you need an id column to identify the rows later. Then, calculate the 90% quantile of all temperature values. At the end subset data witch any row cells exceeding q.
DailyT <- cbind(id=rownames(DailyT), DailyT) # to identify rows later
q <- quantile(as.matrix(DailyT[, -(1:3)]), .9, na.rm = T, type = 6) # 293.7003
DailyT.q <- DailyT[which(sapply(1:nrow(DailyT), function(x) any(DailyT[x, -(1:2)] >= q))), ]
Yields
> DailyT.q
id x y X1988.05.01 X1988.05.02 X1988.05.03 X1988.05.04 X1988.05.05
1 1 34.000 33 291.7603 291.8044 291.6158 292.9659 293.7032
2 2 34.125 33 291.7240 291.7951 291.5439 292.9451 293.7017
Edit:
To get the quantile rowwise use apply()
q90 <- apply(DailyT[, 4:8], MARGIN=1, quantile, .9,na.rm = T, type = 6)
> data.frame(DailyT, q90=q90)
id x y X1988.05.01 X1988.05.02 X1988.05.03 X1988.05.04 X1988.05.05 q90
1 1 34.000 33 291.7603 291.8044 291.6158 292.9659 293.7032 293.7032
2 2 34.125 33 291.7240 291.7951 291.5439 292.9451 293.7017 293.7017
3 3 34.250 33 291.6884 291.7866 291.4721 292.9250 293.7001 293.7001
4 4 34.375 33 291.6521 291.7781 291.4010 292.9049 293.6986 293.6986
Data
> dput(DailyT)
structure(list(x = c(34, 34.125, 34.25, 34.375), y = c(33L, 33L,
33L, 33L), X1988.05.01 = c(291.7603, 291.724, 291.6884, 291.6521
), X1988.05.02 = c(291.8044, 291.7951, 291.7866, 291.7781), X1988.05.03 = c(291.6158,
291.5439, 291.4721, 291.401), X1988.05.04 = c(292.9659, 292.9451,
292.925, 292.9049), X1988.05.05 = c(293.7032, 293.7017, 293.7001,
293.6986)), class = "data.frame", row.names = c(NA, -4L))
answered Nov 22 '18 at 9:21
jay.sfjay.sf
5,51631739
First, you need an id column to identify the rows later. Then, calculate the 90% quantile of all temperature values. At the end subset data witch any row cells exceeding q.
DailyT <- cbind(id=rownames(DailyT), DailyT) # to identify rows later
q <- quantile(as.matrix(DailyT[, -(1:3)]), .9, na.rm = T, type = 6) # 293.7003
DailyT.q <- DailyT[which(sapply(1:nrow(DailyT), function(x) any(DailyT[x, -(1:2)] >= q))), ]
Yields
> DailyT.q
id x y X1988.05.01 X1988.05.02 X1988.05.03 X1988.05.04 X1988.05.05
1 1 34.000 33 291.7603 291.8044 291.6158 292.9659 293.7032
2 2 34.125 33 291.7240 291.7951 291.5439 292.9451 293.7017
Edit:
To get the quantile rowwise use apply()
q90 <- apply(DailyT[, 4:8], MARGIN=1, quantile, .9,na.rm = T, type = 6)
> data.frame(DailyT, q90=q90)
id x y X1988.05.01 X1988.05.02 X1988.05.03 X1988.05.04 X1988.05.05 q90
1 1 34.000 33 291.7603 291.8044 291.6158 292.9659 293.7032 293.7032
2 2 34.125 33 291.7240 291.7951 291.5439 292.9451 293.7017 293.7017
3 3 34.250 33 291.6884 291.7866 291.4721 292.9250 293.7001 293.7001
4 4 34.375 33 291.6521 291.7781 291.4010 292.9049 293.6986 293.6986
Data
> dput(DailyT)
structure(list(x = c(34, 34.125, 34.25, 34.375), y = c(33L, 33L,
33L, 33L), X1988.05.01 = c(291.7603, 291.724, 291.6884, 291.6521
), X1988.05.02 = c(291.8044, 291.7951, 291.7866, 291.7781), X1988.05.03 = c(291.6158,
291.5439, 291.4721, 291.401), X1988.05.04 = c(292.9659, 292.9451,
292.925, 292.9049), X1988.05.05 = c(293.7032, 293.7017, 293.7001,
293.6986)), class = "data.frame", row.names = c(NA, -4L))
answered Nov 22 '18 at 9:21
jay.sfjay.sf
5,51631739
edited Nov 24 '18 at 11:24
answered Nov 22 '18 at 9:21
jay.sfjay.sf
5,51631739
answered Nov 22 '18 at 9:21
jay.sfjay.sf
5,51631739
answered Nov 22 '18 at 9:21
jay.sfjay.sf
5,51631739
5,51631739
Thanks, I need to calculate the 90% quantile of each row not of all data.
– Ali
Nov 24 '18 at 11:09
Aha, please see my edit.
– jay.sf
Nov 24 '18 at 11:24
Worked.... Many thanks
– Ali
Nov 24 '18 at 13:32
Very good! - Please mark the question as answered when you're satisfied with the given answer and win +2 reputation. This stops people spending time on answering a question that has already been answered.
– jay.sf
Nov 24 '18 at 13:58
add a comment |
Thanks, I need to calculate the 90% quantile of each row not of all data.
– Ali
Nov 24 '18 at 11:09
Aha, please see my edit.
– jay.sf
Nov 24 '18 at 11:24
Worked.... Many thanks
– Ali
Nov 24 '18 at 13:32
Very good! - Please mark the question as answered when you're satisfied with the given answer and win +2 reputation. This stops people spending time on answering a question that has already been answered.
– jay.sf
Nov 24 '18 at 13:58
Thanks, I need to calculate the 90% quantile of each row not of all data.
– Ali
Nov 24 '18 at 11:09
Thanks, I need to calculate the 90% quantile of each row not of all data.
– Ali
Nov 24 '18 at 11:09
Aha, please see my edit.
– jay.sf
Nov 24 '18 at 11:24
Aha, please see my edit.
– jay.sf
Nov 24 '18 at 11:24
Worked.... Many thanks
– Ali
Nov 24 '18 at 13:32
Worked.... Many thanks
– Ali
Nov 24 '18 at 13:32
Very good! - Please mark the question as answered when you're satisfied with the given answer and win +2 reputation. This stops people spending time on answering a question that has already been answered.
– jay.sf
Nov 24 '18 at 13:58
Very good! - Please mark the question as answered when you're satisfied with the given answer and win +2 reputation. This stops people spending time on answering a question that has already been answered.
– jay.sf
Nov 24 '18 at 13:58
add a comment |
draft saved
draft discarded
Thanks for contributing an answer to Stack Overflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53427098%2fsubset-using-percentile-for-gridded-data%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Maybe you find this helpful.
– A. Suliman
Nov 22 '18 at 9:07