Change cell value in one raster based on another raster
up vote
2
down vote
favorite
I have two raster maps from two points in time (t1 and t2) with two land-cover categories in each (LC1, LC2). I want impose a rule that a LC2-cell in t1 cannot change to LC1-cell in t2, i.e., only LC1 can change to LC2 through time but not the other way around. I am having a hard time coming up with a rule for that in R. What I had in mind was something like this:
#create test rasters
r <- raster(nrows=25, ncols=25, vals=round(rnorm(625, 3), 0)) #land-use/cover raster
r[ r > 2 ] <- 2
r[ r < 1 ] <- 1
r2 <- r
plot(r2) #r2 is t2
r <- raster(nrows=25, ncols=25, vals=round(rnorm(625, 3), 0)) #land-use/cover raster
r[ r > 2 ] <- 2
r[ r < 1 ] <- 1
plot(r) #r is t1
r_fix <- overlay(r, r2, fun = function(x, y)
if (x[ x==2 ] & y[ y==1 ]) #1 is LC1, 2 is LC2
x[ x==2 ] <- 1
return(x)
)
But it returns an error (because of they way I am using the if statement with rasters?):
Error in (function (x, fun, filename = "", recycle = TRUE, forcefun = FALSE, :
cannot use this formula, probably because it is not vectorized
I wonder if there is a simple way to implement something similar to that that works with rasters? Thank you in advance.
r if-statement raster
add a comment |
up vote
2
down vote
favorite
I have two raster maps from two points in time (t1 and t2) with two land-cover categories in each (LC1, LC2). I want impose a rule that a LC2-cell in t1 cannot change to LC1-cell in t2, i.e., only LC1 can change to LC2 through time but not the other way around. I am having a hard time coming up with a rule for that in R. What I had in mind was something like this:
#create test rasters
r <- raster(nrows=25, ncols=25, vals=round(rnorm(625, 3), 0)) #land-use/cover raster
r[ r > 2 ] <- 2
r[ r < 1 ] <- 1
r2 <- r
plot(r2) #r2 is t2
r <- raster(nrows=25, ncols=25, vals=round(rnorm(625, 3), 0)) #land-use/cover raster
r[ r > 2 ] <- 2
r[ r < 1 ] <- 1
plot(r) #r is t1
r_fix <- overlay(r, r2, fun = function(x, y)
if (x[ x==2 ] & y[ y==1 ]) #1 is LC1, 2 is LC2
x[ x==2 ] <- 1
return(x)
)
But it returns an error (because of they way I am using the if statement with rasters?):
Error in (function (x, fun, filename = "", recycle = TRUE, forcefun = FALSE, :
cannot use this formula, probably because it is not vectorized
I wonder if there is a simple way to implement something similar to that that works with rasters? Thank you in advance.
r if-statement raster
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I have two raster maps from two points in time (t1 and t2) with two land-cover categories in each (LC1, LC2). I want impose a rule that a LC2-cell in t1 cannot change to LC1-cell in t2, i.e., only LC1 can change to LC2 through time but not the other way around. I am having a hard time coming up with a rule for that in R. What I had in mind was something like this:
#create test rasters
r <- raster(nrows=25, ncols=25, vals=round(rnorm(625, 3), 0)) #land-use/cover raster
r[ r > 2 ] <- 2
r[ r < 1 ] <- 1
r2 <- r
plot(r2) #r2 is t2
r <- raster(nrows=25, ncols=25, vals=round(rnorm(625, 3), 0)) #land-use/cover raster
r[ r > 2 ] <- 2
r[ r < 1 ] <- 1
plot(r) #r is t1
r_fix <- overlay(r, r2, fun = function(x, y)
if (x[ x==2 ] & y[ y==1 ]) #1 is LC1, 2 is LC2
x[ x==2 ] <- 1
return(x)
)
But it returns an error (because of they way I am using the if statement with rasters?):
Error in (function (x, fun, filename = "", recycle = TRUE, forcefun = FALSE, :
cannot use this formula, probably because it is not vectorized
I wonder if there is a simple way to implement something similar to that that works with rasters? Thank you in advance.
r if-statement raster
I have two raster maps from two points in time (t1 and t2) with two land-cover categories in each (LC1, LC2). I want impose a rule that a LC2-cell in t1 cannot change to LC1-cell in t2, i.e., only LC1 can change to LC2 through time but not the other way around. I am having a hard time coming up with a rule for that in R. What I had in mind was something like this:
#create test rasters
r <- raster(nrows=25, ncols=25, vals=round(rnorm(625, 3), 0)) #land-use/cover raster
r[ r > 2 ] <- 2
r[ r < 1 ] <- 1
r2 <- r
plot(r2) #r2 is t2
r <- raster(nrows=25, ncols=25, vals=round(rnorm(625, 3), 0)) #land-use/cover raster
r[ r > 2 ] <- 2
r[ r < 1 ] <- 1
plot(r) #r is t1
r_fix <- overlay(r, r2, fun = function(x, y)
if (x[ x==2 ] & y[ y==1 ]) #1 is LC1, 2 is LC2
x[ x==2 ] <- 1
return(x)
)
But it returns an error (because of they way I am using the if statement with rasters?):
Error in (function (x, fun, filename = "", recycle = TRUE, forcefun = FALSE, :
cannot use this formula, probably because it is not vectorized
I wonder if there is a simple way to implement something similar to that that works with rasters? Thank you in advance.
r if-statement raster
r if-statement raster
edited Nov 10 at 20:50
Julius Vainora
30.1k75878
30.1k75878
asked Nov 10 at 20:34
Thales West
3362417
3362417
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
You were really close,
overlay(r, r2, fun = function(x, y) x[x == 2 & y == 1] <- 1; x)
seems to do the job.
In terms of your solution,
x[x == 2] <- 1
doesn't cause any errors, although it's not exactly what you want to use in your case either. However,
if (x[x == 2] & y[y == 1])
is a problem because x[x == 2] & y[y == 1]
returns a matrix, while if
wants just a single logical input. Subsetting, on the other hand, can handle logical matrices, which is exactly what is happening in x[x == 2 & y == 1]
.
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',
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
);
);
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%2f53243163%2fchange-cell-value-in-one-raster-based-on-another-raster%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
up vote
2
down vote
accepted
You were really close,
overlay(r, r2, fun = function(x, y) x[x == 2 & y == 1] <- 1; x)
seems to do the job.
In terms of your solution,
x[x == 2] <- 1
doesn't cause any errors, although it's not exactly what you want to use in your case either. However,
if (x[x == 2] & y[y == 1])
is a problem because x[x == 2] & y[y == 1]
returns a matrix, while if
wants just a single logical input. Subsetting, on the other hand, can handle logical matrices, which is exactly what is happening in x[x == 2 & y == 1]
.
add a comment |
up vote
2
down vote
accepted
You were really close,
overlay(r, r2, fun = function(x, y) x[x == 2 & y == 1] <- 1; x)
seems to do the job.
In terms of your solution,
x[x == 2] <- 1
doesn't cause any errors, although it's not exactly what you want to use in your case either. However,
if (x[x == 2] & y[y == 1])
is a problem because x[x == 2] & y[y == 1]
returns a matrix, while if
wants just a single logical input. Subsetting, on the other hand, can handle logical matrices, which is exactly what is happening in x[x == 2 & y == 1]
.
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
You were really close,
overlay(r, r2, fun = function(x, y) x[x == 2 & y == 1] <- 1; x)
seems to do the job.
In terms of your solution,
x[x == 2] <- 1
doesn't cause any errors, although it's not exactly what you want to use in your case either. However,
if (x[x == 2] & y[y == 1])
is a problem because x[x == 2] & y[y == 1]
returns a matrix, while if
wants just a single logical input. Subsetting, on the other hand, can handle logical matrices, which is exactly what is happening in x[x == 2 & y == 1]
.
You were really close,
overlay(r, r2, fun = function(x, y) x[x == 2 & y == 1] <- 1; x)
seems to do the job.
In terms of your solution,
x[x == 2] <- 1
doesn't cause any errors, although it's not exactly what you want to use in your case either. However,
if (x[x == 2] & y[y == 1])
is a problem because x[x == 2] & y[y == 1]
returns a matrix, while if
wants just a single logical input. Subsetting, on the other hand, can handle logical matrices, which is exactly what is happening in x[x == 2 & y == 1]
.
answered Nov 10 at 20:43
Julius Vainora
30.1k75878
30.1k75878
add a comment |
add a comment |
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f53243163%2fchange-cell-value-in-one-raster-based-on-another-raster%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