在相同尺寸 R 的两个光栅中有条件地替换 NA
Conditional replacement of NAs in two rasters of same dimensions R
这个问题与 有关，但我想在栅格上做同样的事情，而不是将 df1 和 df2 转换为数据帧。
给定两个尺寸相等的栅格 (df1,df2)：
df1: dimensions : 510, 1068, 544680, 1 (nrow, ncol, ncell, nlayers)
df2: dimensions : 510, 1068, 544680, 1 (nrow, ncol, ncell, nlayers)
如何找到 df1 中所有行都是 NAs 的所有列，转到 df2 并将匹配列中的所有值设置为 NAs？
鉴于所报告的示例数据，如果我很好地理解了这个问题，这应该可以解决问题：
library(raster)
r <- raster(nrow=5, ncol=5)
df1 <- stack( sapply(1:5, function(i) setValues(r, rnorm(ncell(r), i, 3))))
df2<- stack( sapply(1:5, function(i) setValues(r, rnorm(ncell(r), i, 3))))
df1[1,]<-NA
head(df1)
sumrows <- raster::rowSums(df1, na.rm = T) # if all NA in row, the sum is 0
where_NA <- which(sumrows[1,] == 0) # find 0 sum rows
df2[1,where_NA] <- NA
head(df2)
但是，如果 "rows" 全部为零，这将失败。如果这可能发生（如下面修改后的示例数据），您可以这样做：
library(raster)
r <- raster(nrow=5, ncol=5)
df1 <- stack( sapply(1:5, function(i) setValues(r, rnorm(ncell(r), i, 3))))
df2<- stack( sapply(1:5, function(i) setValues(r, rnorm(ncell(r), i, 3))))
df1[1,]<-NA
df1[1,5] <- 0
head(df1)
sumrows <- raster::rowSums(df1, na.rm = T) # if all NA in row, the sum is 0
where_sum0 <- which(sumrows [1,] == 0) # rows adding up to 0
# Find rows adding up to 0 not due to all NAs
min_df1 <- min(df1, na.rm = T)
max_df1 <- max(df1, na.rm = T)
where_sum0_nonNA <- which(sumrows [1,] == 0 & (!is.na(min_df1[1,]) & !is.na(max_df1[1,])))
# Find rows to delete
where_del <- setdiff(where_sum0, where_sum0_nonNA)
# find "rows" with 0 sum and min and max not zero
df2[1,where_del] <- NA # replace in df2
head(df2)
然而这实际上应该回答问题：
给定多波段栅格 df1，在 所有 波段中识别具有 NA 的 像素。然后将第二个栅格 df2 中相同像素的所有波段设置为 NA。
，这是 OP 似乎要问的（也给出了他的最后评论）
HTH!
首先，OP在他对dracodoc的评论中生成示例数据的方式与OP的问题不一致：
library(raster)
r <- raster(nrow=5, ncol=5)
df1 <- stack( sapply(1:5, function(i) setValues(r, rnorm(ncell(r), i, 3))))
df2<- stack( sapply(1:5, function(i) setValues(r, rnorm(ncell(r), i, 3))))
df1[1,]<-NA
这会将 df1 的每一层的第一个 行 设置为 NA。那么问题应该是：
How can one find all rows in df1 where all columns are NAs, go to df2 and set all values in matching rows to NAs?
我先回答这个问题，再解决补码问题：
How can one find all columns in df1 where all rows are NAs, go to df2 and set all values in matching columns to NAs?
这是 OP 的原始问题。
将 df1 中 NA 的行设置为 df2
中的 NA
为了回答第一个问题，模拟另一个具有不同层数、行数和列数的示例数据集；然后，设置
不同行在df1到NA的每一层：
set.seed(123) ## only for full reproducibility
library(raster)
r <- raster(nrow=7, ncol=5) ## 7 rows to differentiate from 5 columns
nlayers <- 6
df1 <- stack(sapply(1:nlayers, function(i) setValues(r, rnorm(ncell(r), i, 3))))
df2<- stack(sapply(1:nlayers, function(i) setValues(r, rnorm(ncell(r), i, 3))))
df1[[1]][1,]<-NA ## set row 1 of layer 1 to NA
df1[[1]][2,]<-NA ## set row 2 of layer 1 to NA
df1[[2]][2,]<-NA ## set row 2 of layer 2 to NA
df1[[3]][2,]<-NA ## set row 2 of layer 3 to NA
df1[[4]][5,]<-NA ## set row 5 of layer 4 to NA
df1[[5]][5,]<-NA ## set row 5 of layer 5 to NA
df1[[6]][3,]<-NA ## set row 3 of layer 6 to NA
请注意 df1 看起来像：
df1[]
## layer.1 layer.2 layer.3 layer.4 layer.5 layer.6
## [1,] NA 4.0659208 1.5269065 3.8649168 7.1053530 2.712011
## [2,] NA 3.6617530 -3.9275066 1.6452866 4.2134075 6.113365
## [3,] NA 1.8142649 6.0172156 -1.0038258 0.2835675 6.931442
## [4,] NA 1.0821120 0.8723977 2.8593204 0.4559970 7.309570
## [5,] NA 0.8585870 0.9359742 6.7569898 0.1953915 4.624904
## [6,] NA NA NA 2.2739591 3.4072804 2.810022
## [7,] NA NA NA 5.8238930 0.6147332 9.789556
## [8,] NA NA NA -0.8536481 7.0637503 4.951049
## [9,] NA NA NA 3.8333141 11.3003268 3.403461
##[10,] NA NA NA 5.5582216 1.1389086 5.291161
##[11,] 4.6722454 -1.3693257 3.0172926 4.9034601 7.3632165 NA
##[12,] 2.0794415 0.7913455 4.1558412 4.3170286 7.3071267 NA
##[13,] 2.2023144 0.6000339 1.8880199 2.0778820 5.9966077 NA
##[14,] 1.3320481 4.3398954 4.9331296 1.4508870 1.9748702 NA
##[15,] -0.6675234 1.7498928 2.3385403 0.9276136 4.6416422 NA
##[16,] 6.3607394 2.7599555 3.9953459 4.3529398 4.1588140 6.643336
##[17,] 2.4935514 1.9143597 6.2905170 1.1575762 6.6889686 5.025942
##[18,] -4.8998515 1.8713886 4.3055445 2.5283277 3.8826837 6.283751
##[19,] 3.1040677 6.1058069 2.0222052 3.2317234 7.9309202 3.313910
##[20,] -0.4183742 1.3226870 6.4464229 9.5315860 3.8762574 2.067595
##[21,] -2.2034711 6.5494118 5.9805116 NA NA 11.991640
##[22,] 0.3460753 -2.6462584 4.6451909 NA NA 7.802126
##[23,] -2.0780133 3.7538412 3.7161952 NA NA 2.246186
##[24,] -1.1866737 2.3715627 1.1162818 NA NA 4.166502
##[25,] -0.8751178 2.6478247 7.0819573 NA NA 2.443560
##[26,] -4.0600799 3.1389184 1.1992212 8.3336526 5.8946828 12.596431
##[27,] 3.5133611 0.4930296 9.5619990 5.3545122 6.9097090 9.937239
##[28,] 1.4601194 1.0003778 7.5978319 4.1236988 3.5486581 5.204565
##[29,] -2.4144108 -1.0557261 2.2928989 2.7325095 6.5505861 7.629582
##[30,] 4.7614448 -1.2153737 -0.0792627 -2.1597417 6.1068936 4.756980
##[31,] 2.2793927 2.9105859 0.8687803 7.3940116 4.3538585 4.571259
##[32,] 0.1147856 3.3446293 3.7706511 -0.3819202 5.1958791 3.634191
##[33,] 3.6853770 2.1590127 2.2599244 6.2198425 4.8977982 4.216148
##[34,] 3.6344005 4.7668024 1.9573722 9.7273107 11.3853557 10.952722
##[35,] 3.4647432 8.1502541 0.1451443 -0.3316795 2.7759917 5.837916
请注意，现在可以清楚地看到此显示中的每个图层列都是按行优先顺序排列的栅格。
为了找到 df1 中每一层都是 NA 的行，我们从 OP 的链接 SO question/answer 中借用了 akrun 的解决方案，它同样适用于 Raster* 个对象：
inds <- which(!rowSums(!is.na(df1)), arr.ind=TRUE)
## row col
##[1,] 1 1
##[2,] 2 1
##[3,] 2 2
##[4,] 2 3
##[5,] 5 4
##[6,] 5 5
##[7,] 3 6
which(..., arr.ind=TRUE) returns 作为其 第一行 列 行 并作为其 第二列df中的层都是NA。然后我们可以使用层循环将这些行设置为 NA:
子集 df2
## loop over layers and set the appropriate row for each layer
for (i in seq_len(nrow(inds))) df2[[inds[i,2]]][inds[i,1],] <- NA
df2[]
## layer.1 layer.2 layer.3 layer.4 layer.5 layer.6
## [1,] NA 7.707085465 -2.0024253 6.11057171 8.0448295 10.135710
## [2,] NA 1.697075344 5.2094879 3.68298600 -0.9782455 7.368709
## [3,] NA -2.079522111 4.1580797 0.22405412 3.7181621 2.593235
## [4,] NA 0.005691694 2.2030451 9.05330712 5.3499119 4.693064
## [5,] NA 3.456379937 3.3544335 6.73417388 2.3203773 7.038311
## [6,] NA NA NA 4.71229082 6.0017088 4.058863
## [7,] NA NA NA 7.65432583 6.2342898 -0.472939
## [8,] NA NA NA -0.01632286 4.9008915 8.652752
## [9,] NA NA NA 5.98246089 -2.3976946 3.511567
##[10,] NA NA NA 2.43126287 12.7143744 4.279319
##[11,] -0.72192044 1.734304664 6.5051516 6.05123657 4.3841022 NA
##[12,] 2.85395745 5.242398488 6.1625431 3.81753414 6.9535798 NA
##[13,] 4.32954442 3.892262347 6.4357893 5.89888214 5.8212995 NA
##[14,] 3.12276506 1.659080313 1.2675960 8.00655285 8.0740197 NA
##[15,] -0.09097189 -2.598706009 9.0074482 4.02187027 7.4529783 NA
##[16,] 1.17924981 0.436648047 3.2001026 7.05267591 4.3706205 5.779332
##[17,] -1.11378939 0.530388641 8.6005555 0.43469789 6.1345033 2.494046
##[18,] -1.15165448 2.141463298 -1.0527081 1.83518668 2.1637735 4.095755
##[19,] 3.65395150 5.900596033 3.0629508 8.55765313 7.5707690 5.913475
##[20,] -2.04677774 8.879236921 6.7497437 5.13216392 3.6168850 8.012088
##[21,] 6.86588190 6.642743177 0.8542734 NA NA 1.048360
##[22,] 0.72904122 1.600547107 0.7419331 NA NA 4.950737
##[23,] 1.64361648 -3.269582187 0.1843839 NA NA 8.269219
##[24,] -1.21558311 0.833660408 -0.1575398 NA NA 4.383573
##[25,] -0.72316607 2.267621669 1.6885214 NA NA 6.681876
##[26,] -2.95104840 4.535039012 3.9935375 5.87256242 3.2315569 7.476686
##[27,] 0.45122383 4.887583905 -3.0426315 6.87701613 2.0096578 6.803505
##[28,] 2.25694721 4.052928288 3.6359413 9.01316449 5.4334271 7.959773
##[29,] 1.97291303 -2.185823049 6.7100251 4.16805020 4.9570778 5.631874
##[30,] -1.34460946 4.548929137 9.1127221 3.84405428 -0.3708437 4.758970
##[31,] -1.36586591 0.660328351 6.9035280 -1.25971208 5.1036532 -1.929447
##[32,] -0.50659616 2.524408100 5.2703243 4.29798278 5.5706909 5.721177
##[33,] 5.48818201 2.223653532 -2.1801912 2.28444983 5.5241792 7.290854
##[34,] -2.41191086 3.284500295 1.1954799 1.07797125 1.8349489 7.606197
##[35,] 0.46284522 2.074024948 1.9438606 3.46028131 6.4283998 4.334165
将 df1 中的 NA 列设置为 df2
中的 NA
对于补题，模拟数据集为：
set.seed(123) ## only for full reproducibility
library(raster)
r <- raster(nrow=7, ncol=5)
nlayers <- 6
df1 <- stack(sapply(1:nlayers, function(i) setValues(r, rnorm(ncell(r), i, 3))))
df2<- stack(sapply(1:nlayers, function(i) setValues(r, rnorm(ncell(r), i, 3))))
df1[[1]][,1]<-NA ## set column 1 of layer 1 to NA
df1[[1]][,2]<-NA ## set column 2 of layer 1 to NA
df1[[2]][,2]<-NA ## set column 2 of layer 2 to NA
df1[[3]][,2]<-NA ## set column 2 of layer 3 to NA
df1[[4]][,5]<-NA ## set column 5 of layer 4 to NA
df1[[5]][,5]<-NA ## set column 5 of layer 5 to NA
df1[[6]][,3]<-NA ## set column 3 of layer 6 to NA
请注意 df1 现在看起来像：
df1[]
## layer.1 layer.2 layer.3 layer.4 layer.5 layer.6
## [1,] NA 4.06592076 1.5269065 3.8649168 7.1053530 2.712011
## [2,] NA NA NA 1.6452866 4.2134075 6.113365
## [3,] 5.6761249 1.81426487 6.0172156 -1.0038258 0.2835675 NA
## [4,] 1.2115252 1.08211201 0.8723977 2.8593204 0.4559970 7.309570
## [5,] 1.3878632 0.85858700 0.9359742 NA NA 4.624904
## [6,] NA -0.08412094 6.0767141 2.2739591 3.4072804 2.810022
## [7,] NA NA NA 5.8238930 0.6147332 9.789556
## [8,] -2.7951837 -1.79618905 -0.6621531 -0.8536481 7.0637503 NA
## [9,] -1.0605586 8.50686790 3.5439104 3.8333141 11.3003268 3.403461
##[10,] -0.3369859 5.62388599 2.5833259 NA NA 5.291161
##[11,] NA -1.36932575 3.0172926 4.9034601 7.3632165 5.408472
##[12,] NA NA NA 4.3170286 7.3071267 9.329761
##[13,] 2.2023144 0.60003394 1.8880199 2.0778820 5.9966077 NA
##[14,] 1.3320481 4.33989536 4.9331296 1.4508870 1.9748702 8.262161
##[15,] -0.6675234 1.74989280 2.3385403 NA NA 4.502124
##[16,] NA 2.75995554 3.9953459 4.3529398 4.1588140 6.643336
##[17,] NA NA NA 1.1575762 6.6889686 5.025942
##[18,] -4.8998515 1.87138863 4.3055445 2.5283277 3.8826837 NA
##[19,] 3.1040677 6.10580685 2.0222052 3.2317234 7.9309202 3.313910
##[20,] -0.4183742 1.32268704 6.4464229 NA NA 2.067595
##[21,] NA 6.54941181 5.9805116 2.0441503 8.1581344 11.991640
##[22,] NA NA NA 4.7061597 1.8524690 7.802126
##[23,] -2.0780133 3.75384125 3.7161952 4.2338825 1.2195343 NA
##[24,] -1.1866737 2.37156273 1.1162818 1.1144301 14.7231198 4.166502
##[25,] -0.8751178 2.64782471 7.0819573 NA NA 2.443560
##[26,] NA 3.13891845 1.1992212 8.3336526 5.8946828 12.596431
##[27,] NA NA NA 5.3545122 6.9097090 9.937239
##[28,] 1.4601194 1.00037785 7.5978319 4.1236988 3.5486581 NA
##[29,] -2.4144108 -1.05572615 2.2928989 2.7325095 6.5505861 7.629582
##[30,] 4.7614448 -1.21537368 -0.0792627 NA NA 4.756980
##[31,] NA 2.91058592 0.8687803 7.3940116 4.3538585 4.571259
##[32,] NA NA NA -0.3819202 5.1958791 3.634191
##[33,] 3.6853770 2.15901268 2.2599244 6.2198425 4.8977982 NA
##[34,] 3.6344005 4.76680240 1.9573722 9.7273107 11.3853557 10.952722
##[35,] 3.4647432 8.15025406 0.1451443 NA NA 5.837916
解决方案是一样的，只是现在我们使用colSums而不是rowSums，并且我们为每一层设置列而不是行：
inds <- which(!colSums(!is.na(df1)), arr.ind=TRUE)
## loop over layers and set the appropriate column for each layer
for (i in seq_len(nrow(inds))) df2[[inds[i,2]]][,inds[i,1]] <- NA
df2[]
## layer.1 layer.2 layer.3 layer.4 layer.5 layer.6
## [1,] NA 7.707085465 -2.0024253 6.11057171 8.04482952 10.135710
## [2,] NA NA NA 3.68298600 -0.97824547 7.368709
## [3,] 4.69742764 -2.079522111 4.1580797 0.22405412 3.71816214 NA
## [4,] -0.54819149 0.005691694 2.2030451 9.05330712 5.34991185 4.693064
## [5,] -1.97752145 3.456379937 3.3544335 NA NA 7.038311
## [6,] NA 0.873191385 3.4021159 4.71229082 6.00170883 4.058863
## [7,] NA NA NA 7.65432583 6.23428976 -0.472939
## [8,] -1.16919791 0.968248298 7.9225385 -0.01632286 4.90089152 NA
## [9,] -2.70881936 2.271489941 2.3428489 5.98246089 -2.39769458 3.511567
##[10,] -2.85414717 6.795526313 3.5041962 NA NA 4.279319
##[11,] NA 1.734304664 6.5051516 6.05123657 4.38410223 10.511702
##[12,] NA NA NA 3.81753414 6.95357984 3.677565
##[13,] 4.32954442 3.892262347 6.4357893 5.89888214 5.82129947 NA
##[14,] 3.12276506 1.659080313 1.2675960 8.00655285 8.07401970 2.217951
##[15,] -0.09097189 -2.598706009 9.0074482 NA NA 4.936373
##[16,] NA 0.436648047 3.2001026 7.05267591 4.37062049 5.779332
##[17,] NA NA NA 0.43469789 6.13450332 2.494046
##[18,] -1.15165448 2.141463298 -1.0527081 1.83518668 2.16377351 NA
##[19,] 3.65395150 5.900596033 3.0629508 8.55765313 7.57076903 5.913475
##[20,] -2.04677774 8.879236921 6.7497437 NA NA 8.012088
##[21,] NA 6.642743177 0.8542734 -2.15666846 12.25032006 1.048360
##[22,] NA NA NA -0.09211236 0.04685331 4.950737
##[23,] 1.64361648 -3.269582187 0.1843839 3.39765695 3.60803827 NA
##[24,] -1.21558311 0.833660408 -0.1575398 6.59733821 7.47613959 4.383573
##[25,] -0.72316607 2.267621669 1.6885214 NA NA 6.681876
##[26,] NA 4.535039012 3.9935375 5.87256242 3.23155688 7.476686
##[27,] NA NA NA 6.87701613 2.00965777 6.803505
##[28,] 2.25694721 4.052928288 3.6359413 9.01316449 5.43342711 NA
##[29,] 1.97291303 -2.185823049 6.7100251 4.16805020 4.95707776 5.631874
##[30,] -1.34460946 4.548929137 9.1127221 NA NA 4.758970
##[31,] NA 0.660328351 6.9035280 -1.25971208 5.10365320 -1.929447
##[32,] NA NA NA 4.29798278 5.57069095 5.721177
##[33,] 5.48818201 2.223653532 -2.1801912 2.28444983 5.52417919 NA
##[34,] -2.41191086 3.284500295 1.1954799 1.07797125 1.83494887 7.606197
##[35,] 0.46284522 2.074024948 1.9438606 NA NA 4.334165
这个问题与 df1 和 df2 转换为数据帧。
给定两个尺寸相等的栅格 (df1,df2)：
df1: dimensions : 510, 1068, 544680, 1 (nrow, ncol, ncell, nlayers)
df2: dimensions : 510, 1068, 544680, 1 (nrow, ncol, ncell, nlayers)
如何找到 df1 中所有行都是 NAs 的所有列，转到 df2 并将匹配列中的所有值设置为 NAs？
鉴于所报告的示例数据，如果我很好地理解了这个问题，这应该可以解决问题：
library(raster)
r <- raster(nrow=5, ncol=5)
df1 <- stack( sapply(1:5, function(i) setValues(r, rnorm(ncell(r), i, 3))))
df2<- stack( sapply(1:5, function(i) setValues(r, rnorm(ncell(r), i, 3))))
df1[1,]<-NA
head(df1)
sumrows <- raster::rowSums(df1, na.rm = T) # if all NA in row, the sum is 0
where_NA <- which(sumrows[1,] == 0) # find 0 sum rows
df2[1,where_NA] <- NA
head(df2)
但是，如果 "rows" 全部为零，这将失败。如果这可能发生（如下面修改后的示例数据），您可以这样做：
library(raster)
r <- raster(nrow=5, ncol=5)
df1 <- stack( sapply(1:5, function(i) setValues(r, rnorm(ncell(r), i, 3))))
df2<- stack( sapply(1:5, function(i) setValues(r, rnorm(ncell(r), i, 3))))
df1[1,]<-NA
df1[1,5] <- 0
head(df1)
sumrows <- raster::rowSums(df1, na.rm = T) # if all NA in row, the sum is 0
where_sum0 <- which(sumrows [1,] == 0) # rows adding up to 0
# Find rows adding up to 0 not due to all NAs
min_df1 <- min(df1, na.rm = T)
max_df1 <- max(df1, na.rm = T)
where_sum0_nonNA <- which(sumrows [1,] == 0 & (!is.na(min_df1[1,]) & !is.na(max_df1[1,])))
# Find rows to delete
where_del <- setdiff(where_sum0, where_sum0_nonNA)
# find "rows" with 0 sum and min and max not zero
df2[1,where_del] <- NA # replace in df2
head(df2)
然而这实际上应该回答问题：
给定多波段栅格 df1，在 所有 波段中识别具有 NA 的 像素。然后将第二个栅格 df2 中相同像素的所有波段设置为 NA。
，这是 OP 似乎要问的（也给出了他的最后评论）
HTH!
首先，OP在他对dracodoc的评论中生成示例数据的方式与OP的问题不一致：
library(raster)
r <- raster(nrow=5, ncol=5)
df1 <- stack( sapply(1:5, function(i) setValues(r, rnorm(ncell(r), i, 3))))
df2<- stack( sapply(1:5, function(i) setValues(r, rnorm(ncell(r), i, 3))))
df1[1,]<-NA
这会将 df1 的每一层的第一个 行 设置为 NA。那么问题应该是：
How can one find all rows in
df1where all columns areNAs, go todf2and set all values in matching rows toNAs?
我先回答这个问题，再解决补码问题：
How can one find all columns in
df1where all rows areNAs, go todf2and set all values in matching columns toNAs?
这是 OP 的原始问题。
将 df1 中 NA 的行设置为 df2
中的 NA
为了回答第一个问题，模拟另一个具有不同层数、行数和列数的示例数据集；然后，设置
不同行在df1到NA的每一层：
set.seed(123) ## only for full reproducibility
library(raster)
r <- raster(nrow=7, ncol=5) ## 7 rows to differentiate from 5 columns
nlayers <- 6
df1 <- stack(sapply(1:nlayers, function(i) setValues(r, rnorm(ncell(r), i, 3))))
df2<- stack(sapply(1:nlayers, function(i) setValues(r, rnorm(ncell(r), i, 3))))
df1[[1]][1,]<-NA ## set row 1 of layer 1 to NA
df1[[1]][2,]<-NA ## set row 2 of layer 1 to NA
df1[[2]][2,]<-NA ## set row 2 of layer 2 to NA
df1[[3]][2,]<-NA ## set row 2 of layer 3 to NA
df1[[4]][5,]<-NA ## set row 5 of layer 4 to NA
df1[[5]][5,]<-NA ## set row 5 of layer 5 to NA
df1[[6]][3,]<-NA ## set row 3 of layer 6 to NA
请注意 df1 看起来像：
df1[]
## layer.1 layer.2 layer.3 layer.4 layer.5 layer.6
## [1,] NA 4.0659208 1.5269065 3.8649168 7.1053530 2.712011
## [2,] NA 3.6617530 -3.9275066 1.6452866 4.2134075 6.113365
## [3,] NA 1.8142649 6.0172156 -1.0038258 0.2835675 6.931442
## [4,] NA 1.0821120 0.8723977 2.8593204 0.4559970 7.309570
## [5,] NA 0.8585870 0.9359742 6.7569898 0.1953915 4.624904
## [6,] NA NA NA 2.2739591 3.4072804 2.810022
## [7,] NA NA NA 5.8238930 0.6147332 9.789556
## [8,] NA NA NA -0.8536481 7.0637503 4.951049
## [9,] NA NA NA 3.8333141 11.3003268 3.403461
##[10,] NA NA NA 5.5582216 1.1389086 5.291161
##[11,] 4.6722454 -1.3693257 3.0172926 4.9034601 7.3632165 NA
##[12,] 2.0794415 0.7913455 4.1558412 4.3170286 7.3071267 NA
##[13,] 2.2023144 0.6000339 1.8880199 2.0778820 5.9966077 NA
##[14,] 1.3320481 4.3398954 4.9331296 1.4508870 1.9748702 NA
##[15,] -0.6675234 1.7498928 2.3385403 0.9276136 4.6416422 NA
##[16,] 6.3607394 2.7599555 3.9953459 4.3529398 4.1588140 6.643336
##[17,] 2.4935514 1.9143597 6.2905170 1.1575762 6.6889686 5.025942
##[18,] -4.8998515 1.8713886 4.3055445 2.5283277 3.8826837 6.283751
##[19,] 3.1040677 6.1058069 2.0222052 3.2317234 7.9309202 3.313910
##[20,] -0.4183742 1.3226870 6.4464229 9.5315860 3.8762574 2.067595
##[21,] -2.2034711 6.5494118 5.9805116 NA NA 11.991640
##[22,] 0.3460753 -2.6462584 4.6451909 NA NA 7.802126
##[23,] -2.0780133 3.7538412 3.7161952 NA NA 2.246186
##[24,] -1.1866737 2.3715627 1.1162818 NA NA 4.166502
##[25,] -0.8751178 2.6478247 7.0819573 NA NA 2.443560
##[26,] -4.0600799 3.1389184 1.1992212 8.3336526 5.8946828 12.596431
##[27,] 3.5133611 0.4930296 9.5619990 5.3545122 6.9097090 9.937239
##[28,] 1.4601194 1.0003778 7.5978319 4.1236988 3.5486581 5.204565
##[29,] -2.4144108 -1.0557261 2.2928989 2.7325095 6.5505861 7.629582
##[30,] 4.7614448 -1.2153737 -0.0792627 -2.1597417 6.1068936 4.756980
##[31,] 2.2793927 2.9105859 0.8687803 7.3940116 4.3538585 4.571259
##[32,] 0.1147856 3.3446293 3.7706511 -0.3819202 5.1958791 3.634191
##[33,] 3.6853770 2.1590127 2.2599244 6.2198425 4.8977982 4.216148
##[34,] 3.6344005 4.7668024 1.9573722 9.7273107 11.3853557 10.952722
##[35,] 3.4647432 8.1502541 0.1451443 -0.3316795 2.7759917 5.837916
请注意，现在可以清楚地看到此显示中的每个图层列都是按行优先顺序排列的栅格。
为了找到 df1 中每一层都是 NA 的行，我们从 OP 的链接 SO question/answer 中借用了 akrun 的解决方案，它同样适用于 Raster* 个对象：
inds <- which(!rowSums(!is.na(df1)), arr.ind=TRUE)
## row col
##[1,] 1 1
##[2,] 2 1
##[3,] 2 2
##[4,] 2 3
##[5,] 5 4
##[6,] 5 5
##[7,] 3 6
which(..., arr.ind=TRUE) returns 作为其 第一行 列 行 并作为其 第二列df中的层都是NA。然后我们可以使用层循环将这些行设置为 NA:
df2
## loop over layers and set the appropriate row for each layer
for (i in seq_len(nrow(inds))) df2[[inds[i,2]]][inds[i,1],] <- NA
df2[]
## layer.1 layer.2 layer.3 layer.4 layer.5 layer.6
## [1,] NA 7.707085465 -2.0024253 6.11057171 8.0448295 10.135710
## [2,] NA 1.697075344 5.2094879 3.68298600 -0.9782455 7.368709
## [3,] NA -2.079522111 4.1580797 0.22405412 3.7181621 2.593235
## [4,] NA 0.005691694 2.2030451 9.05330712 5.3499119 4.693064
## [5,] NA 3.456379937 3.3544335 6.73417388 2.3203773 7.038311
## [6,] NA NA NA 4.71229082 6.0017088 4.058863
## [7,] NA NA NA 7.65432583 6.2342898 -0.472939
## [8,] NA NA NA -0.01632286 4.9008915 8.652752
## [9,] NA NA NA 5.98246089 -2.3976946 3.511567
##[10,] NA NA NA 2.43126287 12.7143744 4.279319
##[11,] -0.72192044 1.734304664 6.5051516 6.05123657 4.3841022 NA
##[12,] 2.85395745 5.242398488 6.1625431 3.81753414 6.9535798 NA
##[13,] 4.32954442 3.892262347 6.4357893 5.89888214 5.8212995 NA
##[14,] 3.12276506 1.659080313 1.2675960 8.00655285 8.0740197 NA
##[15,] -0.09097189 -2.598706009 9.0074482 4.02187027 7.4529783 NA
##[16,] 1.17924981 0.436648047 3.2001026 7.05267591 4.3706205 5.779332
##[17,] -1.11378939 0.530388641 8.6005555 0.43469789 6.1345033 2.494046
##[18,] -1.15165448 2.141463298 -1.0527081 1.83518668 2.1637735 4.095755
##[19,] 3.65395150 5.900596033 3.0629508 8.55765313 7.5707690 5.913475
##[20,] -2.04677774 8.879236921 6.7497437 5.13216392 3.6168850 8.012088
##[21,] 6.86588190 6.642743177 0.8542734 NA NA 1.048360
##[22,] 0.72904122 1.600547107 0.7419331 NA NA 4.950737
##[23,] 1.64361648 -3.269582187 0.1843839 NA NA 8.269219
##[24,] -1.21558311 0.833660408 -0.1575398 NA NA 4.383573
##[25,] -0.72316607 2.267621669 1.6885214 NA NA 6.681876
##[26,] -2.95104840 4.535039012 3.9935375 5.87256242 3.2315569 7.476686
##[27,] 0.45122383 4.887583905 -3.0426315 6.87701613 2.0096578 6.803505
##[28,] 2.25694721 4.052928288 3.6359413 9.01316449 5.4334271 7.959773
##[29,] 1.97291303 -2.185823049 6.7100251 4.16805020 4.9570778 5.631874
##[30,] -1.34460946 4.548929137 9.1127221 3.84405428 -0.3708437 4.758970
##[31,] -1.36586591 0.660328351 6.9035280 -1.25971208 5.1036532 -1.929447
##[32,] -0.50659616 2.524408100 5.2703243 4.29798278 5.5706909 5.721177
##[33,] 5.48818201 2.223653532 -2.1801912 2.28444983 5.5241792 7.290854
##[34,] -2.41191086 3.284500295 1.1954799 1.07797125 1.8349489 7.606197
##[35,] 0.46284522 2.074024948 1.9438606 3.46028131 6.4283998 4.334165
将 df1 中的 NA 列设置为 df2
中的 NA
对于补题，模拟数据集为：
set.seed(123) ## only for full reproducibility
library(raster)
r <- raster(nrow=7, ncol=5)
nlayers <- 6
df1 <- stack(sapply(1:nlayers, function(i) setValues(r, rnorm(ncell(r), i, 3))))
df2<- stack(sapply(1:nlayers, function(i) setValues(r, rnorm(ncell(r), i, 3))))
df1[[1]][,1]<-NA ## set column 1 of layer 1 to NA
df1[[1]][,2]<-NA ## set column 2 of layer 1 to NA
df1[[2]][,2]<-NA ## set column 2 of layer 2 to NA
df1[[3]][,2]<-NA ## set column 2 of layer 3 to NA
df1[[4]][,5]<-NA ## set column 5 of layer 4 to NA
df1[[5]][,5]<-NA ## set column 5 of layer 5 to NA
df1[[6]][,3]<-NA ## set column 3 of layer 6 to NA
请注意 df1 现在看起来像：
df1[]
## layer.1 layer.2 layer.3 layer.4 layer.5 layer.6
## [1,] NA 4.06592076 1.5269065 3.8649168 7.1053530 2.712011
## [2,] NA NA NA 1.6452866 4.2134075 6.113365
## [3,] 5.6761249 1.81426487 6.0172156 -1.0038258 0.2835675 NA
## [4,] 1.2115252 1.08211201 0.8723977 2.8593204 0.4559970 7.309570
## [5,] 1.3878632 0.85858700 0.9359742 NA NA 4.624904
## [6,] NA -0.08412094 6.0767141 2.2739591 3.4072804 2.810022
## [7,] NA NA NA 5.8238930 0.6147332 9.789556
## [8,] -2.7951837 -1.79618905 -0.6621531 -0.8536481 7.0637503 NA
## [9,] -1.0605586 8.50686790 3.5439104 3.8333141 11.3003268 3.403461
##[10,] -0.3369859 5.62388599 2.5833259 NA NA 5.291161
##[11,] NA -1.36932575 3.0172926 4.9034601 7.3632165 5.408472
##[12,] NA NA NA 4.3170286 7.3071267 9.329761
##[13,] 2.2023144 0.60003394 1.8880199 2.0778820 5.9966077 NA
##[14,] 1.3320481 4.33989536 4.9331296 1.4508870 1.9748702 8.262161
##[15,] -0.6675234 1.74989280 2.3385403 NA NA 4.502124
##[16,] NA 2.75995554 3.9953459 4.3529398 4.1588140 6.643336
##[17,] NA NA NA 1.1575762 6.6889686 5.025942
##[18,] -4.8998515 1.87138863 4.3055445 2.5283277 3.8826837 NA
##[19,] 3.1040677 6.10580685 2.0222052 3.2317234 7.9309202 3.313910
##[20,] -0.4183742 1.32268704 6.4464229 NA NA 2.067595
##[21,] NA 6.54941181 5.9805116 2.0441503 8.1581344 11.991640
##[22,] NA NA NA 4.7061597 1.8524690 7.802126
##[23,] -2.0780133 3.75384125 3.7161952 4.2338825 1.2195343 NA
##[24,] -1.1866737 2.37156273 1.1162818 1.1144301 14.7231198 4.166502
##[25,] -0.8751178 2.64782471 7.0819573 NA NA 2.443560
##[26,] NA 3.13891845 1.1992212 8.3336526 5.8946828 12.596431
##[27,] NA NA NA 5.3545122 6.9097090 9.937239
##[28,] 1.4601194 1.00037785 7.5978319 4.1236988 3.5486581 NA
##[29,] -2.4144108 -1.05572615 2.2928989 2.7325095 6.5505861 7.629582
##[30,] 4.7614448 -1.21537368 -0.0792627 NA NA 4.756980
##[31,] NA 2.91058592 0.8687803 7.3940116 4.3538585 4.571259
##[32,] NA NA NA -0.3819202 5.1958791 3.634191
##[33,] 3.6853770 2.15901268 2.2599244 6.2198425 4.8977982 NA
##[34,] 3.6344005 4.76680240 1.9573722 9.7273107 11.3853557 10.952722
##[35,] 3.4647432 8.15025406 0.1451443 NA NA 5.837916
解决方案是一样的，只是现在我们使用colSums而不是rowSums，并且我们为每一层设置列而不是行：
inds <- which(!colSums(!is.na(df1)), arr.ind=TRUE)
## loop over layers and set the appropriate column for each layer
for (i in seq_len(nrow(inds))) df2[[inds[i,2]]][,inds[i,1]] <- NA
df2[]
## layer.1 layer.2 layer.3 layer.4 layer.5 layer.6
## [1,] NA 7.707085465 -2.0024253 6.11057171 8.04482952 10.135710
## [2,] NA NA NA 3.68298600 -0.97824547 7.368709
## [3,] 4.69742764 -2.079522111 4.1580797 0.22405412 3.71816214 NA
## [4,] -0.54819149 0.005691694 2.2030451 9.05330712 5.34991185 4.693064
## [5,] -1.97752145 3.456379937 3.3544335 NA NA 7.038311
## [6,] NA 0.873191385 3.4021159 4.71229082 6.00170883 4.058863
## [7,] NA NA NA 7.65432583 6.23428976 -0.472939
## [8,] -1.16919791 0.968248298 7.9225385 -0.01632286 4.90089152 NA
## [9,] -2.70881936 2.271489941 2.3428489 5.98246089 -2.39769458 3.511567
##[10,] -2.85414717 6.795526313 3.5041962 NA NA 4.279319
##[11,] NA 1.734304664 6.5051516 6.05123657 4.38410223 10.511702
##[12,] NA NA NA 3.81753414 6.95357984 3.677565
##[13,] 4.32954442 3.892262347 6.4357893 5.89888214 5.82129947 NA
##[14,] 3.12276506 1.659080313 1.2675960 8.00655285 8.07401970 2.217951
##[15,] -0.09097189 -2.598706009 9.0074482 NA NA 4.936373
##[16,] NA 0.436648047 3.2001026 7.05267591 4.37062049 5.779332
##[17,] NA NA NA 0.43469789 6.13450332 2.494046
##[18,] -1.15165448 2.141463298 -1.0527081 1.83518668 2.16377351 NA
##[19,] 3.65395150 5.900596033 3.0629508 8.55765313 7.57076903 5.913475
##[20,] -2.04677774 8.879236921 6.7497437 NA NA 8.012088
##[21,] NA 6.642743177 0.8542734 -2.15666846 12.25032006 1.048360
##[22,] NA NA NA -0.09211236 0.04685331 4.950737
##[23,] 1.64361648 -3.269582187 0.1843839 3.39765695 3.60803827 NA
##[24,] -1.21558311 0.833660408 -0.1575398 6.59733821 7.47613959 4.383573
##[25,] -0.72316607 2.267621669 1.6885214 NA NA 6.681876
##[26,] NA 4.535039012 3.9935375 5.87256242 3.23155688 7.476686
##[27,] NA NA NA 6.87701613 2.00965777 6.803505
##[28,] 2.25694721 4.052928288 3.6359413 9.01316449 5.43342711 NA
##[29,] 1.97291303 -2.185823049 6.7100251 4.16805020 4.95707776 5.631874
##[30,] -1.34460946 4.548929137 9.1127221 NA NA 4.758970
##[31,] NA 0.660328351 6.9035280 -1.25971208 5.10365320 -1.929447
##[32,] NA NA NA 4.29798278 5.57069095 5.721177
##[33,] 5.48818201 2.223653532 -2.1801912 2.28444983 5.52417919 NA
##[34,] -2.41191086 3.284500295 1.1954799 1.07797125 1.83494887 7.606197
##[35,] 0.46284522 2.074024948 1.9438606 NA NA 4.334165