如何从for循环结果填充R中的空矩阵？

我正在计算25年内所有49681个地点的NDVI年平均值。我创建了一个for循环，但我不知道如何填充一个空的49681 x 25矩阵。我的代码现在只填充矩阵的第一列。有没有关于如何解决这个问题的建议

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除了for循环之外，还有更好的方法来实现这一点，但是对于初学者来说，您尝试将26组值分配到25列中。您还可以直接告诉R使用

i

行数填充单个列，26次，其中25次覆盖前面的列值。我也很困惑你使用的范围，因为它是一个长度为'23'的范围。抛开这些不谈，按原样回答您的问题，您可以编写一个for循环，如下所示：

for(i in 1:49681){
    yearly.avg[i,1] <- mean(as.numeric(veg.data[i,4:603]))
    yearly.avg[i,2] <- mean(as.numeric(veg.data[i,4:27]))
    ...
}

for（i在1:49681中）{
年平均值[i，1]除了for循环之外，还有更好的方法来实现这一点，但对于初学者来说，您尝试将26组值分配到25列中。您还实际告诉R使用
i
行数填充单个列，26次，其中25次覆盖以前的列值。我还对范围感到非常困惑您使用的是长度范围为“23”的for循环。除此之外，要按原样回答您的问题，您可以编写如下for循环代码：

for(i in 1:49681){
    yearly.avg[i,1] <- mean(as.numeric(veg.data[i,4:603]))
    yearly.avg[i,2] <- mean(as.numeric(veg.data[i,4:27]))
    ...
}

for（i在1:49681中）{
根据annual.avg[i，1]，您可以这样做：

# make some fake data and set up the structure
siteCount <- 49681
yearCount <- 25
monthCount <- 12
obsPerMonth <- 2
obsCount <- yearCount * monthCount * obsPerMonth
fakeData <- sample(100:500, size=siteCount * obsCount, replace=TRUE)
veg.data <- matrix(fakeData, nrow=siteCount, ncol=obsCount)

sitenum <- sprintf("N%05d", (1:49681)+9000)
lat <- seq(from=40.90952, by=0.08, length.out=length(sitenum))
long <- seq(from=2.755276, by=0.08, length.out=length(sitenum))

veg.data <- as.data.frame(cbind(sitenum, lat, long, veg.data), stringsAsFactors=FALSE)
v <- expand.grid(c('A', 'B'), sprintf("%02d", 1:monthCount), 1982:2006)
dataColNames <- paste('Y', v[, 3], '.', v[, 2], v[, 1], sep='')

colnames(veg.data) <- c('sitenum', 'x', 'y', dataColNames)

###
# we now have the sample data, we can calculate yearly means
###

# First, get a numeric matrix of just the veg data
veg.data2 <- as.matrix(veg.data[, 4:ncol(veg.data)])
storage.mode(veg.data2) <- "numeric"

# change the column headings to be just the year, so that we can average based on year
colnames(veg.data2) <- substring(colnames(veg.data2), 1, 5)

# now, calculate yearly averages
yearly.avg <- sapply(unique(colnames(veg.data2)), function(x) 
      rowMeans(veg.data2[,colnames(veg.data2)== x,drop=FALSE], na.rm=TRUE))

# have a look
head(yearly.avg)
         1982     1983     1984     1985     1986     1987     1988     1989     1990     1991     1992     1993
[1,] 325.2083 363.7500 283.6250 315.6667 289.7500 260.7917 297.0000 301.5833 285.9167 299.2083 264.9167 311.2083
[2,] 307.6250 287.7500 281.3750 296.5833 330.7083 268.2917 331.5417 309.6667 275.7917 300.5833 287.9583 291.2500
[3,] 272.5000 295.9167 302.1250 314.7083 270.6667 340.2917 287.1250 336.3333 309.2500 266.7500 273.5000 254.2917
[4,] 288.9167 280.7083 299.1667 279.5833 301.4583 283.7917 274.6667 295.6250 238.6250 324.7917 302.2083 283.1667
[5,] 280.8750 282.5833 294.7083 276.0417 303.2917 266.5000 324.9583 301.5417 266.2917 327.0417 295.7083 262.7917
[6,] 275.0833 321.5833 305.1250 308.5417 266.7917 304.2083 304.1250 290.1667 312.9167 266.5000 273.7500 314.2917
         1994     1995     1996     1997     1998     1999     2000     2001     2002     2003     2004     2005
[1,] 320.7083 339.9583 288.9167 329.3750 303.6667 290.0417 288.3333 299.0417 290.3333 315.2500 272.5833 303.1667
[2,] 336.7500 295.0000 301.7917 303.0000 294.7917 337.5417 328.1250 284.5417 301.3333 300.6667 302.7083 288.7917
[3,] 314.2083 313.7500 325.0417 290.2917 276.6250 262.7500 315.7500 267.9167 301.8750 312.3333 288.1667 308.5000
[4,] 283.1667 278.8750 300.3333 278.3333 291.7500 358.2500 326.5833 311.7500 248.8750 250.8333 316.5000 324.0417
[5,] 286.9167 290.7500 331.7500 330.2500 317.5417 326.0417 297.8750 307.4583 371.9583 323.9583 320.5833 290.3750
[6,] 290.5000 306.0833 238.0833 304.7083 300.0417 252.3333 261.1250 253.9167 274.2083 282.8750 326.8750 306.1250
         2006
[1,] 308.0000
[2,] 298.4167
[3,] 293.2083
[4,] 308.0417
[5,] 305.6250
[6,] 297.1667

# Manually calculate average for 1982 to check result
d <- as.matrix(veg.data[, 4:27])
storage.mode(d) <- "numeric"
head(rowMeans(d))
[1] 325.2083 307.6250 272.5000 288.9167 280.8750 275.0833

head(rowMeans(d)) == head(yearly.avg[, 'Y1982'])
[1] TRUE TRUE TRUE TRUE TRUE TRUE

#制作一些假数据并设置结构
siteCount基于，您可以执行以下操作：

# make some fake data and set up the structure
siteCount <- 49681
yearCount <- 25
monthCount <- 12
obsPerMonth <- 2
obsCount <- yearCount * monthCount * obsPerMonth
fakeData <- sample(100:500, size=siteCount * obsCount, replace=TRUE)
veg.data <- matrix(fakeData, nrow=siteCount, ncol=obsCount)

sitenum <- sprintf("N%05d", (1:49681)+9000)
lat <- seq(from=40.90952, by=0.08, length.out=length(sitenum))
long <- seq(from=2.755276, by=0.08, length.out=length(sitenum))

veg.data <- as.data.frame(cbind(sitenum, lat, long, veg.data), stringsAsFactors=FALSE)
v <- expand.grid(c('A', 'B'), sprintf("%02d", 1:monthCount), 1982:2006)
dataColNames <- paste('Y', v[, 3], '.', v[, 2], v[, 1], sep='')

colnames(veg.data) <- c('sitenum', 'x', 'y', dataColNames)

###
# we now have the sample data, we can calculate yearly means
###

# First, get a numeric matrix of just the veg data
veg.data2 <- as.matrix(veg.data[, 4:ncol(veg.data)])
storage.mode(veg.data2) <- "numeric"

# change the column headings to be just the year, so that we can average based on year
colnames(veg.data2) <- substring(colnames(veg.data2), 1, 5)

# now, calculate yearly averages
yearly.avg <- sapply(unique(colnames(veg.data2)), function(x) 
      rowMeans(veg.data2[,colnames(veg.data2)== x,drop=FALSE], na.rm=TRUE))

# have a look
head(yearly.avg)
         1982     1983     1984     1985     1986     1987     1988     1989     1990     1991     1992     1993
[1,] 325.2083 363.7500 283.6250 315.6667 289.7500 260.7917 297.0000 301.5833 285.9167 299.2083 264.9167 311.2083
[2,] 307.6250 287.7500 281.3750 296.5833 330.7083 268.2917 331.5417 309.6667 275.7917 300.5833 287.9583 291.2500
[3,] 272.5000 295.9167 302.1250 314.7083 270.6667 340.2917 287.1250 336.3333 309.2500 266.7500 273.5000 254.2917
[4,] 288.9167 280.7083 299.1667 279.5833 301.4583 283.7917 274.6667 295.6250 238.6250 324.7917 302.2083 283.1667
[5,] 280.8750 282.5833 294.7083 276.0417 303.2917 266.5000 324.9583 301.5417 266.2917 327.0417 295.7083 262.7917
[6,] 275.0833 321.5833 305.1250 308.5417 266.7917 304.2083 304.1250 290.1667 312.9167 266.5000 273.7500 314.2917
         1994     1995     1996     1997     1998     1999     2000     2001     2002     2003     2004     2005
[1,] 320.7083 339.9583 288.9167 329.3750 303.6667 290.0417 288.3333 299.0417 290.3333 315.2500 272.5833 303.1667
[2,] 336.7500 295.0000 301.7917 303.0000 294.7917 337.5417 328.1250 284.5417 301.3333 300.6667 302.7083 288.7917
[3,] 314.2083 313.7500 325.0417 290.2917 276.6250 262.7500 315.7500 267.9167 301.8750 312.3333 288.1667 308.5000
[4,] 283.1667 278.8750 300.3333 278.3333 291.7500 358.2500 326.5833 311.7500 248.8750 250.8333 316.5000 324.0417
[5,] 286.9167 290.7500 331.7500 330.2500 317.5417 326.0417 297.8750 307.4583 371.9583 323.9583 320.5833 290.3750
[6,] 290.5000 306.0833 238.0833 304.7083 300.0417 252.3333 261.1250 253.9167 274.2083 282.8750 326.8750 306.1250
         2006
[1,] 308.0000
[2,] 298.4167
[3,] 293.2083
[4,] 308.0417
[5,] 305.6250
[6,] 297.1667

# Manually calculate average for 1982 to check result
d <- as.matrix(veg.data[, 4:27])
storage.mode(d) <- "numeric"
head(rowMeans(d))
[1] 325.2083 307.6250 272.5000 288.9167 280.8750 275.0833

head(rowMeans(d)) == head(yearly.avg[, 'Y1982'])
[1] TRUE TRUE TRUE TRUE TRUE TRUE

#制作一些假数据并设置结构
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