2

我有一个我创建的人工数据集：

x<-rnorm(100,10,10)
y<-rnorm(100,20,10)
Location<-c((rep("AB", 40)),(rep("TA", 30)),(rep("OP", 30)))
Year<-c((rep("1999", 10)),(rep("2000", 9)),(rep("2001", 12)),(rep("2002", 9)),(rep("1999", 7)),(rep("2000", 6)),(rep("2001", 6)),(rep("2002", 11)),(rep("1999", 12)),(rep("2000", 8)),(rep("2001", 5)),(rep("2002", 5)))
Data<-cbind(x,y,Location,Year)

> head(Data)
       x                  y                  Location Year  
[1,] "1.8938661556415"  "19.851256070398"  "AB"     "1999"
[2,] "21.0735971323312" "17.4993965352294" "AB"     "1999"
[3,] "30.8347289164302" "7.63333686308105" "AB"     "1999"
[4,] "8.913993138201"   "14.7085296541221" "AB"     "1999"
[5,] "20.8309225677419" "12.0888505284667" "AB"     "1999"
[6,] "25.3978549194374" "20.47154776064"   "AB"     "1999"

我想取每个 x 和 y 的 arc2tan，例如：

 Theta<-atan2(y[i+1]-y[i],x[i+1]-x[i])

但我只想在年份位置内的每一年执行此操作，这意味着我不想在 1999 年至 2000 年之间或 2001 年至 2002 年等之间找到 theta。仅在同一位置的同一年的 x 和 y 点之间。

我最初编写了一个执行上述操作的循环（我不想这样做），我想知道是否有人知道如何更改它，以便循环每年都会停止并自行重置。原始循环提供如下：

for (i in 1:length(x)-1){
  Theta[i]<-atan2(y[i+1]-y[i],x[i+1]-x[i])
}

有帮手吗？

4

1 回答 1

1

你可以试试这个。

# a smaller test data set
x <- rnorm(24, 10, 10)
y <- rnorm(24, 20, 10)
loc <- rep(c("A", "B"), each = 4)
year <- rep(1999:2001, each = 8)
df <- data.frame(x, y, loc, year)

df

# apply function on subsets defined by location and year
# use tail and head to 'lag' y and x
by(df, df[ , c("loc", "year")], function(x){
with(x, atan2(y = tail(y, - 1) - head(y, -1), x = tail(x, -1) - head(x, - 1)))
})

# loc: A
# year: 1999
# [1]  2.306794 -2.363359  1.065151
# --------------------------------------------------------------------------- 
# loc: B
# year: 1999
# [1] -1.077345  1.161944 -2.101823
# --------------------------------------------------------------------------- 
# loc: A
# year: 2000
# [1] -1.76557207  1.79463661 -0.05251002
# --------------------------------------------------------------------------- 
# loc: B
# year: 2000
# [1]  2.753115 -1.468055 -1.624389
# ...snip...

一个dplyr替代方案。因为在这种情况下，每个组内函数结果的长度不等于组大小或 1，dplyr所以根本不乐意咀嚼数据帧（请参阅此处和此处）。一种解决方法是dplyr使用data.table. 当然，data.table这里唯一的解决方案将是最干净的。我把它留给比我更熟悉的人data.table...

library(data.table)
library(dplyr)
dt <- data.table(df)
dt %.%
  group_by(loc, year) %.%
  mutate(
    atan = atan2(lead(y, default = NULL) - lag(y, default = NULL),
          lead(x, default = NULL) - lag(x, default = NULL)))

#            x         y loc year        atan
# 1  19.826573 18.354265   A 1999  2.30679446
# 2  11.856696 27.153197   A 1999 -2.36335869
# 3  -3.362242 12.150775   A 1999  1.06515149
# 4  11.126841 38.320662   A 1999  2.30679446
# 5  12.616396 31.782969   A 2000 -1.76557207
# 6   8.492305 10.877870   A 2000  1.79463661
# 7   4.921766 26.561845   A 2000 -0.05251002
# 8  14.398730 26.063752   A 2000 -1.76557207
# 9  11.800173 30.215422   A 2001 -2.74907150
# 10 -6.473259 22.650127   A 2001  0.11997030
# 11  6.528055 24.217425   A 2001 -1.71122202
# 12  4.951238 13.062497   A 2001 -2.74907150
# 13  1.640049 19.886848   B 1999 -1.07734532
# 14  4.123603 15.269110   B 1999  1.16194418
# 15 14.548780 39.330885   B 1999 -2.10182331
# 16  6.925468 26.350556   B 1999 -1.07734532
# ...snip...
于 2014-03-06T20:52:13.363 回答