Finding values by linear interpolation in r

Question

I have huge data that I need to find the values of several variables at a standard height. I want to interpolate linearly the values of the other variables at Height=c(0,100,200,250,400,500) and add them as new columns to the existing data. Here is what I tried to get a value for one variable as the standard Height=c(0,100,200,250,400,500) . This is just for one variable:

data2<-approx(data2$Height,data2$ozone,xout=c(0,100,200,250,400,500))

The expected result should be a data frame with 18 rows and 4 columns.

Here is sample data (data2) :

  ozone     Height      Temp        Wind
23.224833   0.000000    253.005798  3.631531
23.750044   35.218689   253.299332  5.178889
24.589071   70.661133   253.538574  6.892455
25.619747   106.267334  253.492661  8.050934
26.443541   142.014648  253.279053  8.648781
27.235945   213.897034  252.815262  9.263882
27.698713   286.280518  252.10556   9.269853
27.865248   359.172363  251.390045  9.3006
28.361752   432.788086  251.379913  8.90488
30.279163   507.276733  251.849655  7.817647
23.048151   0.000000    251.528275  4.174027
23.477306   34.998413   251.6698    5.630364
24.16725    70.187622   251.759369  7.237537
25.239206   105.544006  251.744934  8.859097
26.319073   141.05011   251.601654  9.928196
27.409718   212.47052   251.214279  10.75243
27.825275   284.45282   250.738007  10.812123
28.214966   357.184631  250.87706   9.980968
29.726873   430.919983  251.84964   9.139032
32.482925   505.574097  252.471924  8.063484
22.369734   0.000000    250.876144  3.82036
22.916582   34.908447   251.044205  5.281044
23.732521   70.014038   251.170456  6.970277
24.998178   105.296021  251.221603  8.801399
26.30809    140.736084  251.133591  10.039667
27.572966   212.052795  250.852631  11.118568
28.233795   283.998474  250.61908   10.677624
29.079391   356.812012  251.179962  9.466641
31.244007   430.597534  252.042175  9.016301
33.636559   505.305542  252.659393  8.103294

Thank you in advance for your help.

UPDATE

Here is the desired answer:

 Height    ozone     Temp      Wind
       0 23.22483 253.0058  3.631531
     100 25.43833 253.5007  7.847021
     200 27.08275 252.9049  9.144964
     300 27.73006 251.9709  9.275640
     400 28.14061 251.3844  9.081132
     500 30.09185 251.8038  7.923858
       0 23.04815 251.5283  4.174027
     100 25.07112 251.7472  8.604831
     200 27.21928 251.2819 10.608513
     300 27.90858 250.7677 10.634455
     400 29.09287 251.4418  9.492087
     500 32.27714 252.4255  8.143790
     0   22.36973 250.8761  3.820360
     100 24.80820 251.2139  8.526537
     200 27.35920 250.9001 10.936230
     300 28.41962 250.7423 10.411498
     400 30.34638 251.6846  9.203049
     500 33.46665 252.6156  8.168133

Answer 1

You just work through columns using lapply . Plus, you can't append your interpolated values to your data2 . data2 has 30 rows, while xout has length 6. You need another data frame to hold interpolation result.

cbind.data.frame(data.frame(Height = 0:5 * 100),
                 lapply(data2[-2], function (u) approx(data2[[2]], u, 0:5 * 100)$y))

#  Height    ozone     Temp      Wind
#1      0 22.88091 251.8034  3.875306
#2    100 24.93562 251.5759  8.509502
#3    200 27.37860 251.2702 10.693545
#4    300 27.96728 251.9255  9.308131
#5    400 29.79659 251.7628  9.138091
#6    500 33.25064 252.5658  8.161940

---

Follow up

The original data is model output for 3 days, and I want to keep it to some standard heights for comparing with other data. So each data frame represents one-day data. So I merge them in one big data frame data2 , with the same height as the other variables vary each day.

Alright, your data2 has time attributes and every 10 rows correspond to one day's data. Well, you shouldn't stack data frome different days by row. If you do this, you should add a new column, say day to highlight this kind of block / group structure.

So, what you really need is an independent linear interpolation for each data. My original answer is doing a unified interpolation using all three days' data. Since you have tied values on Height , it is actually interpolating the mean of ozone , Temp and Wind over 3 days. The following code gets you what you expect.

## change my previous code to a function
result_per_day <- function (dat) {
  cbind.data.frame(data.frame(Height = 0:5 * 100),
                   lapply(dat[-2], function (u) approx(dat[[2]], u, 0:5 * 100)$y))
  }

datalst <- split(data2, gl(3, 10, labels = 1:3))
do.call(rbind.data.frame, lapply(datalst, result_per_day))

#    Height    ozone     Temp      Wind
#1.1      0 23.22483 253.0058  3.631531
#1.2    100 25.43833 253.5007  7.847021
#1.3    200 27.08275 252.9049  9.144964
#1.4    300 27.73006 251.9709  9.275640
#1.5    400 28.14061 251.3844  9.081132
#1.6    500 30.09185 251.8038  7.923858
#2.1      0 23.04815 251.5283  4.174027
#2.2    100 25.07112 251.7472  8.604831
#2.3    200 27.21928 251.2819 10.608513
#2.4    300 27.90858 250.7677 10.634455
#2.5    400 29.09287 251.4418  9.492087
#2.6    500 32.27714 252.4255  8.143790
#3.1      0 22.36973 250.8761  3.820360
#3.2    100 24.80820 251.2139  8.526537
#3.3    200 27.35920 250.9001 10.936230
#3.4    300 28.41962 250.7423 10.411498
#3.5    400 30.34638 251.6846  9.203049
#3.6    500 33.46665 252.6156  8.168133

The row names of this final data frame is quite explanatory. "1.1" to "1.6" are for day 1, while "2.1" to "2.6" are for day 2, etc.