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I am at a complete loss on what I am doing wrong here.

I am trying to build a variogram. No matter what i do i get errors.

My Data:

     id       lon      lat                  datetime         m1      m2       m3
1      1 -121.3501 41.86009 2016-03-01 00:00:00-08:00    43.200  3993.071  3993.071
2      2 -121.3412 41.94472 2016-03-01 00:00:00-08:00    38.800  4036.954  4036.954
3      3 -121.5138 41.99801 2016-03-01 00:00:00-08:00    24.600  4023.301  4023.301

After converting to a Spatial DataFrame

class       : SpatialPointsDataFrame 
features    : 3196 
extent      : -124.2696, -116.1304, 33.11097, 42.01195  (xmin, xmax, ymin, ymax)
coord. ref. : NA 
variables   : 5
names       : id,                  datetime,      m1,      m2,        m3 
min values  :   1, 2016-03-01 00:00:00-08:00,     0.000,     0.020,  0.0003920282 
max values  :   3196, 2016-03-01 00:00:00-08:00, 14360.500, -7470.500, -3.4202715765 

This is just a small piece of the data. About 3000 rows in total

My dataframe setup

df <- read.table("input.csv", header=TRUE, sep=",")
df = na.omit(df)
coordinates(df) = ~lon+lat

Some samples of what I have tried and the error messages:

v = variogram(m3~1, data=df)

Error in variogram(m3 ~ 1, data = df) : unused argument (data = df)


v = variogram(df$m3~1, data=df, dX=0)

Error in variogram(df$m3 ~ 1, data = df, dX = 0) : 
  unused arguments (data = df, dX = 0)

Since it keeps telling me these parameters are unused...I tried this:

v = variogram(df$m3~1)

Error in variogram(df$m3 ~ 1) : 
  argument "X" is missing, with no default

All I am trying to do at this point is create a basic variogram on the m2 or m3 column of data. m3 column is the m2 column with a normal score transformation done on it. I just left the code out for that.

What am I doing wrong?

  • Have you tried v = variogram(m3~1, locations = df ? – Jot eN Sep 13 '18 at 9:10
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If you look at the help for variogram, you'll see:

 library(sp)
 data(meuse)
 # no trend:
 coordinates(meuse) = ~x+y
 variogram(log(zinc)~1, meuse)
 # residual variogram w.r.t. a linear trend:
 variogram(log(zinc)~x+y, meuse)

The first argument is a formula in the column names of the second argument, the data. Neither argument are named in the call. So for your case:

 variogram(m3~1, df)

should work. I don't understand why variogram(m3~1, data=df) doesn't work for you, since variogram(zinc~1, data=meuse) does....

If the above still fails, we might need to check out your data personally.

  • Thanks for your help here. I was doing the formula wrong-ish. I was trying to do log(m3)~1 but my data has negative values. I think this was causing the error. Arcgis helped find that issue trying to plot a histogram on the data. It is working just find using variogram(m3~1, data=df). I did not realize i could do it like that. So thank you very much. – Chris Sep 16 '18 at 4:50

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