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Calculation of Geometric Mean of Data that includes NAs
Geometric Mean: is there a built-in?Aggregate function in R with multiple function argumentsHow to join (merge) data frames (inner, outer, left, right)Geometric Mean: is there a built-in?Drop data frame columns by nameHow to skip NA when applying geometric-mean functioncalculating weighted geometric mean of a large list of numbersMATLAB: cumulative geometric meanGeometric mean filter with opencvcalculate the harmonic mean and geometric mean without exceeding double max value?Calculating geometric mean every 10 min using dplyr or aggregte functionOptimizing calculation for Weighted Geometric Mean of a big set of data using GPU
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EDIT: The problem was not within the geoMean function, but with a wrong use of aggregate(), as explained in the comments
I am trying to calculate the geometric mean of multiple measurements for several different species, which includes NAs. An example of my data looks like this:
species <- c("Ae", "Ae", "Ae", "Be", "Be")
phen <- c(2, NA, 3, 1, 2)
hveg <- c(NA, 15, 12, 60, 59)
df <- data.frame(species, phen, hveg)
When I try to calculate the geometric mean for the species Ae with the built-in function geoMean from the package EnvStats like this
library("EnvStats")
aggregate(df[, 3:3], list(df1$Sp), geoMean, na.rm=TRUE)
it works wonderful and skips the NAs to give me the geometric means per species.
Group.1 phen hveg
1 Ae 4.238536 50.555696
2 Be 1.414214 1.414214
When I do this with my large dataset, however, the function stumbles over NAs and returns NA as result even though there are e.g 10 numerical values and only one NA. This happens for example with the column SLA_mm2/mg.
My large data set looks like this:
> str(cut2trait1)
Classes ‘tbl_df’, ‘tbl’ and 'data.frame': 22 obs. of 19 variables:
$ Cut : chr "15_08" "15_08" "15_08" "15_08" ...
$ Block : num 1 1 1 1 1 1 1 1 1 1 ...
$ ID : num 451 512 431 531 591 432 551 393 511 452 ...
$ Plot : chr "1_1" "1_1" "1_1" "1_1" ...
$ Grazing : chr "n" "n" "n" "n" ...
$ Acro : chr "Leuc.vulg" "Dact.glom" "Cirs.arve" "Trif.prat" ...
$ Sp : chr "Lv" "Dg" "Ca" "Tp" ...
$ Label_neu : chr "Lv021" "Dg022" "Ca021" "Tp021" ...
$ PlantFunctionalType: chr "forb" "grass" "forb" "forb" ...
$ PlotClimate : chr "AC" "AC" "AC" "AC" ...
$ Season : chr "Aug" "Aug" "Aug" "Aug" ...
$ Year : num 2015 2015 2015 2015 2015 ...
$ Tiller : num 6 3 3 5 6 8 5 2 1 7 ...
$ Hveg : num 25 38 70 36 68 65 23 58 71 27 ...
$ Hrep : num 39 54 77 38 76 70 65 88 98 38 ...
$ Phen : num 8 8 7 8 8 7 6.5 8 8 8 ...
$ SPAD : num 40.7 42.4 48.7 43 31.3 ...
$ TDW_in_g : num 4.62 4.85 11.86 5.82 8.99 ...
$ SLA_mm2/mg : num 19.6 19.8 20.3 21.2 21.7 ...
and the result of my code
gm_cut2trait1 <- aggregate(cut2trait1[, 13:19], list(cut2trait1$Sp), geoMean, na.rm=TRUE)
is (only the first two rows):
Group.1 Tiller Hveg Hrep Phen SPAD TDW_in_g SLA_mm2/mg
1 Ae 13.521721 73.43485 106.67933 NA 28.17698 1.2602475 NA
2 Be 8.944272 43.95452 72.31182 5.477226 20.08880 0.7266361 9.309672
Here, the geometric mean of SLA for Ae is NA, even though there are 9 numeric measurements and only one NA in the column used to calculate the geometric mean.
I tried to use the geometric mean function suggested here:
Geometric Mean: is there a built-in?
But instead of NAs, this returned the value 1.000 when used with my big dataset, which doesn't solve my problem.
So my question is: What is the difference between my example df and the big dataset that throws the geoMean function off the rails?
r geometric-mean
asked Mar 27 at 9:46
Hedera.HelixHedera.Helix
62 bronze badges
add a comment |
EDIT: The problem was not within the geoMean function, but with a wrong use of aggregate(), as explained in the comments
I am trying to calculate the geometric mean of multiple measurements for several different species, which includes NAs. An example of my data looks like this:
species <- c("Ae", "Ae", "Ae", "Be", "Be")
phen <- c(2, NA, 3, 1, 2)
hveg <- c(NA, 15, 12, 60, 59)
df <- data.frame(species, phen, hveg)
When I try to calculate the geometric mean for the species Ae with the built-in function geoMean from the package EnvStats like this
library("EnvStats")
aggregate(df[, 3:3], list(df1$Sp), geoMean, na.rm=TRUE)
it works wonderful and skips the NAs to give me the geometric means per species.
Group.1 phen hveg
1 Ae 4.238536 50.555696
2 Be 1.414214 1.414214
When I do this with my large dataset, however, the function stumbles over NAs and returns NA as result even though there are e.g 10 numerical values and only one NA. This happens for example with the column SLA_mm2/mg.
My large data set looks like this:
> str(cut2trait1)
Classes ‘tbl_df’, ‘tbl’ and 'data.frame': 22 obs. of 19 variables:
$ Cut : chr "15_08" "15_08" "15_08" "15_08" ...
$ Block : num 1 1 1 1 1 1 1 1 1 1 ...
$ ID : num 451 512 431 531 591 432 551 393 511 452 ...
$ Plot : chr "1_1" "1_1" "1_1" "1_1" ...
$ Grazing : chr "n" "n" "n" "n" ...
$ Acro : chr "Leuc.vulg" "Dact.glom" "Cirs.arve" "Trif.prat" ...
$ Sp : chr "Lv" "Dg" "Ca" "Tp" ...
$ Label_neu : chr "Lv021" "Dg022" "Ca021" "Tp021" ...
$ PlantFunctionalType: chr "forb" "grass" "forb" "forb" ...
$ PlotClimate : chr "AC" "AC" "AC" "AC" ...
$ Season : chr "Aug" "Aug" "Aug" "Aug" ...
$ Year : num 2015 2015 2015 2015 2015 ...
$ Tiller : num 6 3 3 5 6 8 5 2 1 7 ...
$ Hveg : num 25 38 70 36 68 65 23 58 71 27 ...
$ Hrep : num 39 54 77 38 76 70 65 88 98 38 ...
$ Phen : num 8 8 7 8 8 7 6.5 8 8 8 ...
$ SPAD : num 40.7 42.4 48.7 43 31.3 ...
$ TDW_in_g : num 4.62 4.85 11.86 5.82 8.99 ...
$ SLA_mm2/mg : num 19.6 19.8 20.3 21.2 21.7 ...
and the result of my code
gm_cut2trait1 <- aggregate(cut2trait1[, 13:19], list(cut2trait1$Sp), geoMean, na.rm=TRUE)
is (only the first two rows):
Group.1 Tiller Hveg Hrep Phen SPAD TDW_in_g SLA_mm2/mg
1 Ae 13.521721 73.43485 106.67933 NA 28.17698 1.2602475 NA
2 Be 8.944272 43.95452 72.31182 5.477226 20.08880 0.7266361 9.309672
Here, the geometric mean of SLA for Ae is NA, even though there are 9 numeric measurements and only one NA in the column used to calculate the geometric mean.
I tried to use the geometric mean function suggested here:
Geometric Mean: is there a built-in?
But instead of NAs, this returned the value 1.000 when used with my big dataset, which doesn't solve my problem.
So my question is: What is the difference between my example df and the big dataset that throws the geoMean function off the rails?
r geometric-mean
asked Mar 27 at 9:46
Hedera.HelixHedera.Helix
62 bronze badges
It is difficult to say what the difference is between the example dataset and your other dataset, for obvious reasons. Since you are computing the geometric mean per group, have you checked that there is at least one observation for each measurement in each group? Are all measurements positive?
– dipetkov
Mar 27 at 10:16
Thanks you for your suggestion. Yes, I checked for that and all observations are positive, and there's at least one for each group.
– Hedera.Helix
Mar 27 at 10:22
Then perhaps you can try to find a subset of your big dataset that still gives the error but it is small enough to post here?
– dipetkov
Mar 27 at 10:26
I couldn't do that, which led me to the original mistake:gm_cut2trait1 <- aggregate(cut2trait1[, 13:19], list(cut2trait1$Sp), geoMean, na.rm=TRUE)uses the argumentna.rm=TRUEfor the aggregate function, not for the geoMean function. Because of that, geoMean defaulted tona.rm=FALSE.
– Hedera.Helix
Mar 27 at 11:09
I found the answer to my problem (using an argument for a function within aggregate) here
– Hedera.Helix
Mar 27 at 11:15
add a comment |
EDIT: The problem was not within the geoMean function, but with a wrong use of aggregate(), as explained in the comments
I am trying to calculate the geometric mean of multiple measurements for several different species, which includes NAs. An example of my data looks like this:
species <- c("Ae", "Ae", "Ae", "Be", "Be")
phen <- c(2, NA, 3, 1, 2)
hveg <- c(NA, 15, 12, 60, 59)
df <- data.frame(species, phen, hveg)
When I try to calculate the geometric mean for the species Ae with the built-in function geoMean from the package EnvStats like this
library("EnvStats")
aggregate(df[, 3:3], list(df1$Sp), geoMean, na.rm=TRUE)
it works wonderful and skips the NAs to give me the geometric means per species.
Group.1 phen hveg
1 Ae 4.238536 50.555696
2 Be 1.414214 1.414214
When I do this with my large dataset, however, the function stumbles over NAs and returns NA as result even though there are e.g 10 numerical values and only one NA. This happens for example with the column SLA_mm2/mg.
My large data set looks like this:
> str(cut2trait1)
Classes ‘tbl_df’, ‘tbl’ and 'data.frame': 22 obs. of 19 variables:
$ Cut : chr "15_08" "15_08" "15_08" "15_08" ...
$ Block : num 1 1 1 1 1 1 1 1 1 1 ...
$ ID : num 451 512 431 531 591 432 551 393 511 452 ...
$ Plot : chr "1_1" "1_1" "1_1" "1_1" ...
$ Grazing : chr "n" "n" "n" "n" ...
$ Acro : chr "Leuc.vulg" "Dact.glom" "Cirs.arve" "Trif.prat" ...
$ Sp : chr "Lv" "Dg" "Ca" "Tp" ...
$ Label_neu : chr "Lv021" "Dg022" "Ca021" "Tp021" ...
$ PlantFunctionalType: chr "forb" "grass" "forb" "forb" ...
$ PlotClimate : chr "AC" "AC" "AC" "AC" ...
$ Season : chr "Aug" "Aug" "Aug" "Aug" ...
$ Year : num 2015 2015 2015 2015 2015 ...
$ Tiller : num 6 3 3 5 6 8 5 2 1 7 ...
$ Hveg : num 25 38 70 36 68 65 23 58 71 27 ...
$ Hrep : num 39 54 77 38 76 70 65 88 98 38 ...
$ Phen : num 8 8 7 8 8 7 6.5 8 8 8 ...
$ SPAD : num 40.7 42.4 48.7 43 31.3 ...
$ TDW_in_g : num 4.62 4.85 11.86 5.82 8.99 ...
$ SLA_mm2/mg : num 19.6 19.8 20.3 21.2 21.7 ...
and the result of my code
gm_cut2trait1 <- aggregate(cut2trait1[, 13:19], list(cut2trait1$Sp), geoMean, na.rm=TRUE)
is (only the first two rows):
Group.1 Tiller Hveg Hrep Phen SPAD TDW_in_g SLA_mm2/mg
1 Ae 13.521721 73.43485 106.67933 NA 28.17698 1.2602475 NA
2 Be 8.944272 43.95452 72.31182 5.477226 20.08880 0.7266361 9.309672
Here, the geometric mean of SLA for Ae is NA, even though there are 9 numeric measurements and only one NA in the column used to calculate the geometric mean.
I tried to use the geometric mean function suggested here:
Geometric Mean: is there a built-in?
But instead of NAs, this returned the value 1.000 when used with my big dataset, which doesn't solve my problem.
So my question is: What is the difference between my example df and the big dataset that throws the geoMean function off the rails?
r geometric-mean
asked Mar 27 at 9:46
Hedera.HelixHedera.Helix
62 bronze badges
EDIT: The problem was not within the geoMean function, but with a wrong use of aggregate(), as explained in the comments
I am trying to calculate the geometric mean of multiple measurements for several different species, which includes NAs. An example of my data looks like this:
species <- c("Ae", "Ae", "Ae", "Be", "Be")
phen <- c(2, NA, 3, 1, 2)
hveg <- c(NA, 15, 12, 60, 59)
df <- data.frame(species, phen, hveg)
When I try to calculate the geometric mean for the species Ae with the built-in function geoMean from the package EnvStats like this
library("EnvStats")
aggregate(df[, 3:3], list(df1$Sp), geoMean, na.rm=TRUE)
it works wonderful and skips the NAs to give me the geometric means per species.
Group.1 phen hveg
1 Ae 4.238536 50.555696
2 Be 1.414214 1.414214
When I do this with my large dataset, however, the function stumbles over NAs and returns NA as result even though there are e.g 10 numerical values and only one NA. This happens for example with the column SLA_mm2/mg.
My large data set looks like this:
> str(cut2trait1)
Classes ‘tbl_df’, ‘tbl’ and 'data.frame': 22 obs. of 19 variables:
$ Cut : chr "15_08" "15_08" "15_08" "15_08" ...
$ Block : num 1 1 1 1 1 1 1 1 1 1 ...
$ ID : num 451 512 431 531 591 432 551 393 511 452 ...
$ Plot : chr "1_1" "1_1" "1_1" "1_1" ...
$ Grazing : chr "n" "n" "n" "n" ...
$ Acro : chr "Leuc.vulg" "Dact.glom" "Cirs.arve" "Trif.prat" ...
$ Sp : chr "Lv" "Dg" "Ca" "Tp" ...
$ Label_neu : chr "Lv021" "Dg022" "Ca021" "Tp021" ...
$ PlantFunctionalType: chr "forb" "grass" "forb" "forb" ...
$ PlotClimate : chr "AC" "AC" "AC" "AC" ...
$ Season : chr "Aug" "Aug" "Aug" "Aug" ...
$ Year : num 2015 2015 2015 2015 2015 ...
$ Tiller : num 6 3 3 5 6 8 5 2 1 7 ...
$ Hveg : num 25 38 70 36 68 65 23 58 71 27 ...
$ Hrep : num 39 54 77 38 76 70 65 88 98 38 ...
$ Phen : num 8 8 7 8 8 7 6.5 8 8 8 ...
$ SPAD : num 40.7 42.4 48.7 43 31.3 ...
$ TDW_in_g : num 4.62 4.85 11.86 5.82 8.99 ...
$ SLA_mm2/mg : num 19.6 19.8 20.3 21.2 21.7 ...
and the result of my code
gm_cut2trait1 <- aggregate(cut2trait1[, 13:19], list(cut2trait1$Sp), geoMean, na.rm=TRUE)
is (only the first two rows):
Group.1 Tiller Hveg Hrep Phen SPAD TDW_in_g SLA_mm2/mg
1 Ae 13.521721 73.43485 106.67933 NA 28.17698 1.2602475 NA
2 Be 8.944272 43.95452 72.31182 5.477226 20.08880 0.7266361 9.309672
Here, the geometric mean of SLA for Ae is NA, even though there are 9 numeric measurements and only one NA in the column used to calculate the geometric mean.
I tried to use the geometric mean function suggested here:
Geometric Mean: is there a built-in?
But instead of NAs, this returned the value 1.000 when used with my big dataset, which doesn't solve my problem.
So my question is: What is the difference between my example df and the big dataset that throws the geoMean function off the rails?
r geometric-mean
r geometric-mean
asked Mar 27 at 9:46
Hedera.HelixHedera.Helix
62 bronze badges
asked Mar 27 at 9:46
Hedera.HelixHedera.Helix
62 bronze badges
edited Mar 27 at 11:21
Hedera.Helix
asked Mar 27 at 9:46
Hedera.HelixHedera.Helix
62 bronze badges
asked Mar 27 at 9:46
Hedera.HelixHedera.Helix
62 bronze badges
asked Mar 27 at 9:46
Hedera.HelixHedera.Helix
62 bronze badges
62 bronze badges
It is difficult to say what the difference is between the example dataset and your other dataset, for obvious reasons. Since you are computing the geometric mean per group, have you checked that there is at least one observation for each measurement in each group? Are all measurements positive?
– dipetkov
Mar 27 at 10:16
Thanks you for your suggestion. Yes, I checked for that and all observations are positive, and there's at least one for each group.
– Hedera.Helix
Mar 27 at 10:22
Then perhaps you can try to find a subset of your big dataset that still gives the error but it is small enough to post here?
– dipetkov
Mar 27 at 10:26
I couldn't do that, which led me to the original mistake:gm_cut2trait1 <- aggregate(cut2trait1[, 13:19], list(cut2trait1$Sp), geoMean, na.rm=TRUE)uses the argumentna.rm=TRUEfor the aggregate function, not for the geoMean function. Because of that, geoMean defaulted tona.rm=FALSE.
– Hedera.Helix
Mar 27 at 11:09
I found the answer to my problem (using an argument for a function within aggregate) here
– Hedera.Helix
Mar 27 at 11:15
add a comment |
It is difficult to say what the difference is between the example dataset and your other dataset, for obvious reasons. Since you are computing the geometric mean per group, have you checked that there is at least one observation for each measurement in each group? Are all measurements positive?
– dipetkov
Mar 27 at 10:16
Thanks you for your suggestion. Yes, I checked for that and all observations are positive, and there's at least one for each group.
– Hedera.Helix
Mar 27 at 10:22
Then perhaps you can try to find a subset of your big dataset that still gives the error but it is small enough to post here?
– dipetkov
Mar 27 at 10:26
I couldn't do that, which led me to the original mistake:gm_cut2trait1 <- aggregate(cut2trait1[, 13:19], list(cut2trait1$Sp), geoMean, na.rm=TRUE)uses the argumentna.rm=TRUEfor the aggregate function, not for the geoMean function. Because of that, geoMean defaulted tona.rm=FALSE.
– Hedera.Helix
Mar 27 at 11:09
I found the answer to my problem (using an argument for a function within aggregate) here
– Hedera.Helix
Mar 27 at 11:15
It is difficult to say what the difference is between the example dataset and your other dataset, for obvious reasons. Since you are computing the geometric mean per group, have you checked that there is at least one observation for each measurement in each group? Are all measurements positive?
– dipetkov
Mar 27 at 10:16
It is difficult to say what the difference is between the example dataset and your other dataset, for obvious reasons. Since you are computing the geometric mean per group, have you checked that there is at least one observation for each measurement in each group? Are all measurements positive?
– dipetkov
Mar 27 at 10:16
Thanks you for your suggestion. Yes, I checked for that and all observations are positive, and there's at least one for each group.
– Hedera.Helix
Mar 27 at 10:22
Thanks you for your suggestion. Yes, I checked for that and all observations are positive, and there's at least one for each group.
– Hedera.Helix
Mar 27 at 10:22
Then perhaps you can try to find a subset of your big dataset that still gives the error but it is small enough to post here?
– dipetkov
Mar 27 at 10:26
Then perhaps you can try to find a subset of your big dataset that still gives the error but it is small enough to post here?
– dipetkov
Mar 27 at 10:26
I couldn't do that, which led me to the original mistake:
gm_cut2trait1 <- aggregate(cut2trait1[, 13:19], list(cut2trait1$Sp), geoMean, na.rm=TRUE) uses the argument na.rm=TRUE for the aggregate function, not for the geoMean function. Because of that, geoMean defaulted to na.rm=FALSE.– Hedera.Helix
Mar 27 at 11:09
I couldn't do that, which led me to the original mistake:
gm_cut2trait1 <- aggregate(cut2trait1[, 13:19], list(cut2trait1$Sp), geoMean, na.rm=TRUE) uses the argument na.rm=TRUE for the aggregate function, not for the geoMean function. Because of that, geoMean defaulted to na.rm=FALSE.– Hedera.Helix
Mar 27 at 11:09
I found the answer to my problem (using an argument for a function within aggregate) here
– Hedera.Helix
Mar 27 at 11:15
I found the answer to my problem (using an argument for a function within aggregate) here
– Hedera.Helix
Mar 27 at 11:15
add a comment |
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It is difficult to say what the difference is between the example dataset and your other dataset, for obvious reasons. Since you are computing the geometric mean per group, have you checked that there is at least one observation for each measurement in each group? Are all measurements positive?
– dipetkov
Mar 27 at 10:16
Thanks you for your suggestion. Yes, I checked for that and all observations are positive, and there's at least one for each group.
– Hedera.Helix
Mar 27 at 10:22
Then perhaps you can try to find a subset of your big dataset that still gives the error but it is small enough to post here?
– dipetkov
Mar 27 at 10:26
I couldn't do that, which led me to the original mistake:
gm_cut2trait1 <- aggregate(cut2trait1[, 13:19], list(cut2trait1$Sp), geoMean, na.rm=TRUE)uses the argumentna.rm=TRUEfor the aggregate function, not for the geoMean function. Because of that, geoMean defaulted tona.rm=FALSE.– Hedera.Helix
Mar 27 at 11:09
I found the answer to my problem (using an argument for a function within aggregate) here
– Hedera.Helix
Mar 27 at 11:15