As an R beginner, I have found it surprisingly difficult to figure out how to compute descriptive statistics on multiply imputed data (more so than running some of the other basic analyses, such as correlations and regressions).
These types of questions are prefaced with apologies (Descriptive statistics (Means, StdDevs) using multiply imputed data: R) but have have not been answered (https://stats.stackexchange.com/questions/296193/pooling-basic-descriptives-from-several-multiply-imputed-datasets-using-mice) or are quickly cast a down vote.
Here is a description of a miceadds function(https://www.rdocumentation.org/packages/miceadds/versions/2.10-14/topics/stats0), which I find difficult to follow with data that has been stored in the mids format.
I have gotten some output such as mean, median, min, max using the summary(complete(imp)) but would love to know how to get additional summary output (e.g., skew/kurtosis, standard deviation, variance).
Illustration borrowed from a previous poster above:
> imp <- mice(nhanes, seed = 23109)
iter imp variable
1 1 bmi hyp chl
1 2 bmi hyp chl
1 3 bmi hyp chl
1 4 bmi hyp chl
1 5 bmi hyp chl
2 1 bmi hyp chl
2 2 bmi hyp chl
2 3 bmi hyp chl
> summary(complete(imp))
age bmi hyp chl
1:12 Min. :20.40 1:18 Min. :113
2: 7 1st Qu.:24.90 2: 7 1st Qu.:186
3: 6 Median :27.40 Median :199
Mean :27.37 Mean :194
3rd Qu.:30.10 3rd Qu.:218
Max. :35.30 Max. :284
Would someone kindly take the time to illustrate how one might take the mids object to get the basic descriptives?
Below are some steps you can do to better understand what happens with R objects after each step. I would also recommend to look at this tutorial:
https://gerkovink.github.io/miceVignettes/
library(mice)
# nhanes object is just a simple dataframe:
data(nhanes)
str(nhanes)
#'data.frame': 25 obs. of 4 variables:
# $ age: num 1 2 1 3 1 3 1 1 2 2 ...
#$ bmi: num NA 22.7 NA NA 20.4 NA 22.5 30.1 22 NA ...
#$ hyp: num NA 1 1 NA 1 NA 1 1 1 NA ...
#$ chl: num NA 187 187 NA 113 184 118 187 238 NA ...
# you can generate multivariate imputation using mice() function
imp <- mice(nhanes, seed=23109)
#The output variable is an object of class "mids" which you can explore using str() function
str(imp)
# List of 17
# $ call : language mice(data = nhanes)
# $ data :'data.frame': 25 obs. of 4 variables:
# ..$ age: num [1:25] 1 2 1 3 1 3 1 1 2 2 ...
# ..$ bmi: num [1:25] NA 22.7 NA NA 20.4 NA 22.5 30.1 22 NA ...
# ..$ hyp: num [1:25] NA 1 1 NA 1 NA 1 1 1 NA ...
# ..$ chl: num [1:25] NA 187 187 NA 113 184 118 187 238 NA ...
# $ m : num 5
# ...
# $ imp :List of 4
#..$ age: NULL
#..$ bmi:'data.frame': 9 obs. of 5 variables:
#.. ..$ 1: num [1:9] 28.7 30.1 22.7 24.9 30.1 35.3 27.5 29.6 33.2
#.. ..$ 2: num [1:9] 27.2 30.1 27.2 25.5 29.6 26.3 26.3 30.1 30.1
#.. ..$ 3: num [1:9] 22.5 30.1 20.4 22.5 27.4 22 26.3 27.4 35.3
#.. ..$ 4: num [1:9] 27.2 22 22.7 21.7 25.5 27.2 24.9 30.1 22
#.. ..$ 5: num [1:9] 28.7 28.7 20.4 21.7 25.5 22.5 22.5 25.5 22.7
#...
#You can extract individual components of this object using $, for example
#To view the actual imputation for bmi column
imp$imp$bmi
# 1 2 3 4 5
# 1 28.7 27.2 22.5 27.2 28.7
# 3 30.1 30.1 30.1 22.0 28.7
# 4 22.7 27.2 20.4 22.7 20.4
# 6 24.9 25.5 22.5 21.7 21.7
# 10 30.1 29.6 27.4 25.5 25.5
# 11 35.3 26.3 22.0 27.2 22.5
# 12 27.5 26.3 26.3 24.9 22.5
# 16 29.6 30.1 27.4 30.1 25.5
# 21 33.2 30.1 35.3 22.0 22.7
# The above output is again just a regular dataframe:
str(imp$imp$bmi)
# 'data.frame': 9 obs. of 5 variables:
# $ 1: num 28.7 30.1 22.7 24.9 30.1 35.3 27.5 29.6 33.2
# $ 2: num 27.2 30.1 27.2 25.5 29.6 26.3 26.3 30.1 30.1
# $ 3: num 22.5 30.1 20.4 22.5 27.4 22 26.3 27.4 35.3
# $ 4: num 27.2 22 22.7 21.7 25.5 27.2 24.9 30.1 22
# $ 5: num 28.7 28.7 20.4 21.7 25.5 22.5 22.5 25.5 22.7
# complete() function returns imputed dataset:
mat <- complete(imp)
# The output of this function is a regular data frame:
str(mat)
# 'data.frame': 25 obs. of 4 variables:
# $ age: num 1 2 1 3 1 3 1 1 2 2 ...
# $ bmi: num 28.7 22.7 30.1 22.7 20.4 24.9 22.5 30.1 22 30.1 ...
# $ hyp: num 1 1 1 2 1 2 1 1 1 1 ...
# $ chl: num 199 187 187 204 113 184 118 187 238 229 ...
# So you can run any descriptive statistics you need with this object
# Just like you would do with a regular dataframe:
> summary(mat)
# age bmi hyp chl
# Min. :1.00 Min. :20.40 Min. :1.00 Min. :113.0
# 1st Qu.:1.00 1st Qu.:24.90 1st Qu.:1.00 1st Qu.:187.0
# Median :2.00 Median :27.50 Median :1.00 Median :204.0
# Mean :1.76 Mean :27.48 Mean :1.24 Mean :204.9
# 3rd Qu.:2.00 3rd Qu.:30.10 3rd Qu.:1.00 3rd Qu.:229.0
# Max. :3.00 Max. :35.30 Max. :2.00 Max. :284.0
There are several mistakes in both your code and the answer from Katia and the link provided by Katia is no longer available.
To compute simple statistics after multiple imputation, you must follow Rubin's Rule, which is the method used in mice for a selected bunch of model fits.
When using
library(mice)
imp <- mice(nhanes, seed = 23109)
mat <- complete(imp)
mat
age bmi hyp chl
1 1 28.7 1 199
2 2 22.7 1 187
3 1 30.1 1 187
4 3 22.7 2 204
5 1 20.4 1 113
6 3 24.9 2 184
7 1 22.5 1 118
8 1 30.1 1 187
9 2 22.0 1 238
10 2 30.1 1 229
11 1 35.3 1 187
12 2 27.5 1 229
13 3 21.7 1 206
14 2 28.7 2 204
15 1 29.6 1 238
16 1 29.6 1 238
17 3 27.2 2 284
18 2 26.3 2 199
19 1 35.3 1 218
20 3 25.5 2 206
21 1 33.2 1 238
22 1 33.2 1 229
23 1 27.5 1 131
24 3 24.9 1 284
25 2 27.4 1 186
You only return the first imputed dataset, whereas you imputed five by default. See ?mice::complete for more informations "The default is action = 1L returns the first imputed data set."
To get the five imputed datasets, you have to specify the action argument of mice::complete
mat2 <- complete(imp, "long")
mat2
.imp .id age bmi hyp chl
1 1 1 1 28.7 1 199
2 1 2 2 22.7 1 187
3 1 3 1 30.1 1 187
4 1 4 3 22.7 2 204
5 1 5 1 20.4 1 113
6 1 6 3 24.9 2 184
7 1 7 1 22.5 1 118
8 1 8 1 30.1 1 187
9 1 9 2 22.0 1 238
10 1 10 2 30.1 1 229
...
115 5 15 1 29.6 1 187
116 5 16 1 25.5 1 187
117 5 17 3 27.2 2 284
118 5 18 2 26.3 2 199
119 5 19 1 35.3 1 218
120 5 20 3 25.5 2 218
121 5 21 1 22.7 1 238
122 5 22 1 33.2 1 229
123 5 23 1 27.5 1 131
124 5 24 3 24.9 1 186
125 5 25 2 27.4 1 186
Both summary(mat) and summary(mat2) are false.
Let's focus on bmi. The first one provides the mean bmi over the first imputed dataset. The second one provides the mean of an artifical m times larger dataset. The second dataset also has inappropriately low variance.
mean(mat$bmi)
27.484
mean(mat2$bmi)
26.5192
I have not found a better solution than applying manually Rubin's rule to the mean estimate. The correct estimate is simply the mean of estimates accross all imputed datasets
res <- with(imp, mean(bmi)) #get the mean for each imputed dataset, stored in res$analyses
do.call(sum, res$analyses) / 5 #compute mean over m = 5 mean estimations
26.5192
The variance / standard deviation has to be calculated appropriately. You can use Rubin's rule to compute any simple statistic that you wish. You can find the way of doing so here https://bookdown.org/mwheymans/bookmi/rubins-rules.html
Hope this helps.
Related
I am trying to build a decison tree model in R having both categorical and numerical variables.Some categorical variables have 3 levels , so can I just use as.factor and then use in my model? I tried to use model.matrix but my doubt is model.matrix converts the variable in numeric values of 0s and 1s and splitting happens on basis of these numeric values. For eg if Color has 3 level- blue,red,green, the splitting rule will look like color_green < 0.5 instead it should always take 0s and 1s only.
If you are asking whether you can use factors to build an rpart decision tree. Then yes. See below example from the documentation. Note that there are a lot of possible packages for decision trees.
library(rpart)
rpart(Reliability ~ ., data=car90)
#> n=76 (35 observations deleted due to missingness)
#>
#> node), split, n, loss, yval, (yprob)
#> * denotes terminal node
#>
#> 1) root 76 53 average (0.2 0.12 0.3 0.11 0.28)
#> 2) Country=Germany,Korea,Mexico,Sweden,USA 49 29 average (0.31 0.18 0.41 0.1 0)
#> 4) Tires=145,155/80,165/80,185/80,195/60,195/65,195/70,205/60,215/65,225/75,275/40 17 9 Much worse (0.47 0.29 0 0.24 0) *
#> 5) Tires=175/70,185/65,185/70,185/75,195/75,205/70,205/75,215/70 32 12 average (0.22 0.12 0.62 0.031 0)
#> 10) HP.revs< 4650 13 7 Much worse (0.46 0.23 0.31 0 0) *
#> 11) HP.revs>=4650 19 3 average (0.053 0.053 0.84 0.053 0) *
#> 3) Country=Japan,Japan/USA 27 6 Much better (0 0 0.11 0.11 0.78) *
str(car90)
#> 'data.frame': 111 obs. of 34 variables:
#> $ Country : Factor w/ 10 levels "Brazil","England",..: 5 5 4 4 4 4 10 10 10 NA ...
#> $ Disp : num 112 163 141 121 152 209 151 231 231 189 ...
#> $ Disp2 : num 1.8 2.7 2.3 2 2.5 3.5 2.5 3.8 3.8 3.1 ...
#> $ Eng.Rev : num 2935 2505 2775 2835 2625 ...
#> $ Front.Hd : num 3.5 2 2.5 4 2 3 4 6 5 5.5 ...
#> $ Frt.Leg.Room: num 41.5 41.5 41.5 42 42 42 42 42 41 41 ...
#> $ Frt.Shld : num 53 55.5 56.5 52.5 52 54.5 56.5 58.5 59 58 ...
#> $ Gear.Ratio : num 3.26 2.95 3.27 3.25 3.02 2.8 NA NA NA NA ...
#> $ Gear2 : num 3.21 3.02 3.25 3.25 2.99 2.85 2.84 1.99 1.99 2.33 ...
#> $ HP : num 130 160 130 108 168 208 110 165 165 101 ...
#> $ HP.revs : num 6000 5900 5500 5300 5800 5700 5200 4800 4800 4400 ...
#> $ Height : num 47.5 50 51.5 50.5 49.5 51 49.5 50.5 51 50.5 ...
#> $ Length : num 177 191 193 176 175 186 189 197 197 192 ...
#> $ Luggage : num 16 14 17 10 12 12 16 16 16 15 ...
#> $ Mileage : num NA 20 NA 27 NA NA 21 NA 23 NA ...
#> $ Model2 : Factor w/ 21 levels ""," Turbo 4 (3)",..: 1 1 1 1 1 1 1 14 13 1 ...
#> $ Price : num 11950 24760 26900 18900 24650 ...
#> $ Rear.Hd : num 1.5 2 3 1 1 2.5 2.5 4.5 3.5 3.5 ...
#> $ Rear.Seating: num 26.5 28.5 31 28 25.5 27 28 30.5 28.5 27.5 ...
#> $ RearShld : num 52 55.5 55 52 51.5 55.5 56 58.5 58.5 56.5 ...
#> $ Reliability : Ord.factor w/ 5 levels "Much worse"<"worse"<..: 5 5 NA NA 4 NA 3 3 3 NA ...
#> $ Rim : Factor w/ 6 levels "R12","R13","R14",..: 3 4 4 3 3 4 3 3 3 3 ...
#> $ Sratio.m : num NA NA NA NA NA NA NA NA NA NA ...
#> $ Sratio.p : num 0.86 0.96 0.97 0.71 0.88 0.78 0.76 0.83 0.87 0.88 ...
#> $ Steering : Factor w/ 3 levels "manual","power",..: 2 2 2 2 2 2 2 2 2 2 ...
#> $ Tank : num 13.2 18 21.1 15.9 16.4 21.1 15.7 18 18 16.5 ...
#> $ Tires : Factor w/ 30 levels "145","145/80",..: 16 20 20 8 17 28 13 23 23 22 ...
#> $ Trans1 : Factor w/ 4 levels "","man.4","man.5",..: 3 3 3 3 3 3 1 1 1 1 ...
#> $ Trans2 : Factor w/ 4 levels "","auto.3","auto.4",..: 3 3 2 2 3 3 2 3 3 3 ...
#> $ Turning : num 37 42 39 35 35 39 41 43 42 41 ...
#> $ Type : Factor w/ 6 levels "Compact","Large",..: 4 3 3 1 1 3 3 2 2 NA ...
#> $ Weight : num 2700 3265 2935 2670 2895 ...
#> $ Wheel.base : num 102 109 106 100 101 109 105 111 111 108 ...
#> $ Width : num 67 69 71 67 65 69 69 72 72 71 ...
I was trying to forecast a time series problem using lm() and my data looks like below
Customer_key date sales
A35 2018-05-13 31
A35 2018-05-20 20
A35 2018-05-27 43
A35 2018-06-03 31
BH22 2018-05-13 60
BH22 2018-05-20 67
BH22 2018-05-27 78
BH22 2018-06-03 55
Converted my df to a list format by
df <- dcast(df, date ~ customer_key,value.var = c("sales"))
df <- subset(df, select = -c(dt))
demandWithKey <- as.list(df)
Trying to write a function such that applying this function across all customers
my_fun <- function(x) {
fit <- lm(ds_load ~ date, data=df) ## After changing to list ds_load and date column names
## are no longer available for formula
fit_b <- forecast(fit$fitted.values, h=20) ## forecast using lm()
return(data.frame(c(fit$fitted.values, fit_b[["mean"]])))
}
fcast <- lapply(df, my_fun)
I know the above function doesn't work, but basically I'm looking for getting both the fitted values and forecasted values for a grouped data.
But I've tried all other methods using tslm() (converting into time series data) and so on but no luck I can get the lm() work somehow on just one customer though. Also many questions/posts were on just fitting the model but I would like to forecast too at same time.
lm() is for a regression model
but here you have a time serie so for forecasting the serie you have to use one of the time serie model (ARMA ARCH GARCH...)
so you can use the function in r : auto.arima() in "forecast" package
I don't know what you're up to exactly, but you could make this less complicated.
Using by avoids the need to reshape your data, it splits your data e.g. by customer ID as in your case and applies a function on the subsets (i.e. it's a combination of split and lapply; see ?by).
Since you want to compare fitted and forecasted values somehow in your result, you probably need predict rather than $fitted.values, otherwise the values won't be of same length. Because your independent variable is a date in weekly intervals, you may use seq.Date and take the first date as a starting value; the sequence has length actual values (nrow each customer) plus h= argument of the forecast.
For demonstration purposes I add the fitted values as first column in the following.
res <- by(dat, dat$cus_key, function(x) {
H <- 20 ## globally define 'h'
fit <- lm(sales ~ date, x)
fitted <- fit$fitted.values
pred <- predict(fit, newdata=data.frame(
date=seq(x$date[1], length.out= nrow(x) + H, by="week")))
fcst <- c(fitted, forecast(fitted, h=H)$mean)
fit.na <- `length<-`(unname(fitted), length(pred)) ## for demonstration
return(cbind(fit.na, pred, fcst))
})
Result
res
# dat$cus_key: A28
# fit.na pred fcst
# 1 41.4 41.4 41.4
# 2 47.4 47.4 47.4
# 3 53.4 53.4 53.4
# 4 59.4 59.4 59.4
# 5 65.4 65.4 65.4
# 6 NA 71.4 71.4
# 7 NA 77.4 77.4
# 8 NA 83.4 83.4
# 9 NA 89.4 89.4
# 10 NA 95.4 95.4
# 11 NA 101.4 101.4
# 12 NA 107.4 107.4
# 13 NA 113.4 113.4
# 14 NA 119.4 119.4
# 15 NA 125.4 125.4
# 16 NA 131.4 131.4
# 17 NA 137.4 137.4
# 18 NA 143.4 143.4
# 19 NA 149.4 149.4
# 20 NA 155.4 155.4
# 21 NA 161.4 161.4
# 22 NA 167.4 167.4
# 23 NA 173.4 173.4
# 24 NA 179.4 179.4
# 25 NA 185.4 185.4
# ----------------------------------------------------------------
# dat$cus_key: B16
# fit.na pred fcst
# 1 49.0 49.0 49.0
# 2 47.7 47.7 47.7
# 3 46.4 46.4 46.4
# 4 45.1 45.1 45.1
# 5 43.8 43.8 43.8
# 6 NA 42.5 42.5
# 7 NA 41.2 41.2
# 8 NA 39.9 39.9
# 9 NA 38.6 38.6
# 10 NA 37.3 37.3
# 11 NA 36.0 36.0
# 12 NA 34.7 34.7
# 13 NA 33.4 33.4
# 14 NA 32.1 32.1
# 15 NA 30.8 30.8
# 16 NA 29.5 29.5
# 17 NA 28.2 28.2
# 18 NA 26.9 26.9
# 19 NA 25.6 25.6
# 20 NA 24.3 24.3
# 21 NA 23.0 23.0
# 22 NA 21.7 21.7
# 23 NA 20.4 20.4
# 24 NA 19.1 19.1
# 25 NA 17.8 17.8
# ----------------------------------------------------------------
# dat$cus_key: C12
# fit.na pred fcst
# 1 56.4 56.4 56.4
# 2 53.2 53.2 53.2
# 3 50.0 50.0 50.0
# 4 46.8 46.8 46.8
# 5 43.6 43.6 43.6
# 6 NA 40.4 40.4
# 7 NA 37.2 37.2
# 8 NA 34.0 34.0
# 9 NA 30.8 30.8
# 10 NA 27.6 27.6
# 11 NA 24.4 24.4
# 12 NA 21.2 21.2
# 13 NA 18.0 18.0
# 14 NA 14.8 14.8
# 15 NA 11.6 11.6
# 16 NA 8.4 8.4
# 17 NA 5.2 5.2
# 18 NA 2.0 2.0
# 19 NA -1.2 -1.2
# 20 NA -4.4 -4.4
# 21 NA -7.6 -7.6
# 22 NA -10.8 -10.8
# 23 NA -14.0 -14.0
# 24 NA -17.2 -17.2
# 25 NA -20.4 -20.4
As you can see, prediction and forecast yield the same values, since both methods are based on the same single explanatory variable date in this case.
Toy data:
set.seed(42)
dat <- transform(expand.grid(cus_key=paste0(LETTERS[1:3], sample(12:43, 3)),
date=seq.Date(as.Date("2018-05-13"), length.out=5, by="week")),
sales=sample(20:80, 15, replace=TRUE))
I have a function that takes in a dataframe, a percentile threshold, and the name of a given column, and computes all values that are above this threshold in the given column as a new column (0 for <, and 1 for >=). However, it won't allow me to do the df$column_name inside the quantile function because column_name is not actually a column name, but a variable storing the actual column name. Therefore df$column_name will return NULL. Is there any way to work around this and keep the code forma somewhat similar to what it is currently? Or do I have to specify the actual numerical column value instead of the name? While I can do this, it is definitely not as convenient/comprehensible as just passing in the column name.
func1 <- function(df, threshold, column_name) {
threshold_value <- quantile(df$column_name, c(threshold))
new_df <- df %>%
mutate(ifelse(column_name > threshold_value, 1, 0))
return(new_df)
}
Thank you so much for your help!
I modified your function as follows. Now the function can take a data frame, a threshold, and a column name. This function only needs the base R.
# Modified function
func1 <- function(df, threshold, column_name) {
threshold_value <- quantile(df[[column_name]], threshold)
new_df <- df
new_df[["new_col"]] <- ifelse(df[[column_name]] > threshold_value, 1, 0)
return(new_df)
}
# Take the trees data frame as an example
head(trees)
# Girth Height Volume
# 1 8.3 70 10.3
# 2 8.6 65 10.3
# 3 8.8 63 10.2
# 4 10.5 72 16.4
# 5 10.7 81 18.8
# 6 10.8 83 19.7
# Apply the function
func1(trees, 0.5, "Volume")
# Girth Height Volume new_col
# 1 8.3 70 10.3 0
# 2 8.6 65 10.3 0
# 3 8.8 63 10.2 0
# 4 10.5 72 16.4 0
# 5 10.7 81 18.8 0
# 6 10.8 83 19.7 0
# 7 11.0 66 15.6 0
# 8 11.0 75 18.2 0
# 9 11.1 80 22.6 0
# 10 11.2 75 19.9 0
# 11 11.3 79 24.2 0
# 12 11.4 76 21.0 0
# 13 11.4 76 21.4 0
# 14 11.7 69 21.3 0
# 15 12.0 75 19.1 0
# 16 12.9 74 22.2 0
# 17 12.9 85 33.8 1
# 18 13.3 86 27.4 1
# 19 13.7 71 25.7 1
# 20 13.8 64 24.9 1
# 21 14.0 78 34.5 1
# 22 14.2 80 31.7 1
# 23 14.5 74 36.3 1
# 24 16.0 72 38.3 1
# 25 16.3 77 42.6 1
# 26 17.3 81 55.4 1
# 27 17.5 82 55.7 1
# 28 17.9 80 58.3 1
# 29 18.0 80 51.5 1
# 30 18.0 80 51.0 1
# 31 20.6 87 77.0 1
If you still want to use dplyr, it is essential to learn how to deal with non-standard evaluation. Please see this to learn more (https://cran.r-project.org/web/packages/dplyr/vignettes/programming.html). The following code will also works.
library(dplyr)
func2 <- function(df, threshold, column_name) {
col_en <- enquo(column_name)
threshold_value <- quantile(df %>% pull(!!col_en), threshold)
new_df <- df %>%
mutate(new_col := ifelse(!!col_en >= threshold_value, 1, 0))
return(new_df)
}
func2(trees, 0.5, Volume)
# Girth Height Volume new_col
# 1 8.3 70 10.3 0
# 2 8.6 65 10.3 0
# 3 8.8 63 10.2 0
# 4 10.5 72 16.4 0
# 5 10.7 81 18.8 0
# 6 10.8 83 19.7 0
# 7 11.0 66 15.6 0
# 8 11.0 75 18.2 0
# 9 11.1 80 22.6 0
# 10 11.2 75 19.9 0
# 11 11.3 79 24.2 1
# 12 11.4 76 21.0 0
# 13 11.4 76 21.4 0
# 14 11.7 69 21.3 0
# 15 12.0 75 19.1 0
# 16 12.9 74 22.2 0
# 17 12.9 85 33.8 1
# 18 13.3 86 27.4 1
# 19 13.7 71 25.7 1
# 20 13.8 64 24.9 1
# 21 14.0 78 34.5 1
# 22 14.2 80 31.7 1
# 23 14.5 74 36.3 1
# 24 16.0 72 38.3 1
# 25 16.3 77 42.6 1
# 26 17.3 81 55.4 1
# 27 17.5 82 55.7 1
# 28 17.9 80 58.3 1
# 29 18.0 80 51.5 1
# 30 18.0 80 51.0 1
# 31 20.6 87 77.0 1
I try to make KDA (Kernel discriminant analysis) for carc data, but when I call command X<-data.frame(scale(X)); r shows error:
"Error in colMeans(x, na.rm = TRUE) : 'x' must be numeric"
I tried to use as.numeric(as.matrix(carc)) and carc<-na.omit(carc), but it does not help either
library(ks);library(MASS);library(klaR);library(FSelector)
install.packages("klaR")
install.packages("FSelector")
library(ks);library(MASS);library(klaR);library(FSelector)
attach("carc.rda")
data<-load("carc.rda")
data
carc<-na.omit(carc)
head(carc)
class(carc) # check for its class
class(as.matrix(carc)) # change class, and
as.numeric(as.matrix(carc))
XX<-carc
X<-XX[,1:12];X.class<-XX[,13];
X<-data.frame(scale(X));
fit.pc<-princomp(X,scores=TRUE);
plot(fit.pc,type="line")
X.new<-fit.pc$scores[,1:5]; X.new<-data.frame(X.new);
cfs(X.class~.,cbind(X.new,X.class))
X.new<-fit.pc$scores[,c(1,4)]; X.new<-data.frame(X.new);
fit.kda1<-Hkda(x=X.new,x.group=X.class,pilot="samse",
bw="plugin",pre="sphere")
kda.fit1 <- kda(x=X.new, x.group=X.class, Hs=fit.kda1)
Can you help to resolve this problem and make this analysis?
Added:The car data set( Chambers, kleveland, Kleiner & Tukey 1983)
> head(carc)
P M R78 R77 H R Tr W L T D G C
AMC_Concord 4099 22 3 2 2.5 27.5 11 2930 186 40 121 3.58 US
AMC_Pacer 4749 17 3 1 3.0 25.5 11 3350 173 40 258 2.53 US
AMC_Spirit 3799 22 . . 3.0 18.5 12 2640 168 35 121 3.08 US
Audi_5000 9690 17 5 2 3.0 27.0 15 2830 189 37 131 3.20 Europe
Audi_Fox 6295 23 3 3 2.5 28.0 11 2070 174 36 97 3.70 Europe
Here is a small dataset with similar characteristics to what you describe
in order to answer this error:
"Error in colMeans(x, na.rm = TRUE) : 'x' must be numeric"
carc <- data.frame(type1=rep(c('1','2'), each=5),
type2=rep(c('5','6'), each=5),
x = rnorm(10,1,2)/10, y = rnorm(10))
This should be similar to your data.frame
str(carc)
# 'data.frame': 10 obs. of 3 variables:
# $ type1: Factor w/ 2 levels "1","2": 1 1 1 1 1 2 2 2 2 2
# $ type2: Factor w/ 2 levels "5","6": 1 1 1 1 1 2 2 2 2 2
# $ x : num -0.1177 0.3443 0.1351 0.0443 0.4702 ...
# $ y : num -0.355 0.149 -0.208 -1.202 -1.495 ...
scale(carc)
# Similar error
# Error in colMeans(x, na.rm = TRUE) : 'x' must be numeric
Using set()
require(data.table)
DT <- data.table(carc)
cols_fix <- c("type1", "type2")
for (col in cols_fix) set(DT, j=col, value = as.numeric(as.character(DT[[col]])))
str(DT)
# Classes ‘data.table’ and 'data.frame': 10 obs. of 4 variables:
# $ type1: num 1 1 1 1 1 2 2 2 2 2
# $ type2: num 5 5 5 5 5 6 6 6 6 6
# $ x : num 0.0465 0.1712 0.1582 0.1684 0.1183 ...
# $ y : num 0.155 -0.977 -0.291 -0.766 -1.02 ...
# - attr(*, ".internal.selfref")=<externalptr>
The first column(s) of your data set may be factors. Taking the data from corrgram:
library(corrgram)
carc <- auto
str(carc)
# 'data.frame': 74 obs. of 14 variables:
# $ Model : Factor w/ 74 levels "AMC Concord ",..: 1 2 3 4 5 6 7 8 9 10 ...
# $ Origin: Factor w/ 3 levels "A","E","J": 1 1 1 2 2 2 1 1 1 1 ...
# $ Price : int 4099 4749 3799 9690 6295 9735 4816 7827 5788 4453 ...
# $ MPG : int 22 17 22 17 23 25 20 15 18 26 ...
# $ Rep78 : num 3 3 NA 5 3 4 3 4 3 NA ...
# $ Rep77 : num 2 1 NA 2 3 4 3 4 4 NA ...
# $ Hroom : num 2.5 3 3 3 2.5 2.5 4.5 4 4 3 ...
# $ Rseat : num 27.5 25.5 18.5 27 28 26 29 31.5 30.5 24 ...
# $ Trunk : int 11 11 12 15 11 12 16 20 21 10 ...
# $ Weight: int 2930 3350 2640 2830 2070 2650 3250 4080 3670 2230 ...
# $ Length: int 186 173 168 189 174 177 196 222 218 170 ...
# $ Turn : int 40 40 35 37 36 34 40 43 43 34 ...
# $ Displa: int 121 258 121 131 97 121 196 350 231 304 ...
# $ Gratio: num 3.58 2.53 3.08 3.2 3.7 3.64 2.93 2.41 2.73 2.87 ...
So exclude them by trying this:
X<-XX[,3:14]
or this
X<-XX[,-(1:2)]
I have data collected for a few subjects, every 15 seconds over an hour split up by periods. Here's how the dataframe looks like, the time is "Temps", subjects are "Sujet" and the periods are determined by "Palier".
data.frame': 2853 obs. of 22 variables:
$ Temps : Factor w/ 217 levels "00:15","00:30",..: 1 2 3 4 5 6 7 8 9 10 ...
$ Sujet : int 1 1 1 1 1 1 1 1 1 1 ...
$ Test : Factor w/ 3 levels "VO2max","Tlim",..: 1 1 1 1 1 1 1 1 1 1 ...
$ Palier : int 1 1 1 1 1 1 1 1 1 1 ...
$ RPE : int 8 8 8 8 8 8 8 8 8 8 ...
$ Rmec : num 39.1 27.5 23.3 21.5 20.3 21.7 20.5 20.7 20.2 20.1 ...
Here a glimpse of the data.frame:
Temps Sujet Test Palier RPE Rmec Pmec Pchim Fr Vt VE FeO2 FeCO2 VO2 VCO2 RER HR VO2rel VE.VO2 VE.VCO2
1 00:15 1 VO2max 1 8 39.1 185 473.6 19 1854 34.60 16.24 4.48 1353 1268 0.94 121 17.6 0.02557280 0.02728707
2 00:30 1 VO2max 1 8 27.5 185 672.4 17 2602 44.30 15.77 4.78 1921 1731 0.90 124 25.0 0.02306091 0.02559214
3 00:45 1 VO2max 1 8 23.3 185 794.5 18 2793 50.83 15.63 4.85 2270 2015 0.89 131 29.6 0.02239207 0.02522581
4 01:00 1 VO2max 1 8 21.5 185 860.3 20 2756 55.76 15.68 4.88 2458 2224 0.90 137 32.0 0.02268511 0.02507194
5 01:15 1 VO2max 1 8 20.3 185 909.3 23 2709 61.26 15.84 4.88 2598 2446 0.94 139 33.8 0.02357968 0.02504497
6 01:30 1 VO2max 1 8 21.7 185 853.7 21 2899 59.85 16.00 4.89 2439 2395 0.98 140 31.8 0.02453875 0.02498956
Each "Palier" lasts about 5 min and there are from 5 to 10 "Palier". For each subject and "Palier", I need to compute the mean for the last 2 min for all the variables. I haven't figured it out yet with dcast() or ddply(), but I am a newbie!
Any help would be much appreciated!
If you turned it into a data.table (which you'd have to install), you could do this with
library(data.table)
dt = as.data.table(d) # assuming your existing data frame was called d
last.two.min = dt[, mean(tail(Rmec, 9)), by=Sujet]
This assumes that your original data frame was called d, and that you want the last 9 items (since it is every 15 seconds- you might want the last 8 if you want from 58:15 to 60:00).
I assumed that Rmec was the variable you wanted to get the mean for. If there are multiple for which you want to get the mean, you can do something like:
last.two.min = dt[, list(mean.Rmec=mean(tail(Rmec, 9)),
mean.RPE=mean(tail(RPE, 9))), by=Sujet]