Access frequencies of an atomic vector in a tibble data frame - r

I am doing Exploratory Data Analysis on a tibble data frame. I've never used tibble so I'm experiecing some difficulties.
My tibble data frame has this structure:
spec_tbl_df [7,397 x 19] (S3: spec_tbl_df/tbl_df/tbl/data.frame)
$ X1 : num [1:7397] 9617 12179 9905 5745 10067 ...
$ Administrative : num [1:7397] 5 26 4 3 7 16 4 3 2 0 ...
$ Administrative_Duration: num [1:7397] 408 1562 58 103 165 ...
$ Informational : num [1:7397] 2 9 2 0 1 3 4 5 0 0 ...
$ Informational_Duration : num [1:7397] 47.5 503.7 28.5 0 28.5 ...
$ ProductRelated : num [1:7397] 54 183 82 25 115 86 75 23 27 33 ...
$ ProductRelated_Duration: num [1:7397] 1547 9676 4729 1109 3428 ...
$ BounceRates : num [1:7397] 0 0.0111 0 0 0 ...
$ ExitRates : num [1:7397] 0.01733 0.0142 0.01454 0.00167 0.01629 ...
$ PageValues : num [1:7397] 0 19.57 9.06 61.3 4.97 ...
$ SpecialDay : num [1:7397] 0 0 0 0 0 0 0 0 0 0 ...
$ Month : Factor w/ 10 levels "Aug","Dec","Feb",..: 8 8 8 1 8 4 8 7 8 8 ...
$ OperatingSystems : Factor w/ 8 levels "1","2","3","4",..: 2 3 2 2 2 3 3 4 8 2 ...
$ Browser : Factor w/ 13 levels "1","2","3","4",..: 2 2 2 2 2 2 2 1 2 5 ...
$ Region : Factor w/ 9 levels "1","2","3","4",..: 3 2 1 6 4 8 1 1 7 3 ...
$ TrafficType : Factor w/ 19 levels "1","2","3","4",..: 2 12 2 5 10 4 2 4 2 1 ...
$ VisitorType : Factor w/ 3 levels "New_Visitor",..: 3 3 3 1 3 3 3 3 1 3 ...
$ Weekend : Factor w/ 2 levels "FALSE","TRUE": 2 1 1 1 1 1 1 1 1 1 ...
$ Revenue : Factor w/ 2 levels "FALSE","TRUE": 2 2 2 2 2 2 2 2 2 2 ...
Now if I use plot_bar to plot the cathegorical data (using DataExplorer package) I have no problem. I would like, for example, to create a boxplot for the cathegorical variable "Month" where for each month I have a boxplot showing how values are distribuited. The problem is that I can't find a way to access the frequencies. If I do the following:
boxplot(Month)
It creates a single boxplot for all the data (all the months) but it's not helpfull at all. Like this:
I would like the months on the x axis and the frequencies on the y axis and a boxplot for each month.
I've tried to "extract" the feature month, transform it to a matrix and repeat the process but it does not work.
Here is the variable montht taken alone:
> summary(x_Month)
Aug Dec Feb Jul June Mar May Nov Oct Sep
258 1034 123 259 166 1125 2014 1814 327 277
What am I missing ?


Something like this would probably work to create barplots for the frequencies of Month:
library(ggplot2)
spec_tbl_df %>%
ggplot(aes(x = Month)) +
geom_bar()

Related

How to combine training and testing dataset in same format

I am practicing with this dataset: http://archive.ics.uci.edu/ml/datasets/Census+Income
I loaded training & testing data.
# Downloading train and test data
trainFile = "adult.data"; testFile = "adult.test"
if (!file.exists (trainFile))
download.file (url = "http://archive.ics.uci.edu/ml/machine-learning-databases/adult/adult.data",
destfile = trainFile)
if (!file.exists (testFile))
download.file (url = "http://archive.ics.uci.edu/ml/machine-learning-databases/adult/adult.test",
destfile = testFile)
# Assigning column names
colNames = c ("age", "workclass", "fnlwgt", "education",
"educationnum", "maritalstatus", "occupation",
"relationship", "race", "sex", "capitalgain",
"capitalloss", "hoursperweek", "nativecountry",
"incomelevel")
# Reading training data
training = read.table (trainFile, header = FALSE, sep = ",",
strip.white = TRUE, col.names = colNames,
na.strings = "?", stringsAsFactors = TRUE)
# Load the testing data set
testing = read.table (testFile, header = FALSE, sep = ",",
strip.white = TRUE, col.names = colNames,
na.strings = "?", fill = TRUE, stringsAsFactors = TRUE)
I needed to combined two into one. But, there is a problem. I am seeing structure of the two data is not same.
Display structure of the training data
> str (training)
'data.frame': 32561 obs. of 15 variables:
$ age : int 39 50 38 53 28 37 49 52 31 42 ...
$ workclass : Factor w/ 8 levels "Federal-gov",..: 7 6 4 4 4 4 4 6 4 4 ...
$ fnlwgt : int 77516 83311 215646 234721 338409 284582 160187 209642 45781 159449 ...
$ education : Factor w/ 16 levels "10th","11th",..: 10 10 12 2 10 13 7 12 13 10 ...
$ educationnum : int 13 13 9 7 13 14 5 9 14 13 ...
$ maritalstatus: Factor w/ 7 levels "Divorced","Married-AF-spouse",..: 5 3 1 3 3 3 4 3 5 3 ...
$ occupation : Factor w/ 14 levels "Adm-clerical",..: 1 4 6 6 10 4 8 4 10 4 ...
$ relationship : Factor w/ 6 levels "Husband","Not-in-family",..: 2 1 2 1 6 6 2 1 2 1 ...
$ race : Factor w/ 5 levels "Amer-Indian-Eskimo",..: 5 5 5 3 3 5 3 5 5 5 ...
$ sex : Factor w/ 2 levels "Female","Male": 2 2 2 2 1 1 1 2 1 2 ...
$ capitalgain : int 2174 0 0 0 0 0 0 0 14084 5178 ...
$ capitalloss : int 0 0 0 0 0 0 0 0 0 0 ...
$ hoursperweek : int 40 13 40 40 40 40 16 45 50 40 ...
$ nativecountry: Factor w/ 41 levels "Cambodia","Canada",..: 39 39 39 39 5 39 23 39 39 39 ...
$ incomelevel : Factor w/ 2 levels "<=50K",">50K": 1 1 1 1 1 1 1 2 2 2 ...
Display structure of the testing data
> str (testing)
'data.frame': 16282 obs. of 15 variables:
$ age : Factor w/ 74 levels "|1x3 Cross validator",..: 1 10 23 13 29 3 19 14 48 9 ...
$ workclass : Factor w/ 9 levels "","Federal-gov",..: 1 5 5 3 5 NA 5 NA 7 5 ...
$ fnlwgt : int NA 226802 89814 336951 160323 103497 198693 227026 104626 369667 ...
$ education : Factor w/ 17 levels "","10th","11th",..: 1 3 13 9 17 17 2 13 16 17 ...
$ educationnum : int NA 7 9 12 10 10 6 9 15 10 ...
$ maritalstatus: Factor w/ 8 levels "","Divorced",..: 1 6 4 4 4 6 6 6 4 6 ...
$ occupation : Factor w/ 15 levels "","Adm-clerical",..: 1 8 6 12 8 NA 9 NA 11 9 ...
$ relationship : Factor w/ 7 levels "","Husband","Not-in-family",..: 1 5 2 2 2 5 3 6 2 6 ...
$ race : Factor w/ 6 levels "","Amer-Indian-Eskimo",..: 1 4 6 6 4 6 6 4 6 6 ...
$ sex : Factor w/ 3 levels "","Female","Male": 1 3 3 3 3 2 3 3 3 2 ...
$ capitalgain : int NA 0 0 0 7688 0 0 0 3103 0 ...
$ capitalloss : int NA 0 0 0 0 0 0 0 0 0 ...
$ hoursperweek : int NA 40 50 40 40 30 30 40 32 40 ...
$ nativecountry: Factor w/ 41 levels "","Cambodia",..: 1 39 39 39 39 39 39 39 39 39 ...
$ incomelevel : Factor w/ 3 levels "","<=50K.",">50K.": 1 2 2 3 3 2 2 2 3 2 ...
Problem 1:
age has become factor at testing. and all other levels of factor in testing is being increased by 1 than levels of factor in training. This is because first row is an unnecessary row in testing.
|1x3 Cross validator
I tried to get rid of this by re-assigning testing:
testing = testing[-1,]
but, after running str() command again, I don't see any change.
Problem 2:
Like I said at previous, I needed to combine those two data-frame into one data-frame. So, I run this:
combined <- rbind(training , testing)
Besides the problem-1, I can see new a problem after running str()
> str(combined)
'data.frame': 48842 obs. of 15 variables:
$ age : chr "39" "50" "38" "53" ...
$ workclass : Factor w/ 9 levels "Federal-gov",..: 7 6 4 4 4 4 4 6 4 4 ...
$ fnlwgt : int 77516 83311 215646 234721 338409 284582 160187 209642 45781 159449 ...
$ education : Factor w/ 17 levels "10th","11th",..: 10 10 12 2 10 13 7 12 13 10 ...
$ educationnum : int 13 13 9 7 13 14 5 9 14 13 ...
$ maritalstatus: Factor w/ 8 levels "Divorced","Married-AF-spouse",..: 5 3 1 3 3 3 4 3 5 3 ...
$ occupation : Factor w/ 15 levels "Adm-clerical",..: 1 4 6 6 10 4 8 4 10 4 ...
$ relationship : Factor w/ 7 levels "Husband","Not-in-family",..: 2 1 2 1 6 6 2 1 2 1 ...
$ race : Factor w/ 6 levels "Amer-Indian-Eskimo",..: 5 5 5 3 3 5 3 5 5 5 ...
$ sex : Factor w/ 3 levels "Female","Male",..: 2 2 2 2 1 1 1 2 1 2 ...
$ capitalgain : int 2174 0 0 0 0 0 0 0 14084 5178 ...
$ capitalloss : int 0 0 0 0 0 0 0 0 0 0 ...
$ hoursperweek : int 40 13 40 40 40 40 16 45 50 40 ...
$ nativecountry: Factor w/ 42 levels "Cambodia","Canada",..: 39 39 39 39 5 39 23 39 39 39 ...
$ incomelevel : Factor w/ 5 levels "<=50K",">50K",..: 1 1 1 1 1 1 1 2 2 2 ...
factor levels at target variable (incomelevel) in combined data-frame is 5 where it's 2 (which is correct) in the training data-frame and 3 (increased by 1 for problem-1) in testing data-frame. This is because there is a . (dot) after each value at incomelevel in testing data-frame (<=50K., <=50K., >50K.,......). So, I need to remove that .(dot) But, I am not getting idea how to remove it. Is there any function?
I am very in data and r. That's why, facing this type of basic issues. Can you please help me to solve the issue I am facing?

I think you can ignore the first line of test, this will solve the issue of age being a factor, because it seems like a header:
head(readLines(testFile))
[1] "|1x3 Cross validator"
[2] "25, Private, 226802, 11th, 7, Never-married, Machine-op-inspct, Own-child, Black, Male, 0, 0, 40, United-States, <=50K."
[3] "38, Private, 89814, HS-grad, 9, Married-civ-spouse, Farming-fishing, Husband, White, Male, 0, 0, 50, United-States, <=50K."
We run your code, we can use read.csv, with skip=1 for test:
colNames = c ("age", "workclass", "fnlwgt", "education",
"educationnum", "maritalstatus", "occupation",
"relationship", "race", "sex", "capitalgain",
"capitalloss", "hoursperweek", "nativecountry",
"incomelevel")
# Reading training data
training = read.csv (trainFile, header = FALSE, col.names = colNames,stringsAsFactors = TRUE,na.strings = "?",strip.white = TRUE)
testing = read.csv (testFile, header = FALSE, col.names = colNames,na.strings = "?",stringsAsFactors = TRUE,skip=1,strip.white = TRUE)
Now, the income level, unfortunately we have to correct it manually, it's a good thing you check:
testing$incomelevel = factor(gsub("\\.","",as.character(testing$incomelevel)))
We check levels, only difference is native country:
all.equal(sapply(testing,levels) ,sapply(training,levels))
[1] "Component “nativecountry”: Lengths (40, 41) differ (string compare on first 40)"
[2] "Component “nativecountry”: 26 string mismatches"
And I don't think there's much you can do, maybe you have to remove it before / after joining:
setdiff(levels(training$nativecountry),levels(testing$nativecountry))
[1] "Holand-Netherlands"

Extracting complete dataframe from Hmisc package in R

I've used aregImpute to impute the missing values then i used impute.transcan function trying to get complete dataset using the following code.
impute_arg <- aregImpute(~ age + job + marital + education + default +
balance + housing + loan + contact + day + month + duration + campaign +
pdays + previous + poutcome + y , data = mov.miss, n.impute = 10 , nk =0)
imputed <- impute.transcan(impute_arg, imputation=1, data=mov.miss, list.out=TRUE, pr=FALSE, check=FALSE)
y <- completed[names(imputed)]
and when i used str(y) it already gives me a dataframe but with NAs as it is not imputed before, My question is how to get complete dataset without NAs after imputation?
str(y)
'data.frame': 4521 obs. of 17 variables:
$ age : int 30 NA 35 30 NA 35 36 39 41 43 ...
$ job : Factor w/ 12 levels "admin.","blue-collar",..: 11 8 5 5 2 5 7 10 3 8 ...
$ marital : Factor w/ 3 levels "divorced","married",..: 2 2 3 2 2 3 2 2 2 2 ...
$ education: Factor w/ 4 levels "primary","secondary",..: 1 2 3 3 2 3 NA 2 3 1 ...
$ default : Factor w/ 2 levels "no","yes": 1 1 1 1 1 1 NA 1 1 1 ...
$ balance : int NA 4789 1350 1476 0 747 307 147 NA -88 ...
$ housing : Factor w/ 2 levels "no","yes": NA 2 2 2 NA 1 2 2 2 2 ...
$ loan : Factor w/ 2 levels "no","yes": 1 2 1 2 NA 1 1 NA 1 2 ...
$ contact : Factor w/ 3 levels "cellular","telephone",..: 1 1 1 3 3 1 1 1 NA 1 ...
$ day : int 19 NA 16 3 5 23 14 6 14 NA ...
$ month : Factor w/ 12 levels "apr","aug","dec",..: 11 9 1 7 9 4 NA 9 9 1 ...
$ duration : int 79 220 185 199 226 141 341 151 57 313 ...
$ campaign : int 1 1 1 4 1 2 1 2 2 NA ...
$ pdays : int -1 339 330 NA -1 176 330 -1 -1 NA ...
$ previous : int 0 4 NA 0 NA 3 2 0 0 2 ...
$ poutcome : Factor w/ 4 levels "failure","other",..: 4 1 1 4 4 1 2 4 4 1 ...
$ y : Factor w/ 2 levels "no","yes": 1 1 1 1 1 1 1 1 1 1 ...

I have tested your code myself, and it works just fine, except for the last line:
y <- completed[names(imputed)]
I believe there's a type in the above line. Plus, you do not even need the completed function.
Besides, if you want to get a data.frame from the impute.transcan function, then wrap it with as.data.frame:
imputed <- as.data.frame(impute.transcan(impute_arg, imputation=1, data=mov.miss, list.out=TRUE, pr=FALSE, check=FALSE))
Moreover, if you need to test your missing data pattern, you can also use the md.pattern function provided by the mice package.

Contingency table with conditional probability using R

I have those data: http://www.unige.ch/ses/spo/static/simonhug/madi/Mitchell_et_al_1984.csv
> str(dataset)
'data.frame': 135 obs. of 13 variables:
$ CCode : int 2 20 40 41 42 51 52 70 90 91 ...
$ StateAbb : Factor w/ 130 levels "AFG","ALB","ALG",..: 124 19 28 52 33 62 117 75 49 53 ...
$ StateNme : Factor w/ 130 levels "Afghanistan",..: 122 20 27 51 33 62 116 76 47 52 ...
$ prison_score : Factor w/ 5 levels "never","often",..: 1 1 2 4 5 1 NA 4 5 4 ...
$ torture_score : Factor w/ 5 levels "never","often",..: 1 3 1 4 2 1 NA 2 5 2 ...
$ ht_colonial : Factor w/ 10 levels "0. Never colonized by a Western overseas colonial power",..: 1 1 4 8 4 7 7 4 4 4 ...
$ british : int NA NA 0 0 0 1 1 0 0 0 ...
$ british_colony : Factor w/ 2 levels "no","yes": NA NA 1 1 1 2 2 1 1 1 ...
$ continent : Factor w/ 5 levels "Africa","Americas",..: 2 2 2 2 2 2 2 2 2 2 ...
$ region_wb : Factor w/ 19 levels "Australia and New Zealand",..: 10 10 2 2 2 2 2 3 3 3 ...
$ gdppc_l1 : num 25839 23550 10095 1846 4758 ...
$ colonialExperience: chr NA NA "Other Colonial Background" "Other Colonial Background" ...
And have to create a similar result
With this code
# Copy the torture_score in a new col
dataset$torture_score_new = dataset$torture_score
# Add a level to the factor torture_score_new so we can t
levels(dataset$torture_score_new) = c(levels(dataset$torture_score_new), "rarely or never")
### Recode variables
# Torture score
dataset$torture_score_new[dataset$torture_score == "rarely"] = "rarely or never"
dataset$torture_score_new[dataset$torture_score == "never"] = "rarely or never"
dataset$torture_score_new = droplevels(dataset$torture_score_new)
dataset$torture_score_new = ordered(dataset$torture_score_new, levels =c("rarely or never", "somtimes", "often", "very often"))
### Text
dataset$colonialExperience = ifelse(dataset$british_colony == "yes",
"Former British Colony",
"Other Colonial Background")
useOfTortureByColonialExperience = table(dataset$torture_score_new, dataset$colonialExperience)
addmargins(round(prop.table(useOfTortureByColonialExperience)*100,2),1)
and get this result
Former British Colony Other Colonial Background
rarely or never 9.76 20.73
somtimes 10.98 15.85
often 6.10 18.29
very often 10.98 7.32
Sum 37.82 62.19
But I don't understand how to use conditional stat and get the Chi Square.
(I'm a programmer, but a total newbe to R)

Ok it's what I end up doing.
useOfTortureByColonialExperience = table(dataset$torture_score_new, dataset$colonialExperience)
# Get the number of observation
addmargins(useOfTortureByColonialExperience,1);
# Contingency table with conditional probability
useOfTortureByColonialExperienceProp = prop.table(useOfTortureByColonialExperience,2)
print(addmargins(useOfTortureByColonialExperienceProp*100,1),3)
## Chisq
chisq.test(useOfTortureByColonialExperience)
cramersV(useOfTortureByColonialExperience)

multinom (nnet) invalid type (closure) for variable '(weights)'

I am trying to run multinomial logistic regression on my data of supermarket monthly store data. The data looks like this
data.frame': 833233 obs. of 22 variables:
$ ProductId : num 105422 105422 143863 170645 397474 ...
$ Brand : num NA NA NA NA NA NA NA NA NA NA ...
$ Supplier : Factor w/ 788 levels "[00000] 武商量贩",..: 1 113 265 154 99 99 99 99 99 99 ...
$ Mode.of.operations : Factor w/ 3 levels "[1] Distribution",..: 1 1 1 3 2 2 2 2 2 2 ...
$ Category : Factor w/ 27 levels "[01] Fuits and Vegetables",..: 5 5 9 1 22 22 22 22 22 22 ...
$ Name : chr "土腊肉" "土腊肉" "佳品红金龙" "野山笋" ...
$ Packaging : Factor w/ 108 levels "1","2","3","4",..: 1 1 96 1 1 1 1 1 1 1 ...
$ Specs : Factor w/ 3477 levels "(1*2)","（16+5）ml",..: 3466 3466 2678 3466 92 92 92 92 92 92 ...
$ Unit : Factor w/ 72 levels "1*1","kg","Kg",..: 2 2 57 18 8 8 8 8 8 8 ...
$ Origin : Factor w/ 370 levels "409","China",..: 15 15 15 15 15 15 15 15 15 15 ...
$ Price : num 73.5 73.5 4.4 0 6.64 ...
$ Sale.quantity : num 0 0 464 0 1 0 6 0 0 0 ...
$ Sale.revenue : num 0 0 2784 0 8 ...
$ Sale.revenue.wo.tax : num 0 0 2141.54 0 5.68 ...
$ Profit.margin : num 0 0 237.95 0 1.16 ...
$ Profit.margin.percentage : num 0 0 0.1 0 0.17 ...
$ Inventory.turnover.days : num 0 0 30.2 0 1007 ...
$ Purchase.amount.wo.tax : num 0 0 0 0 0 0 0 0 0 0 ...
$ Inventory.leftover.value.wo.tax: num 111.14 0.22 1066.15 0 181.61 ...
$ Month : Factor w/ 23 levels "1","2","3","4",..: 17 17 17 17 17 17 17 17 17 17 ...
$ Adjusted.price : num 0 0 6 0 8 0 15.9 0 0 0 ...
$ Wuhan : Factor w/ 2 levels "0","1": 2 2 2 2 2 2 2 2 2 2 ...
I tried to find multinomial logistic regression by
model1 = multinom(Mode.of.operations ~ Category+Wuhan+Inventory.turnover.days+Adjusted.price, data = wushang, na.omit)
but i ended up with following error
Error in model.frame.default(formula = Mode.of.operations ~ Category + :
invalid type (closure) for variable '(weights)'
I tried to search for an answer to why its happening but couldn't find anything.
If someone could help me figure it out please.
Thanks

Allrighty, i got your error solved but i got another error.
You have to put the arguments as follows:
multinom.glmulti <- function(formula, data,...)
multinom(formula, data=data, maxit=10000,...)
Im using my own formula with my terms.
```{r}
formula_autom = reformulate(variables_autom_0,"clase")
And my own data.
Now we are suposed to get the models.
res <- glmulti(formula_autom, data=clase_training,
level=4, fitfunction=multinom.glmulti, crit="aicc", confsetsize=100,na.action=na.omit)
But i got an error:
Error in nobs.default(object) : no 'nobs' method is available

Error in `row.names<-.data.frame using mlogit in R language

Here are the steps I'm following to do a Multinomial Linear Regression.
> z<-read.table("2008 Racedata.txt", header=TRUE, sep="\t", row.names=NULL)
> head(z)
datekey raceno horseno place winner draw winodds log_odds jwt hwt
1 2008091501 1 8 1 1 2 12.0 2.484907 128 1170
2 2008091501 1 11 2 0 3 8.6 2.151762 123 1135
3 2008091501 1 6 3 0 5 7.0 1.945910 127 1114
4 2008091501 1 12 4 0 10 23.0 3.135494 123 1018
5 2008091501 1 14 5 0 4 11.0 2.397895 113 1027
6 2008091501 1 5 6 0 14 50.0 3.912023 131 972
> x<-mlogit.data(z,choice="winner",shape="long",id.var="datekey",alt.var="horseno")
Error in `row.names<-.data.frame`(`*tmp*`, value = c("1.8", "1.11", "1.6", :
duplicate 'row.names' are not allowed
In addition: Warning message:
non-unique values when setting 'row.names': ‘10.2’, ‘10.4’, ‘10.8’,
‘100.7’, ‘101.12’, ‘102.1’, ‘102.3’, ‘103.2’, ‘103.4’,
‘103.6’, ‘104.12’, ‘104.3’, ‘104.9’, ‘105.1’, ‘105.5’,
‘105.6’, ‘105.8’, ‘106.11’, ‘106.12’, ‘106.13’, ‘106.7’,
‘107.10’, ‘107.14’, ‘107.3’, ‘108.12’, ‘108.2’, ‘108.6’,
‘108.9’, ‘109.1’, ‘109.14’, ‘109.7’, ‘11.12’, ‘11.5’,
‘11.9’, ‘110.2’, ‘110.3’, ‘110.4’, ‘110.9’, ‘111.1’,
‘111.7’, ‘112.12’, ‘112.3’, ‘112.6’, ‘112.8’, ‘113.10’,
‘113.13’, ‘113.7’, ‘114.12’, ‘114.2’, ‘114.9’, ‘115.10’,
‘115.13’, ‘115.5’, ‘116.11’, ‘116.6’, ‘117.14’, ‘117.3’,
‘117.7’, ‘118.1’, ‘118.13’, ‘118.2’, ‘118.9’, ‘119.10’,
‘119.5’, ‘119.6’, ‘119.8’, ‘12.1’, ‘12.10’, ‘12.3’,
‚Äò12.6‚Äô, ‚Äò120.2‚Äô, ‚Äò120.4‚Äô, ‚Äò120.7‚ [... truncated]
>
What step am I missing here? Why the duplicates in row.names?
Thanks,
Walt

Two problems.
You seem to have some problem with encoding since we are seeing lots of umlauts and accent marks in that error message. Furthernore I am wondering if that datekey column got converted into a factor class?
In this case it it referring to an error in construction of the row.names attribute of the new object, x. If you do:
with( z, table( datekey, horseno) )
... you may see an a horse with multiple entries on the same day.
Actually there were no duplicate datekey x horseno combos. Changing to factor for horseno and datekey and then switching the "long" argument to "wide" produces error free result with this result:
z$datekey <- as.character(z$datekey)
z$horseno <- as.character(z$horseno)
x<-mlogit.data(z,choice="winner",shape="wide",id.var="datekey",alt.var="horseno")
str(x)
#----------
Classes ‘mlogit.data’ and 'data.frame': 18312 obs. of 11 variables:
$ datekey : Factor w/ 733 levels "2008091501","2008091502",..: 1 1 1 1 1 1 1 1 1 1 ...
$ raceno : int 1 1 1 1 1 1 1 1 1 1 ...
$ horseno : chr "0" "1" "0" "1" ...
$ place : int 1 1 2 2 3 3 4 4 5 5 ...
$ winner : logi FALSE TRUE TRUE FALSE TRUE FALSE ...
$ draw : int 2 2 3 3 5 5 10 10 4 4 ...
$ winodds : num 12 12 8.6 8.6 7 7 23 23 11 11 ...
$ log_odds: num 2.48 2.48 2.15 2.15 1.95 ...
$ jwt : int 128 128 123 123 127 127 123 123 113 113 ...
$ hwt : int 1170 1170 1135 1135 1114 1114 1018 1018 1027 1027 ...
$ chid : num 1 1 2 2 3 3 4 4 5 5 ...
- attr(*, "index")='data.frame': 18312 obs. of 3 variables:
..$ chid: Factor w/ 9156 levels "1","2","3","4",..: 1 1 2 2 3 3 4 4 5 5 ...
..$ alt : Factor w/ 2 levels "0","1": 1 2 1 2 1 2 1 2 1 2 ...
..$ id : Factor w/ 733 levels "2008091501","2008091502",..: 1 1 1 1 1 1 1 1 1 1 ...
- attr(*, "choice")= chr "winner"

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