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"
I have this table.
'data.frame': 5303 obs. of 9 variables:
$ Metric.ID : num 7156 7220 7220 7220 7220 ...
$ Metric.Name : Factor w/ 99 levels "Avoid accessing data by using the position and length",..: 51 59 59
$ Technical.Criterion: Factor w/ 25 levels "Architecture - Multi-Layers and Data Access",..: 4 9 9 9 9 9 9 9 9 9 ...
$ RT.Snapshot.name : Factor w/ 1 level "2017_RT12": 1 1 1 1 1 1 1 1 1 1 ...
$ Violation.status : Factor w/ 2 levels "Added","Deleted": 2 1 2 2 2 1 1 1 1 1 ...
$ Critical.Y.N : num 0 0 0 0 0 0 0 0 0 0 ...
$ Grouping : Factor w/ 29 levels "281","Bes",..: 27 6 6 6 6 7 7 7 7 7 ...
$ Object.type : Factor w/ 11 levels "Cobol Program",..: 8 7 7 7 7 7 7 7 7 7 ...
$ Object.name : Factor w/ 3771 levels "[S:\\SOURCES\\",..: 3771 3770 3769 3768 3767 3
I want to have a statistic output like this:
For every Technical.Criterion a row with the sum of all rows of Critical.Y.N = 0 and 1
So I have to combine the rows of my database to a new matrix. Using Values of the factor sums ...
But I have no idea how to start...? Any hints?
Thanks
I believe you're asking for a cross-tabulation. Because you did not provide a reproducible sample, I've used mine:
xtabs(~ Sub.Category + Category, retail)
Produce this:
And if you want the value to be say, based on Sales, instead of the count, then you can modify the code to:
xtabs(Sales ~ Sub.Category + Category, retail)
And you will get the following output:
EDIT based on extra information in the OP's comment
If you want to have your tables also share a common title and want to change the name of that title, you can use a combination of names() and dimnames(). An xtab is a cross-tabulation table and if you call dimnames() on it it returns a list of length 2, first one corresponding to the row and second to the column.
dimnames(xtab(dat))
$Technical.Criterion
[1] "TechnicalCrit1" "TechnicalCrit2" "TechnicalCrit3"
$`Object.type`
[1] "Object.type1" "Object.type2" "Object.type3"
So given a data frame, b:
'data.frame': 3 obs. of 9 variables:
$ Metric.ID : int 101 102 103
$ Metric.Name : Factor w/ 3 levels "A","B","C": 1 2 3
$ Technical.Criterion: Factor w/ 3 levels "TechnicalCrit1",..: 1 2 3
$ RT.Snapshot.name : Factor w/ 3 levels "A","B","C": 1 2 3
$ Violation.status : Factor w/ 2 levels "Added","Deleted": 1 2 1
$ Critical.Y.N : num 1 0 1
$ Grouping : Factor w/ 3 levels "A","B","C": 1 2 3
$ Object.type : Factor w/ 3 levels "Object.type1",..: 1 2 3
$ Object.name : Factor w/ 3 levels "A","B","C": 1 2 3
We can use xtab and then change the "common" header right at the top of our table. Since I don't know how many levels are in b$Violation.status, I would use a generic for loop:
for(i in 1:length(unique(b$Violation.status))){
tab[[i]] <- xtabs(Critical.Y.N ~ Technical.Criterion + Object.type, b)
names(dimnames(tab[[i]]))[2] <- paste("Violation.status", i)
}
This produces:
Violation.status 1
Technical.Criterion Object.type1 Object.type2 Object.type3
TechnicalCrit1 1 0 0
TechnicalCrit2 0 0 0
TechnicalCrit3 0 0 1
Which I can now use in my shiny app.
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.
I have a list of data frames:
str(df.list)
List of 34
$ :'data.frame': 506 obs. of 7 variables:
..$ Protocol : Factor w/ 5 levels "P1","P2","P3",..: 1 1 1 1 1 1 1 1 1 1 ...
..$ Time : num [1:506] 0 2 3 0.5 6 1 24 24 24 24 ...
..$ SampleID : Factor w/ 40 levels "P1T0","P1T0.5",..: 1 5 7 2 8 3 6 6 6 6 ...
..$ VolunteerID: Factor w/ 15 levels "ID-02","ID-03",..: 10 10 10 10 10 10 10 11 13 14 ...
..$ Assay : Factor w/ 1 level "ALAT": 1 1 1 1 1 1 1 1 1 1 ...
..$ ResultAssay: int [1:506] 23 23 23 24 25 24 20 34 28 17 ...
..$ Index : Factor w/ 502 levels "P1T0.5VID-02",..: 8 31 37 2 43 19 25 26 28 29 ...
$ :'data.frame': 505 obs. of 7 variables:
..$ Protocol : Factor w/ 5 levels "P1","P2","P3",..: 1 1 1 1 1 1 1 1 1 1 ...
..$ Time : num [1:505] 0 2 3 0.5 6 1 24 24 24 24 ...
..$ SampleID : Factor w/ 40 levels "P1T0","P1T0.5",..: 1 5 7 2 8 3 6 6 6 6 ...
..$ VolunteerID: Factor w/ 15 levels "ID-02","ID-03",..: 10 10 10 10 10 10 10 11 13 14 ...
..$ Assay : Factor w/ 1 level "ALB": 1 1 1 1 1 1 1 1 1 1 ...
..$ ResultAssay: int [1:505] 45 46 47 47 49 47 46 46 44 43 ...
..$ Index : Factor w/ 501 levels "P1T0.5VID-02",..: 8 31 37 2 43 19 25 26 28 29 ..
The list contains 34 data frames with equal variable names. The variables Time and ResultAssay are of the wrong type: I would like to have Time as factor and ResultAssay as numerical.
I am trying to generate a function to use together with lapply to convert the variable type of this list of 34 data frames in one go, but so far i am unsuccessful.
I have tried things in parallel to:
ChangeType <- function(DF){
DF[,2] <- as.factor(DF[,2])
DF[, "ResultAssay"] <- as.numeric(DF[, c("ResultAssay")]
}
lapply(df.list, ChangeType)
What you have tried is nearly correct, but you also need to return the new data.frame and also store it to your existing variable, as so:
ChangeType <- function(DF){
DF[,2] <- as.factor(DF[,2])
DF[, "ResultAssay"] <- as.numeric(DF[, c("ResultAssay")]
DF #return the data.frame
}
# store the returned value to df.list,
# thus updating your existing data.frame
df.list <- lapply(df.list, ChangeType)
I'm trying to find class probabilities of new input vectors with support vector machines in R.
Training the model shows no errors.
fit <-svm(device~.,data=dataframetrain,
kernel="polynomial",probability=TRUE)
But predicting some input vector shows some errors.
predict(fit,dataframetest,probability=prob)
Error in `contrasts<-`(`*tmp*`, value = contr.funs[1 + isOF[nn]]) :
contrasts can be applied only to factors with 2 or more levels
dataframetrain looks like:
> str(dataframetrain)
'data.frame': 24577 obs. of 5 variables:
$ device : Factor w/ 3 levels "mob","pc","tab": 1 1 1 1 1 1 1 1 1 1 ...
$ geslacht : Factor w/ 2 levels "M","V": 1 1 1 1 1 1 1 1 1 1 ...
$ leeftijd : num 77 67 67 66 64 64 63 61 61 58 ...
$ invultijd: num 12 12 12 12 12 12 12 12 12 12 ...
$ type : Factor w/ 8 levels "A","B","C","D",..: 5 5 5 5 5 5 5 5 5 5 ...
and dataframetest looks like:
> str(dataframetest)
'data.frame': 8 obs. of 4 variables:
$ geslacht : Factor w/ 1 level "M": 1 1 1 1 1 1 1 1
$ leeftijd : num 20 60 30 25 36 52 145 25
$ invultijd: num 6 12 2 5 6 8 69 7
$ type : Factor w/ 8 levels "A","B","C","D",..: 1 2 3 4 5 6 7 8
I trained the model with 2 factors for 'geslacht' but sometime I have to predict data with only 1 factor of 'geslacht'.
Is it maybe possible that the class probabilites can be predicted with a test set with only 1 factor of 'geslacht'?
I hope someone can help me!!
Add another level (but not data) to geslacht.
x <- factor(c("A", "A"), levels = c("A", "B"))
x
[1] A A
Levels: A B
or
x <- factor(c("A", "A"))
levels(x) <- c("A", "B")
x
[1] A A
Levels: A B