Let’s continue to the second part – R programming interview questions and answers for data scientists. This part mainly focuses on R coding interview questions and technical interview questions. Now, you will not face any difficulty in your tech round. Do practice all the questions and get ready for the upcoming interviews.<\/p>\n
In between, if you feel you need to revise any concept you are not clear with you can comment below or check the Free R Tutorials Library<\/a><\/strong><\/em><\/p>\n Q.1 What is f(3) where:<\/strong><\/p>\n Ans.<\/strong> The above-written code will return the result of 19. In f In f(3), x is 3, so x^2 is 6. Then, the value is passed to g(3), where x is globally scoped and hence y + x = 3 + 7 = 10. Hence, we obtain f(3) = 9 + 10 = 19 as the final answer.<\/p>\n Q.2 Given the IRIS dataset, consisting of a mix of continuous variables and categorical ones.<\/strong><\/p>\n Where would you apply – <\/strong><\/p>\n Also, explain where do these two types of plots differ with their respective implementations.<\/strong><\/p>\n Ans.<\/strong> Histograms are used for plotting continuous variables. On the contrary, Barplots are used for plotting categorical variables. In the given data, Species column consists of categorical data whereas rest of the data consists of continuous one. We can plot a histogram for the feature Sepal.Width as follows –<\/p>\n Furthermore, for plotting a barplot, we write the following line of code –<\/p>\n Learn in detail about Barplot and Histogram<\/a><\/strong><\/em><\/p>\n Q.3 Suppose you have to work on a dataset consisting of categorical variables. Let this dataset be iris which consists of the column \u201cSpecies\u201d being categorical in nature. You have to implement the XGBoost Algorithm. What problem would you face and how will you resolve it?<\/strong><\/p>\n Ans.<\/strong> The XGBoost Algorithm only works well with the numerical data. Considering that we have a categorical feature – \u201cSpecies\u201d in our dataset, we will first have to convert that into numerical form. This can be achieved by using dummy variables. We can implement this as follows –<\/p>\n Q.4 In the same iris dataset, you have been assigned the task of identifying outliers. You must be able to provide a comprehensive glimpse of the dataset using a suitable visualisation. What method would you choose and why?<\/strong><\/p>\n Ans.<\/strong> Visualising Boxplot is an ideal method of identifying the outliers. The identification of outliers is important because through boxplots, one can obtain a comprehensive idea of the minimum, lower quartile (25th percentile), median (50th percentile), upper quartile (75th percentile), as well as the maximum of a continuous variable. This will help us to get an overall idea of how the data is, and if there are any outliers which may affect the fitting of our model.<\/p>\n Here is a detailed guide for Data Visualization with R<\/a><\/strong><\/em><\/p>\n We can create our boxplot on the iris dataset as follows –<\/p>\n Q.5 Given a dataset, you wish to explore the variations that are present in the features. However, suppose that the dataset consists of 1000 features. It is not easy to represent all of these features in a single plot to gain an idea of the trends contained within. What will you do in this scenario? For simplicity of implementation, consider the mtcars dataset present in the R environment.<\/strong><\/p>\n Ans.<\/strong> This problem statement requires us to implement Principle Component Analysis<\/a><\/strong><\/em>. With the help of this algorithm, one can easily see the overall shape of the data. Through visualisation of the variations in the data, one can identify similar features as well as dissimilar features. In order to implement PCA on this dataset, we must first drop the \u2018vs\u2019 and \u2018am\u2019 variable as they are categorical in nature. We will then plot our PCA graph with rest of the dataset –<\/p>\n Q6. How do correlated variables in a dataset affect the model performance? Given the \u2018boston\u2019 dataset provided by R, diagnose the model to improve its performance.<\/strong><\/p>\n Ans.<\/strong> While carrying out regression, it is assumed that the variables are not correlated to one another. If the variables are however, correlated then our model will suffer from the following problems.<\/p>\n Furthermore, the solution of our regression model will become highly unstable. In order to compute multicollinearity present in our dataset, we make use of the Variance Inflation Score or VIF. This score measures the inflation caused to the correlation coefficient contributed by multicollinearity that is present in the model.<\/p>\n To check for multicollinearity in our model. We will begin by splitting the boston dataset into training and test data.<\/p>\n In the next step, we will build a regression model and evaluate its accuracy.<\/p>\n Now, we will evaluate the VIF score for each variable<\/p>\n From the VIF scores generated above, we observe that the tax variable has the highest score. Therefore, we will remove this variable and evaluate our model again.<\/p>\n From the above observation, we conclude that removing \u2018tax\u2019 variable does not affect the model much. Therefore, the model requires other areas of improvement to achieve a better score.<\/p>\n Q.7 What will be the output of this code? State the reason behind the flow of code execution. <\/strong><\/p>\n Ans.<\/strong> After we run the above code, following things happen in the code –<\/p>\n Therefore, in the above code, generic2_a2() is called.<\/p>\n Q.8 Will the following code return any error? State the reason behind your answer.<\/strong><\/p>\n Ans.<\/strong> The code will not yield any error. In the above code, the index iterates over the vector val having a length of 0. The loop is taken from index = 1 to index = 0. The \u2018:\u2019 in this case counts down. During the first iteration, val[1] will have an NA indexing for the atomic value. This resulting NA will have an empty length give the result[1] – \u201cout of bounds indexing for the lists\u201d<\/p>\n In the following iteration, val[0] will provide us with the numeric(0) index as a returned value. Squaring the value will not have any effect and as a result, numeric(0) is assigned to result[0]. When a vector of length 0 is assigned to a subset of length 0, it does not alter anything. The code will execute as it step involves valid R operations.<\/p>\n Q.9 Import the diamonds dataset that is provided by the ggplot package. Your aim is to create a scatter plot of price vs depth. How will you proceed to find the depth value where most diamonds are found?<\/strong><\/p>\n Ans.<\/strong> Using ggplot2 package, we will plot the scatterplot of the variable depth vs price as follows –<\/p>\n <\/p>\n We can observe that the diamonds are densely concentrated in an area around 60. However, this does not give us a precise area where the highest concentration of diamonds of located. Therefore, we change the transparency of the points to be 1\/100th of their original and mark the x-axis at every 2 units.<\/p>\n From the above visualisation, we observe that the highest concentration of diamonds is between 60-64.<\/p>\n Don’t miss to revise the important concept of R packages<\/a> for your upcoming data science interviews<\/strong><\/em><\/p>\n Q.10 What are some of the errors present in the code below? Fix these errors with their respective correct answers.<\/strong><\/p>\n Ans.<\/strong> The correct versions of these codes are as follows –<\/p>\n Q.11 Suppose that you want to access a graphics device, run the code on it and then close the device. How will you carry out this procedure in the form of a function? The file that you want to save is a pdf file – \u2018data.pdf\u2019<\/strong><\/p>\n Q.12 How will you convert the following code into its corresponding prefix form?<\/strong><\/p>\n Ans.\u00a0<\/strong>Let us rewrite these expressions to match the given code syntax. Since the prefix functions pre-define the order of execution, the parentheses can be therefore, omitted in the second expression.<\/p>\n Important Tip!! If you want to crack your next data science interview so along with these top R programming Interview Questions and Answers you need to practice some R projects.<\/p>\n DataFlair has published many projects, practice here – Top R projects for Data Science interview<\/a><\/strong><\/em><\/p>\n Q.13 How will you create an R6 class that lets you have a bank account you to access your balance, store deposit as well as withdraw money?<\/strong><\/p>\n Ans.<\/strong><\/p>\n Let us test this by creating an instance and getting a balance.<\/p>\n Q.14 How will you create an R6 class that manages the current working directory? This class should have $get and $set() methods.<\/strong><\/p>\n Ans.<\/strong> The R6 class for managing the current working directory be implemented as follows –<\/p>\n Q.15 What takes place when you alter the levels of a factor given below?<\/strong><\/p>\n Ans.<\/strong> The various integer values that lie in the factor remain the same. However, the levels get altered giving the data a changed appearance.<\/p>\n Q16. Suppose you have a character vector containing the strings as follows –<\/strong><\/p>\n You want to utilize a regex operation that would match the strings for the different variations in the pattern \u2018adb\u2019. You have to perform pattern matching in two ways – <\/strong><\/p>\n How will you carry out this operation?<\/strong><\/p>\n Ans.<\/strong> In order to obtain the two outputs, we will make use of the grep() function. This function takes the regex as first argument and input vector as its second argument. In this case, we will pass value = TRUE as we want the grep to return the vector being a copy of the actual elements in the input vector that can be matched. The two regex operations we will utilize for obtaining the desired output are –<\/p>\n These two operations differ in their quantifier that defines how many times an object must be repeated. The \u2018*\u2019 quantifier would match the repetition atleast 0 times and the \u2018+\u2019 quantifier would perform it atleast 1 time. Then, we will implement this in code as follows –<\/p>\n R Functions<\/a><\/strong><\/em> – A very helpful article for R programming interview questions\u00a0<\/strong><\/em><\/p>\n Q.17 You have a task of finding names of all the .csv files in the repository. How will you carry this out using the regex operation? These csv files are stored in the following pattern –<\/strong><\/p>\n [1] “birds.csv” “boston.csv” “data.csv” “grass.csv” “table.csv”<\/strong><\/p>\n Ans.<\/strong> In this, we will make use of the combination of two regex quantifiers – \u2018.\u2019 and \u2018*\u2019. The \u2018.\u2019 quantifier returns the match found in any single character. We will create a character variable \u2018csvlist\u2019 and use the list.files() function with the pattern = \u201c.*.csv\u201d as follows –<\/p>\n Q.18 Suppose that you have string as follows –<\/strong><\/p>\n data_string <- “I learn Big Data for 4 hours daily with DataFlair”<\/strong><\/p>\n You have to perform following operations using various regex sequences –<\/strong><\/p>\n Ans.<\/strong><\/p>\n 1. For matching the digit and replacing it with \u2018_\u2019, we will make use of the gsub() and regmatch() function as follows –<\/p>\n 2. To match the non-digit part of the string and replace it with \u2018_\u2019, we will use the \\D sequence to match the non-digit part of our string and replace it with \u2018_\u2019.<\/p>\n 3. To match the non-word character and replace it with \u2018&\u2019, we will use the sequence \\W as follows –<\/p>\n Q.19 For a given string, you have to extract all the email addresses that are contained in it.<\/strong><\/p>\n Ans.<\/strong> To carry out this operation, we will make use of the POSIX class. In particular, we will use the [[:alnum:]] and [[:alpha:]] classes. The alnum class matches the alphanumeric characters whereas alpha class matches the letters.<\/p>\nR Coding Interview Questions and Answers<\/h2>\n
x <- 7\r\nf <- function(y) { x <- 3; x^2 + g(y) }\r\ng <- function(y) { y + x }<\/pre>\n![\"plotting](\"https:\/\/data-flair.training\/blogs\/wp-content\/uploads\/sites\/2\/2019\/10\/R-IRIS-dataset.jpg\")
<\/a><\/p>\n\n
> ggplot(data = iris,aes(x=Sepal.Width))+geom_histogram(fill=\"steelblue\",col=\"red\")<\/pre>\n
![\"Histogram](\"https:\/\/data-flair.training\/blogs\/wp-content\/uploads\/sites\/2\/2019\/10\/R-histogram.jpg\")
<\/a><\/p>\n> ggplot(data = iris,aes(x=Species))+geom_bar(fill=\"steelblue\",color = 'red') #DataFlair<\/pre>\n
![\"barplots](\"https:\/\/data-flair.training\/blogs\/wp-content\/uploads\/sites\/2\/2019\/10\/R-Barplot.jpg\")
<\/a><\/p>\n<\/a><\/p>\nlibrary(dummies)\r\ndummy(iris$Species)<\/pre>\n
![\"interview](\"https:\/\/data-flair.training\/blogs\/wp-content\/uploads\/sites\/2\/2019\/10\/R-Dummy-variable.jpg\")
<\/a><\/p>\nboxplot(Sepal.Length ~ Species, data=iris, main=\"Box Plot\", xlab=\"Species\", ylab=\"Sepal Length\", col = \"PaleGreen3\", border = \"darkgreen\")<\/pre>\n
![\"R](\"https:\/\/data-flair.training\/blogs\/wp-content\/uploads\/sites\/2\/2019\/10\/R-boxplot-outliers.jpg\")
<\/a><\/p>\n![\"Principle](\"https:\/\/data-flair.training\/blogs\/wp-content\/uploads\/sites\/2\/2019\/10\/mtcars-dataset.jpg\")
<\/a><\/p>\n> pca <- prcomp(mtcars[,c(1:7,10,11)], center = TRUE,scale. = TRUE)\r\n> library(ggbiplot)\r\n> ggbiplot(pca) #DataFlair<\/pre>\n
![\"plotting](\"https:\/\/data-flair.training\/blogs\/wp-content\/uploads\/sites\/2\/2019\/10\/mtcarsPCA-in-R.jpg\")
<\/a><\/p>\n![\"R](\"https:\/\/data-flair.training\/blogs\/wp-content\/uploads\/sites\/2\/2019\/10\/R-boston-dataset.jpg\")
<\/a><\/p>\n\n
> set.seed(123)\r\n> sample_data <- Boston$medv %>%\r\n+ createDataPartition(p = 0.8, list = FALSE)\r\n> training_data <- Boston[sample_data, ]\r\n> testing_data <- Boston[-sample_data, ]\r\n> #DataFlair<\/pre>\n
![\"R](\"https:\/\/data-flair.training\/blogs\/wp-content\/uploads\/sites\/2\/2019\/10\/splitting-boston-dataset.jpg\")
<\/a><\/p>\n![\"regression](\"https:\/\/data-flair.training\/blogs\/wp-content\/uploads\/sites\/2\/2019\/10\/R-regression-model.jpg\")
<\/a><\/p>\ncar::vif(corr_model)<\/pre>\n
![\"R](\"https:\/\/data-flair.training\/blogs\/wp-content\/uploads\/sites\/2\/2019\/10\/evaluating-vif-score.jpg\")
<\/a><\/p>\n> # Building the model without tax variable #DataFlair\r\n> corr_model <- lm(medv ~. -tax, data = training_data)\r\n> # Generate Predictions\r\n> prediction <- corr_model %>% predict(testing_data)\r\n> # Model performance\r\n> data.frame(\r\n+ RMSE = RMSE(prediction, testing_data$medv),\r\n+ R2 = R2(prediction, testing_data$medv)\r\n+ )<\/pre>\n
![\"R](\"https:\/\/data-flair.training\/blogs\/wp-content\/uploads\/sites\/2\/2019\/10\/evaluating-model-in-R.jpg\")
<\/a><\/p>\ngeneric2 <- function(x) UseMethod(\"generic1\")\r\ngeneric2_a1 <- function(x) \"a1\"\r\ngeneric2_a2 <- function(x) \"a2\"\r\ngeneric2_b1 <- function(x) {\r\n class(x) <- \"a1\"\r\n NextMethod()\r\n}\r\n\r\n\r\ngeneric2(structure(list(), class = c(\"b1\", \"a2\")))<\/pre>\n\n
val <- numeric()\r\nresult <- vector(\"list\", length(val))\r\nfor (index in 1:length(val)) {\r\n result[index] <- val[index] ^ 2\r\n}\r\nresult<\/pre>\n![\"importing](\"https:\/\/data-flair.training\/blogs\/wp-content\/uploads\/sites\/2\/2019\/10\/diamond-data.jpg\")
<\/a><\/p>\nggplot(data = diamonds, aes(x = depth,y = price)) +\r\n+ geom_point(alpha= 0.5) #DataFlair<\/pre>\n
![\"ggplot2](\"https:\/\/data-flair.training\/blogs\/wp-content\/uploads\/sites\/2\/2019\/10\/diamond-scatterplot1.jpg\")
<\/a><\/p>\nggplot(data = diamonds, aes(x = depth,y = price)) +\r\n+ geom_point(alpha= 0.01) +\r\n+ scale_x_continuous(breaks = seq(0,80,2))<\/pre>\n
![\"visualization](\"https:\/\/data-flair.training\/blogs\/wp-content\/uploads\/sites\/2\/2019\/10\/diamond-scatterplot2.jpg\")
<\/a><\/p>\n\n
\n
PDF_plot <- function(code) {\r\n pdf(\"data.pdf\")\r\n on.exit(dev.off(), add = TRUE)\r\n code\r\n}<\/pre>\n2 + 3 + 4\r\n2 + (3 + 4)\r\nif (length(x) <= 5) x[[5]] else x[[n]]<\/pre>\n
`+`(`+`(2, 3), 4)\r\n\r\n`+`(2, `(`(`+`(3, 4)))\r\n`+`(2, `+`(3, 4))<\/pre>\n
BankAccount <- R6Class(\"BankAccount\", list(\r\nbal = 0,\r\namount_deposit = function(dpt = 0) {\r\n self$bal = self$bal + dpt\r\n invisible(self)\r\n},\r\nwithdraw = function(amount_draw) {\r\n self$bal = self$bal - amount_draw\r\n invisible(self)\r\n}\r\n))<\/pre>\n![\"R](\"https:\/\/data-flair.training\/blogs\/wp-content\/uploads\/sites\/2\/2019\/10\/R6-Class.jpg\")
<\/a><\/p>\nnew_account <- BankAccount$new()\r\nnew_account$bal\r\nnew_account$\r\n amount_deposit(10)$\r\n withdraw(20)$\r\n bal<\/pre>\n
![\"questions](\"https:\/\/data-flair.training\/blogs\/wp-content\/uploads\/sites\/2\/2019\/10\/R6-Class-Instance.jpg\")
<\/a><\/p>\nWorkingDirectory <- R6Class(\"WorkingDirectory\", list(\r\n get_directory = function() {\r\n getwd()\r\n},\r\nset_directory = function(value) {\r\n setwd(value)\r\n}\r\n))<\/pre>\nfac <- factor(letters)\r\nlevels(fac) <- rev(levels(fac))<\/pre>\n
> fac <- factor(letters[1:10])\r\n> levels(fac) #DataFlair\r\n> fac\r\n> as.integer(fac)\r\n> levels(fac) <- rev(levels(fac))\r\n> levels(fac)\r\n> fac\r\n> as.integer(fac)<\/pre>\n
![\"R](\"https:\/\/data-flair.training\/blogs\/wp-content\/uploads\/sites\/2\/2019\/10\/Factors-in-R.jpg\")
<\/a><\/p>\n(string1 <- c(\"a\", \"ab\", \"adb\", \"addb\", \"adddb\", \"adddb\"))<\/pre>\n
\n
\n
grep(\"ad*b\", string1, value = TRUE)\r\n\r\ngrep(\"ad+b\", string1, value = TRUE)<\/pre>\n
![\"R](\"https:\/\/data-flair.training\/blogs\/wp-content\/uploads\/sites\/2\/2019\/10\/regex-operation-in-R-.jpg\")
<\/a><\/p>\ncsvlist = list.files(pattern = \".*.csv\")<\/pre>\n
![\"R](\"https:\/\/data-flair.training\/blogs\/wp-content\/uploads\/sites\/2\/2019\/10\/R-regex-operations.jpg\")
<\/a><\/p>\n\n
gsub(pattern = \"\\\\d+\",replacement = \"_\",x = data_string) \r\nregmatches(data_string,regexpr(pattern = \"\\\\d+\",text = data_string))<\/pre>\n
![\"interview](\"https:\/\/data-flair.training\/blogs\/wp-content\/uploads\/sites\/2\/2019\/10\/replacing-digit-with-dash.jpg\")
<\/a><\/p>\ngsub(pattern = \"\\\\D+\",replacement = \"_\",x = data_string)<\/pre>\n
![\"R](\"https:\/\/data-flair.training\/blogs\/wp-content\/uploads\/sites\/2\/2019\/10\/replace-non-digit-with-dash.jpg\")
<\/a><\/p>\n> gsub(pattern = \"\\\\W\",replacement = \"&\",x = data_string)<\/pre>\n
![\"R](\"https:\/\/data-flair.training\/blogs\/wp-content\/uploads\/sites\/2\/2019\/10\/string-in-R-.jpg\")
<\/a><\/p>\nemail <- c(\"My email is abc@dataflair.com\",\"my email address is def@datalfair.com\",\"John Wayne\",\"Clint Eastwood\")<\/pre>\n
> email <- c(\"My email is abc@dataflair.com\",\"my email address is def@datalfair.com\",\"John Wayne\",\"Clint Eastwood\")\r\n> unlist(regmatches(x = email, gregexpr(pattern = \"[[:alnum:]]+\\\\@[[:alpha:]]+\\\\.com\",text = email)))<\/pre>\n
![\"R](\"https:\/\/data-flair.training\/blogs\/wp-content\/uploads\/sites\/2\/2019\/10\/extracting-email-address-in-R.jpg\")
<\/a><\/p>\n![\"R](\"https:\/\/data-flair.training\/blogs\/wp-content\/uploads\/sites\/2\/2019\/10\/dataset-for-R-programming.jpg\")
<\/a><\/p>\n![](\"https:\/\/data-flair.training\/blogs\/wp-content\/uploads\/sites\/2\/2019\/10\/abbreviating-functions-in-R.jpg\"){kind=link}