Finding group-wise mean is a common thing but if we go for step-by-step analysis then sum of values are also required when we have a categorical variable in our data set. This can be easily done with the help of group_by and summarise_each function of dplyr package.ExampleConsider the below data frame:Live Demo> Group Salary Emp EmpOutputGroup Salary 1 D 28256 2 B 31092 3 A 23147 4 C 28209 5 B 37676 6 C 33374 7 D 44864 8 B 40152 9 A 25843 10 A 40946 11 D 23321 12 A 42854 13 C 36960 14 A 35285 15 ... Read More

Most of the times, unlisting is used to find the mean of list elements but we can also use double-square brackets for the same purpose. The double-square brackets are basically used to access the values in the elements of the list, thus mean function works with those values directly. Look at the below example to understand how it works.ExampleConsider the below list:Live Demo> x xOutput[1] 3 3 3 5 3 1 4 7 5 4 5 9 9 7 4 3 6 2 4 3 3 4 7 4 4 [26] 4 5 3 4 4 3 5 7 2 ... Read More

How to extract the names of vector values from a named vector in R?The names of vector values are created by using name function and the names can be extracted by using the same function. For example, if we have a vector called x that contains five values(1 to 5) and their names are defined as first, second, third, fourth and fifth then the names of values in x can be extracted by using names(x)[x==1].Example1Live Demo> x1 names(x1) x1Outputone two three four 1 2 3 4Example> names(x1)[x1==1]Output[1] "one"Example> names(x1)[x1==2]Output[1] "two" Example> names(x1)[x1==3]Output[1] "three"Example> names(x1)[x1==4]Output[1] "four" Example2Live Demo> x2 x2Output [1] ... Read More

Suppose we have a number n we have to generate nth term in “Look and Say” sequence. This is a sequence whose few terms are like below −111211211111221The string will be read like1 (One)11 (One 1) So read the previous 1, and say “One 1”21 (Two 1) So read the previous 11, and say “Two 1”1211 (One 2 one 1) So read the previous 21, and say “One 2 one 1”111221 (One 1 one 2 two 1) So read the previous 1211, and say “One 1 one 2 two 1”Suppose we have a number n, 1

Suppose we have a string s where each characters are sorted and we also have a number k, we have to find the length of the longest substring such that every character occurs at least k times.So, if the input is like s = "aabccddeeffghij" k = 2, then the output will be 8, as the longest substring here is "ccddeeff" here every character occurs at least 2 times.To solve this, we will follow these steps −Define a function rc() . This will take lstc := a map with all characters and their occurrencesacc := a new listans := 0valid ... Read More

The cartesian join is the joining of two objects that creates the combination of each value in object with all the values in the other object. For example, if we have a vector x that contains 1, 2, 3 and the other object y contains a, b, c then the cartesian join will be 1a, 2a, 3a, 1b, 2b, 3b, 1c, 2c, and 3c. Check out the below examples to understand how it can be done.Example> library(data.table) > DT1 DT1Output x 1: 1 2: 2 3: 3 4: 4Example> DT2 DT2Output y 1: 25 2: ... Read More

Suppose we have a list of numbers called nums and another value k, we have to find the length of longest sublist where the absolute difference between the largest and smallest element is ≤ k.So, if the input is like nums = [2, 4, 6, 10] k = 4, then the output will be 3, as we can select pick [2, 4, 6] here the absolute difference is 4.To solve this, we will follow these steps −Create two double ended queue maxd, mindi := 0, res := 1for each index j and value a in A, dowhile maxd is not ... Read More

To find the column number of minimum values in each row for a data frame, we can use apply function but if we want to return the output in tabular form then matrix function should be used. For example, if we have a data frame df then our problem can be solved by using the code: as.matrix(apply(df, 1, which.min)).ExampleConsider the below data frame:Live Demo> set.seed(37) > x1 x2 x3 x4 x5 df1 df1Outputx1 x2 x3 x4 x5 1 1 2 4 9 3 2 0 5 8 10 4 3 1 3 8 6 1 4 1 5 5 8 ... Read More

Suppose we have a binary list, so here only 1s and 0s are available and we also have another number k. We can set at most k 0s to 1s, we have to find the length of the longest sublist containing all 1s.So, if the input is like nums = [0, 1, 1, 0, 0, 1, 1] k = 2, then the output will be 6, as we can set the two middle 0s to 1s and then the list becomes [0, 1, 1, 1, 1, 1, 1].To solve this, we will follow these steps −zeros := 0, ans := ... Read More

The easiest way to replace numbers with ordinal strings is using ifelse function. The ifelse function in R works as ifelse(test_expression, x, y). Here, test_expression must be a logical vector or an object that can be coerced to logical). The return value is a vector with the same length as test_expression.Example1Live Demo> x1 x1Output[1] 1 1 1 0 1 0 0 1 1 1 0 0 1 0 1 0 0 1 1 0 0 1 1 1 1 0 0 0 0 1 1 1 0 1 1 0 1 [38] 0 1 1 0 0 1 0 1 ... Read More