There are many data sets available in base R and in different packages of R. The characteristics of these data sets are very different, for example, some data sets are time series data, some have only numerical columns, some have numerical as well as factor columns, some includes character columns with other type of columns. Therefore, it becomes helpful to everyone who want to learn the use of R programming. To get the list of available data sets in base R we can use data() but to get the list of data sets available in a package we first need ... Read More

For a one point of time, we might have multiple time series data, this could be weather for multiple cities, price variation in multiple products, demand expectancy at different locations, or anything that changes with time and measured for multiple things or locations. If we have such type of time series data then we would be needing to plot that data in a single plot and it can be done with the help of geom_line function of ggplot2 package.ExampleConsider the below data frames − Live Demo> x1 y1 df1 df1Output   x1 y1 1 1 -0.1165387 2 2 -0.9084062 3 3 0.4696637 ... Read More

Creating a numeric vector is the first step towards learning R programming and there are many ways to do that but if we want to generate a sequence of number then it is a bit different thing, not totally different. We can create a vector with a sequence of numbers by using − if the sequence of numbers needs to have only the difference of 1, otherwise seq function can be used.Example Live Demo> x1 x1Output[1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 ... Read More

Suppose we have a list of numbers called nums, we have to find the number of times that the list changes from positive-to-negative or negative-to-positive slope.So, if the input is like [2, 4, 10, 18, 6, 11, 13], then the output will be 2, as it changes the direction at 10 (positive-to-negative), and then at 6 (negative-to-positive).To solve this, we will follow these steps −To solve this, we will follow these steps −for i in range 1 to size of nums - 1, doif nums[i-1] < nums[i] > nums[i+1] or nums[i-1] > nums[i] < nums[i+1], thencount := count + 1return ... Read More

Suppose we have a list of numbers called cells; this list is representing sizes of different cells. Now, in each iteration, the two largest cells a and b interact according to these rules: So, If a = b, they both die. Otherwise, the two cells merge and their size becomes floor of ((a + b) / 3). We have to find the size of the last cell or return -1 if there's no cell is remaining.So, if the input is like [20, 40, 40, 30], then the output will be 16, in first iteration, 40 and 40 will die, then ... Read More

Suppose we have a list of words, we have to concatenate them in camel case format.So, if the input is like ["Hello", "World", "Python", "Programming"], then the output will be "helloWorldPythonProgramming"To solve this, we will follow these steps −s := blank stringfor each word in words −make first letter word uppercase and rest lowercaseconcatenate word with sret := s by converting first letter of s as lowercasereturn retLet us see the following implementation to get better understanding −Example Live Democlass Solution:    def solve(self, words):       s = "".join(word[0].upper() + word[1:].lower() for word in words)       return ... Read More

Suppose we have a lowercase alphabet string s, and an offset number say k. We have to replace every letter in s with a letter k positions further along the alphabet. We have to keep in mind that when the letter overflows past a or z, it gets wrapped around the other side.So, if the input is like "hello", k = 3, then the output will be "khoor"To solve this, we will follow these steps −Define a function shift(). This will take ci := ASCII of (c) - ASCII of ('a')i := i + ki := i mod 26return character ... Read More

Suppose we have a list of prices of cars for sale, and we also have a budget k, we have to find the maximum number of cars we can buy.So, if the input is like [80, 20, 10, 30, 80], k = 85, then the output will be 3 as we can buy three cars with prices 20, 10, 40To solve this, we will follow these steps −count := 0sort the list pricesfor i in range 0 to size of prices, doif prices[i]

Suppose we have a binary list called fighters and another list of binary lists called bosses. In fighters list the 1 is representing a fighter. Similarly, in bosses list 1 representing a boss. That fighters can beat a boss’s row if it contains more fighters than bosses. We have to return a new bosses matrix with defeated boss rows removed.So, if the input is like fighters = [0,1,1]011000001011111then the output will be011111To solve this, we will follow these steps −fighter_cnt := sum of all elements of fightersresult := a new listfor each row in bosses, doif fighter_cnt

Suppose we have a list of strings called book, if we page an index (0-indexed) into the book, and page_size, we have to find the list of words on that page. If the page is out of index then simply return an empty list.So, if the input is like book = ["hello", "world", "programming", "language", "python", "c++", "java"] page = 1 page_size = 3, then the output will be ['language', 'python', 'c++']To solve this, we will follow these steps −l:= page*page_sizereturn elements of book from index l to l+page_size - 1Let us see the following implementation to get better understanding ... Read More