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Liabilities2003 (Rs.)2004 (Rs.)Assets2003 (Rs.)2004 (Rs.)Capital54825482Cash and balance with RBI67058025543697Reserve and surplus69963008406754Balance with banks and money at call short notice37816115349090Deposits7354078879299033Investments4226011046435410Borrowings672177262960Advances5794769962502390Other liabilities and provisions15147661632769Fixed assets369499401240Other assets408108640325228272951389606998115145807124264349SOLUTIONThe solution is as follows −ParticularsYear ending 31st marchIncreased/ DecreasedIncreased/ Decreased (%)2003 (Rs.)2004 (Rs.)Amount (Rs)Amount (Rs)AssetsTotal current assetsABA-B=C(C/A)*100Cash and balance with RBI670580255436971162410517.33Balance with banks and money at call and short notice37816115349090-1567479-41.45Total current assets10487413 (X)10892787 (Y)405374 (Y-X=Z)3.86 (Z/X*100)Fixed assetsAB(A-B=C)(A-B=C)Investments4226011046435410-4175300-9.88Advances5794769957947699-4554691-7.86Fixed Assets369499401240-31741-8.59Other Assets408108640325224856411.90Total Fixed Assets104658394 (X’)113371562 (Y’)8713168 (Y’-X’=Z’)8.33 (Z’/X’*100)Total Assets115145807 (x+x’)124264349 (y+y’)91185427.92Current liabilitiesABA-B=C(C/A)*100Borrowings67217726296040921760.88Other liabilities and provisions15147661632769-1180037.79Total Current Liabilities2186943 (X)1895729(Y)29121413.32ABB-A=C(C/A)*100Fixed liability capital54825482--------------------------------------------Reserves surplus69963008406754141045420.16Deposits735407887929903357582457.83Total Fixed Liabilities80542570 (X’)87711269 (Y’)71686998.90X+X’Y+Y’(Y+Y’)-(X+X’)=C’(C’/(X+X’))*100Total liability827295138960699868774858.31Read More
In data analysis, time series is one of the common data we have to deal with and it might also contain dates data along with other variables. We might want to find the difference between two times to check how many days or weeks have changed the time series. This can be easily done with the help of difftime function.Example> difftime(strptime("25/07/2021", format = "%d/%m/%Y"), + strptime("25/07/2020", format = "%d/%m/%Y"), units="weeks") Time difference of 52.14286 weeks > difftime(strptime("25.07.2021", format = "%d.%m.%Y"), + strptime("25.07.2020", format = "%d.%m.%Y"), units="weeks") Time difference of 52.14286 weeks > difftime(strptime("25.07.2021", format = "%d.%m.%Y"), + strptime("25.07.2020", format = ... Read More
Regression analysis output in R gives us so many values but if we believe that our model is good enough, we might want to extract only coefficients, standard errors, and t-scores or p-values because these are the values that ultimately matters, specifically the coefficients as they help us to interpret the model. We can extract these values from the regression model summary with delta $ operator.ExampleConsider the below data −> set.seed(99) > x1 x2 x3 x4 x5 x6 x7 y Regression_Model summary(Regression_Model) Call: lm(formula = y ~ x1 + x2 + x3 + x4 + x5 + x6 + x7) ... Read More
In data analysis, there are many situations we have to deal with and one of them is creating a new column that has the row sums of only some rows. These sums will be repeated so that we get the total number of values equal to the number of rows in the data frame. We can use rowSums with rep function to create such type of columns.ExampleConsider the below data frame −> set.seed(99) > x1 x2 x3 x4 x5 df df x1 x2 x3 x4 x5 1 0.7139625 4 9.321058 0.33297863 4 2 0.9796581 2 4.298837 -1.47926432 11 3 0.5878287 ... Read More
The error “undefined columns selected when subsetting data frame” means that R does not understand the column that you want to use while subsetting the data frame. Generally, this happens when we forget to use comma while subsetting with single square brackets.ExampleConsider the below data frame −> set.seed(99) > x1 x2 x3 x4 x5 df df x1 x2 x3 x4 x5 1 0.7139625 4 9.321058 0.33297863 4 2 0.9796581 2 4.298837 -1.47926432 11 3 0.5878287 3 7.389898 -0.07847958 5 4 0.9438585 4 7.873764 -1.35241100 6 5 0.1371621 2 5.534758 -1.17969925 4 6 0.6226740 4 8.786676 -1.15705659 5 7 -0.3638452 1 ... Read More
When we have multiple lists but they have similar type of data then we might want to combine or merge those lists. This will be helpful to use because we can perform the calculations using one list name instead of applying them on multiple ones. We can combine multiple lists with the help of mapply function.ExampleConsider the below lists −> List1 List1 [[1]] [1] "a" "b" "c" "d" "e" [[2]] [1] 1 2 3 4 5 [[3]] [1] 5 4 3 2 1 [[4]] [1] 25 [[5]] ... Read More
Categorical variables have multiple categories but if the data set is large and the categories are also large in numbers then it becomes a little difficult to recognize them. Therefore, we can extract unique values for categorical variables that will help us to easily recognize the categories of a categorical variable. We can do this by using unique for every column of an R data frame.ExampleConsider the below data frame −> x1 x2 x3 x4 df df x1 x2 x3 x4 1 A 5 India a 2 A 5 India b 3 A ... Read More
An R data frame can have a large number of categorical variables and these categorical form different combinations. For example, one value of a variable could be linked with two or more values of the other variable. Also, one categorical variable can have all unique categories. We can find this unique combination for as many variables as we want and it can be done with the help of unique function.ExampleConsider the below data frame −> x1 x2 x3 x4 df df x1 x2 x3 x4 1 1 A a 5 2 2 A b 5 3 3 A c 10 ... Read More
Creating a data frame with a column as a list is not difficult but we need to use I with the list so that the list elements do not work as an individual column. Here, you will find the common method to create a list which is incorrect if we want to insert that list in our data, also the correct method is mentioned at the end.The incorrect way −Example> x1 x2 df df x1 c.1..1. c.2..2. c.3..3. c.4..4. c.5..5. c.6..6. c.7..7. c.8..8. c.9..9. 1 1 1 2 3 ... Read More
A bar graph plotted with ggplot function of ggplot2 shows horizontal and vertical gridlines. If we are interested only in the bar heights then we might prefer to remove the horizontal gridlines. In this way, we can have X-axis that helps us to look at the different categories we have in our variable of interest and get rid of the unnecessary information. This can be done by setting breaks argument to NULL in scale_y_discrete function.ExampleConsider the below data frame −> x y df library(ggplot2)Creating the plot with all gridlines −> ggplot(df, aes(x, y))+ + geom_bar(stat='identity')OutputCreating the plot without horizontal gridlines ... Read More