Deforming data in R is about changing the way data is organized in rows and columns. Often, data processing in R by adopting data input as a dataframe; 

Deforming data. Data can be easily extracted from rows and columns of a data frame; But there may be situations where the formatting of the data we need is what we find; be different. R many functions to divide; 

Merge and change rows into columns and vice versa; has it.

Add columns and rows to the data frame

Using the cbind () function, we can use multiple vectors together to create a data frame. We can also merge two data frames using the rbind () function.

# Create vector objects.

city ​​<- c (“Tampa”, “Seattle”, “Hartford”, “Denver”)

state <- c (“FL”, “WA”, “CT”, “CO”)

zipcode <- c (33602,98104,06161,80294)

# Combine above three vectors into one data frame.

addresses <- cbind (city, state, zipcode)

# Print a header.

cat (“# # # # The First data frame \ n”)

# Print the data frame.

print (addresses)

# Create another data frame with similar columns

new.address <- data.frame (

city ​​= c (“Lowry”, “Charlotte”),

state = c (“CO”, “FL”),

zipcode = c (“80230 ″,” 33949 “),

stringsAsFactors = FALSE

)

# Print a header.

cat (“# # The Second data frame \ n”)

# Print the data frame.

print (new.address)

# Combine rows form both the data frames.

all.addresses <- rbind (addresses, new.address)

# Print a header.

cat (“# # # The combined data frame \ n”)

# Print the result.

print (all.addresses)

When we run the above code; The following result is obtained:

# # # # The First data frame

city ​​state zipcode

[1,] “Tampa” “FL” “33602”

[2,] “Seattle” “WA” “98104”

[3,] “Hartford” “CT” “6161”

[4,] “Denver” “CO” “80294”

# # # The Second data frame

city ​​state zipcode

1 Lowry CO 80230

2 Charlotte FL 33949

# # # The combined data frame

city ​​state zipcode

1 Tampa FL 33602

2 Seattle WA 98104

3 Hartford CT 6161

4 Denver CO 80294

5 Lowry CO 80230

6 Charlotte FL 33949

Frame data integration

We can merge two data frames using the merge () function  . Frame data must have the same column names in which the merge must be performed.

In the following example, we assume that the Diabetes data set for women in the Indian state of Pima is available in a library called MASS. We combine two datasets based on blood pressure (“bp”) and body mass index (“bmi”). In selecting these two columns to merge; Information that the values ​​of these two variables match in both datasets; Combine to form a single data frame.

library (MASS)

merged.Pima <- merge (x = Pima.te, y = Pima.tr,

by.x = c (“bp”, “bmi”),

by.y = c (“bp”, “bmi”)

)

print (merged.Pima)

nrow (merged.Pima)

When we run the above code; The following result is obtained:

bp bmi npreg.x glu.x skin.x ped.x age.x type.x npreg.y glu.y skin.y ped.y

1 60 33.8 1 117 23 0.466 27 No 2 125 20 0.088

2 64 29.7 2 75 24 0. 370 33 No 2 100 23 0.368

3 64 31.2 5 189 33 0.58 29 Yes 3 158 13 0.295

4 64 33.2 4 117 27 0. 230 24 No 1 96 27 0.289

5 66 38 ٫ 1 3 115 39 0. 150 28 No 1 114 36 0.289

6 68 38.5 2 100 25 0. 324 26 No 7 129 49 0.439

7 70 27.4 1 116 28 0.204 21 No 0 124 20 0.254

8 70 33 ٫ 1 4 91 32 0.446 22 No 9 123 44 0.374

9 70 35.4 9 124 33 0.282 34 No 6 134 23 0.542

10 72 25.6 1 157 21 0.123 24 No 4 99 17 0.294

11 72 37. 7 5 95 33 0. 370 27 No 6 103 32 0.324

12 74 25.9 9 134 33 0.460 81 No 8 126 38 0.162

13 74 25.9 1 95 21 0.673 36 No 8 126 38 0.162

14 78 27.6 5 88 30 0.258 37 No 6 125 31 0.565

15 78 27.6 10 122 31 0.512 45 No 6 125 31 0.565

16 78 39.4 2 112 50 0.175 24 No 4 112 40 0.236

17 88 34.5 1 117 24 0.403 40 Yes 4 127 11 0.598

age.y type.y

1 31 No

۲ ۲۱ No

3 24 No

4 21 No

5 21 No

6 43 Yes

7 36 Yes

8 40 No

9 29 Yes

10 28 No

11 55 No

12 39 No

13 39 No

14 49 Yes

15 49 Yes

16 38 No

17 17 No

[1] 17

One of the most interesting aspects of R programming is the transformation of data into several steps to achieve the desired shape. Functions used in this regard; () is melt and () is cast.

We consider a database in a library called a “MASS”; calls ships:

library (MASS)

print (ships)

When we run the above code; The following result is obtained:

type year period service incidents

1 A 60 60 127 0

2 A 60 75 63 0

3 A 65 60 1095 3

4 A 65 75 1095 4

5 A 70 60 1512 6

………….

………….

8 A 75 75 2244 11

9 B 60 60 44882 39

10 B 60 75 17176 29

11 B 65 60 28609 58

…………

…………

17 C 60 60 1179 1

18 C 60 75 552 1

19 C 65 60 781 0

…………

…………

Melt data (melt)

Now we have melted the data for their so-called organizers and turned all the columns except the type and year into multiple rows:

molten.ships <- melt (ships, id = c (“type”, ”year”))

print (molten.ships)

When we run the above code; The following result is obtained:

type year variable value

1 A 60 period 60

2 A 60 period 75

3 A 65 period 60

4 A 65 period 75

…………

…………

9 B 60 period 60

10 B 60 period 75

11 B 65 period 60

12 B 65 period 75

13 B 70 period 60

……… ..

……… ..

41 A 60 service 127

42 A 60 service 63

43 A 65 service 1095

……… ..

……… ..

70 D 70 service 1208

71 D 75 service 0

72 D 75 service 2051

73 E 60 service 45

74 E 60 service 0

75 E 65 service 789

……… ..

……… ..

101 C 70 incidents 6

102 C 70 incidents 2

103 C 75 incidents 0

104 C 75 incidents 1

105 D 60 incidents 0

106 D 60 incidents 0

……… ..

……… ..

Formatting of fused data

We can format the melted data into a new form in which each type of ship is created for each year. This is done using the cast () function.

recasted.ship <- cast (molten.ships, type + year ~ variable, sum)

print (recasted.ship)

When we run the above code; The following result is obtained:

type year period service incidents

1 A 60 135 190 0

2 A 65 135 2190 7

3 A 70 135 4865 24

4 A 75 135 2244 11

5 B 60 135 62058 68

6 B 65 135 48979 111

7 B 70 135 20163 56

8 B 75 135 7117 18

9 C 60 135 1731 2

10 C 65 135 1457 1

11 C 70 135 2731 8

12 C 75 135 274 1

13 D 60 135 356 0

14 D 65 135 480 0

15 D 70 135 1557 13

16 D 75 135 2051 4

17 E 60 135 45 0

18 E 65 135 1226 14

19 E 70 135 3318 17

20 E 75 135 542 1