sankara.deva2016@gmail.com
bharat sreeram , Ventech It Solutions
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> info =
read.csv("C:/Users/Hadoop/Desktop/info.txt")
> info
id name sal sex city
1 101 Amar 40000 m hyd
2 102 Amala 50000 f del
3 103 sunil 70000 m hyd
4 104 sunitha 80000 f hyd
5 105 ankith 90000 m del
6 106 ankitha 60000 f hyd
>
> class(info)
[1] "data.frame"
> str(info)
-- structure of info.
'data.frame': 6 obs. of 5 variables:
$ id : int 101 102 103 104 105 106
$ name: Factor w/ 6 levels "Amala","Amar",..: 2 1 5 6 3 4
$ sal : int 40000 50000 70000 80000 90000 60000
$ sex : Factor w/ 2 levels "f","m": 2 1 2 1 2 1
$ city: Factor w/ 2 levels "del","hyd": 2 1 2 2 1 2
>
> info$name
[1] Amar Amala sunil sunitha ankith ankitha
Levels: Amala Amar ankith ankitha sunil sunitha
> info$sal
[1] 40000 50000 70000 80000 90000 60000
>
-------------------
Updating Column:
info$sal = info$sal+1000
Generating new Field for Data frame.
info$tax = info$sal*10/100
# conditional transformation,.
info$grade = "C"
info$grade[info$sal>=50000 & info$sal<=80000] = "B"
info$grade[info$sal>80000]="A"
> info
id name sal sex city tax grade
1 101 Amar 41000 m hyd 4100 C
2 102 Amala 51000 f del 5100 B
3 103 sunil 71000 m hyd 7100 B
4 104 sunitha 81000 f hyd 8100 A
5 105 ankith 91000 m del 9100 A
6 106 ankitha 61000 f hyd 6100 B
>
Grouping Aggregations:
--------------------------------
res1 =
aggregate( sal ~ sex, data=info, FUN=sum)
> res1
sex sal
1 f 193000
2 m 203000
>
res2 =
aggregate( sal ~ sex, data=info, FUN=mean)
> res2
sex sal
1 f 64333.33
2 m 67666.67
res3 =
aggregate( sal ~ sex, data=info, FUN=max)
> res3
sex sal
1 f 81000
2 m 91000
res4 =
aggregate( sal ~ sex, data=info, FUN=min)
> res4
sex sal
1 f 51000
2 m 41000
res5 =
aggregate( sal ~ sex, data=info, FUN=length)
> res5
sex sal
1 f 3
2 m 3
---------------------------
res6 = aggregate(sal ~ grade, data=info,
FUN = length)
> res6
grade sal
1 A 2
2 B 3
3 C 1
-----------------------------------
select sex, grade, sum(sal) from info
group by sex, grade;
Grouping by multiple columns of data frame.
res7 =
aggregate( sal ~ sex + grade, data=info, FUN=sum)
> res7
sex grade sal
1 f A 81000
2 m A 91000
3 f B 112000
4 m B 71000
5 m C 41000
-----------------------------------------
select sex, sum(sal), max(sal),
min(sal), avg(sal), count(*)
from info group by sex;
# performing multiple aggregation,
of each data group.
Performing Multiple Aggregations:
r1 = aggregate(sal ~ sex , data=info , FUN=sum)
r2 = aggregate(sal ~ sex , data=info , FUN=max)
r3 = aggregate(sal ~ sex , data=info , FUN=min)
r4 = aggregate(sal ~ sex , data=info , FUN=mean)
r5 = aggregate(sal ~ sex , data=info , FUN=length)
resall = data.frame(sex=r1$sex,
tot=r1$sal, avg=r4$sal,
max= r2$sal, min=r3$sal,
cnt=r5$sal)
> resall
sex tot avg max min cnt
1 f 193000 64333.33 81000 51000 3
2 m 203000 67666.67 91000 41000 3
>
--------------------------------------
Working With Matrices.
v = 1:15
m1 = matrix(v, nrow=3)
m2 = matrix(v, nrow=3 ,byrow=T)
m3 = matrix(v, ncol=5, byrow=T)
m4 = matrix(1:12,4)
Operations of matrices.
m = matrix(c(10,20,30,40), nrow=2)
mx = matrix(c(12,34,56,23), nrow=2)
r1 = m + mx # element to element sum
r2 = m * mx # " " multiplication
r3 = m %*% mx # matrix multiplication
r4 = solve(m) # inverse of matrix.
r5 = t(m) # transpose of matrix.
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