adding new column to data frame in R - r

rate len ADT trks sigs1 slim shld lane acpt itg lwid hwy
1 4.58 4.99 69 8 0.20040080 55 10 8 4.6 1.20 12 FAI
2 2.86 16.11 73 8 0.06207325 60 10 4 4.4 1.43 12 FAI
3 3.02 9.75 49 10 0.10256410 60 10 4 4.7 1.54 12 FAI
4 2.29 10.65 61 13 0.09389671 65 10 6 3.8 0.94 12 FAI
5 1.61 20.01 28 12 0.04997501 70 10 4 2.2 0.65 12 FAI
6 6.87 5.97 30 6 2.00750419 55 10 4 24.8 0.34 12 PA
7 3.85 8.57 46 8 0.81668611 55 8 4 11.0 0.47 12 PA
8 6.12 5.24 25 9 0.57083969 55 10 4 18.5 0.38 12 PA
9 3.29 15.79 43 12 1.45333122 50 4 4 7.5 0.95 12 PA
I got a question in adding a new column, my data frame is called highway1,and i want to add a column named S/N, as slim divided by acpt, what can I do?
Thanks

> mydf$SN <- mydf$slim/mydf$acpt
> mydf
rate len ADT trks sigs1 slim shld lane acpt itg lwid hwy SN
1 4.58 4.99 69 8 0.20040080 55 10 8 4.6 1.20 12 FAI 11.956522
2 2.86 16.11 73 8 0.06207325 60 10 4 4.4 1.43 12 FAI 13.636364
3 3.02 9.75 49 10 0.10256410 60 10 4 4.7 1.54 12 FAI 12.765957
4 2.29 10.65 61 13 0.09389671 65 10 6 3.8 0.94 12 FAI 17.105263
5 1.61 20.01 28 12 0.04997501 70 10 4 2.2 0.65 12 FAI 31.818182
6 6.87 5.97 30 6 2.00750419 55 10 4 24.8 0.34 12 PA 2.217742
7 3.85 8.57 46 8 0.81668611 55 8 4 11.0 0.47 12 PA 5.000000
8 6.12 5.24 25 9 0.57083969 55 10 4 18.5 0.38 12 PA 2.972973
9 3.29 15.79 43 12 1.45333122 50 4 4 7.5 0.95 12 PA 6.666667
I hope an explanation is not necessary for the above.

While $ is the preferred route, you can also consider cbind.
First, create the numeric vector and assign it to SN:
SN <- Data[,6]/Data[,9]
Now you use cbind to append the numeric vector as a column to the existing data frame:
Data <- cbind(Data, SN)
Again, using the dollar operator $ is preferred, but it doesn't hurt seeing what an alternative looks like.

Related

How to sample data non-random

I have weather dataset my data is date-dependent
I want to predict the temperature from 07 May 2008 until 18 May 2008 (which is maybe a total of 10-15 observations) my data size is around 200
I will be using decision tree/RF and SVM & NN to make my prediction
I've never handled data like this so I'm not sure how to sample non random data
I want to sample data 80% train data and 30% test data but I want to sample the data in the original order not randomly. Is that possible ?
install.packages("rattle")
install.packages("RGtk2")
library("rattle")
seed <- 42
set.seed(seed)
fname <- system.file("csv", "weather.csv", package = "rattle")
dataset <- read.csv(fname, encoding = "UTF-8")
dataset <- dataset[1:200,]
dataset <- dataset[order(dataset$Date),]
set.seed(321)
sample_data = sample(nrow(dataset), nrow(dataset)*.8)
test<-dataset[sample_data,] # 30%
train<-dataset[-sample_data,] # 80%
output
> head(dataset)
Date Location MinTemp MaxTemp Rainfall Evaporation Sunshine WindGustDir WindGustSpeed
1 2007-11-01 Canberra 8.0 24.3 0.0 3.4 6.3 NW 30
2 2007-11-02 Canberra 14.0 26.9 3.6 4.4 9.7 ENE 39
3 2007-11-03 Canberra 13.7 23.4 3.6 5.8 3.3 NW 85
4 2007-11-04 Canberra 13.3 15.5 39.8 7.2 9.1 NW 54
5 2007-11-05 Canberra 7.6 16.1 2.8 5.6 10.6 SSE 50
6 2007-11-06 Canberra 6.2 16.9 0.0 5.8 8.2 SE 44
WindDir9am WindDir3pm WindSpeed9am WindSpeed3pm Humidity9am Humidity3pm Pressure9am
1 SW NW 6 20 68 29 1019.7
2 E W 4 17 80 36 1012.4
3 N NNE 6 6 82 69 1009.5
4 WNW W 30 24 62 56 1005.5
5 SSE ESE 20 28 68 49 1018.3
6 SE E 20 24 70 57 1023.8
Pressure3pm Cloud9am Cloud3pm Temp9am Temp3pm RainToday RISK_MM RainTomorrow
1 1015.0 7 7 14.4 23.6 No 3.6 Yes
2 1008.4 5 3 17.5 25.7 Yes 3.6 Yes
3 1007.2 8 7 15.4 20.2 Yes 39.8 Yes
4 1007.0 2 7 13.5 14.1 Yes 2.8 Yes
5 1018.5 7 7 11.1 15.4 Yes 0.0 No
6 1021.7 7 5 10.9 14.8 No 0.2 No
> head(test)
Date Location MinTemp MaxTemp Rainfall Evaporation Sunshine WindGustDir WindGustSpeed
182 2008-04-30 Canberra -1.8 14.8 0.0 1.4 7.0 N 28
77 2008-01-16 Canberra 17.9 33.2 0.0 10.4 8.4 N 59
88 2008-01-27 Canberra 13.2 31.3 0.0 6.6 11.6 WSW 46
58 2007-12-28 Canberra 15.1 28.3 14.4 8.8 13.2 NNW 28
96 2008-02-04 Canberra 18.2 22.6 1.8 8.0 0.0 ENE 33
126 2008-03-05 Canberra 12.0 27.6 0.0 6.0 11.0 E 46
WindDir9am WindDir3pm WindSpeed9am WindSpeed3pm Humidity9am Humidity3pm Pressure9am
182 E N 2 19 80 40 1024.2
77 N NNE 15 20 58 62 1008.5
88 N WNW 4 26 71 28 1013.1
58 NNW NW 6 13 73 44 1016.8
96 SSE ENE 7 13 92 76 1014.4
126 SSE WSW 7 6 69 35 1025.5
Pressure3pm Cloud9am Cloud3pm Temp9am Temp3pm RainToday RISK_MM RainTomorrow
182 1020.5 1 7 5.3 13.9 No 0.0 No
77 1006.1 6 7 24.5 23.5 No 4.8 Yes
88 1009.5 1 4 19.7 30.7 No 0.0 No
58 1013.4 1 5 18.3 27.4 Yes 0.0 No
96 1011.5 8 8 18.5 22.1 Yes 9.0 Yes
126 1022.2 1 1 15.7 26.2 No 0.0 No
> head(train)
Date Location MinTemp MaxTemp Rainfall Evaporation Sunshine WindGustDir WindGustSpeed
7 2007-11-07 Canberra 6.1 18.2 0.2 4.2 8.4 SE 43
9 2007-11-09 Canberra 8.8 19.5 0.0 4.0 4.1 S 48
11 2007-11-11 Canberra 9.1 25.2 0.0 4.2 11.9 N 30
16 2007-11-16 Canberra 12.4 32.1 0.0 8.4 11.1 E 46
22 2007-11-22 Canberra 16.4 19.4 0.4 9.2 0.0 E 26
25 2007-11-25 Canberra 15.4 28.4 0.0 4.4 8.1 ENE 33
WindDir9am WindDir3pm WindSpeed9am WindSpeed3pm Humidity9am Humidity3pm Pressure9am
7 SE ESE 19 26 63 47 1024.6
9 E ENE 19 17 70 48 1026.1
11 SE NW 6 9 74 34 1024.4
16 SE WSW 7 9 70 22 1017.9
22 ENE E 6 11 88 72 1010.7
25 SSE NE 9 15 85 31 1022.4
Pressure3pm Cloud9am Cloud3pm Temp9am Temp3pm RainToday RISK_MM RainTomorrow
7 1022.2 4 6 12.4 17.3 No 0.0 No
9 1022.7 7 7 14.1 18.9 No 16.2 Yes
11 1021.1 1 2 14.6 24.0 No 0.2 No
16 1012.8 0 3 19.1 30.7 No 0.0 No
22 1008.9 8 8 16.5 18.3 No 25.8 Yes
25 1018.6 8 2 16.8 27.3 No 0.0 No
I use mtcars as an example. An option to non-randomly split your data in train and test is to first create a sample size based on the number of rows in your data. After that you can use split to split the data exact at the 80% of your data. You using the following code:
smp_size <- floor(0.80 * nrow(mtcars))
split <- split(mtcars, rep(1:2, each = smp_size))
With the following code you can turn the split in train and test:
train <- split$`1`
test <- split$`2`
Let's check the number of rows:
> nrow(train)
[1] 25
> nrow(test)
[1] 7
Now the data is split in train and test without losing their order.

Scraping an interactive table with rvest

I'm attempting to scrape the second table shown at the URL below, and I'm running into issues which may be related to the interactive nature of the table.
div_stats_standard appears to refer to the table of interest.
The code runs with no errors but returns an empty list.
url <- 'https://fbref.com/en/comps/9/stats/Premier-League-Stats'
data <- url %>%
read_html() %>%
html_nodes(xpath = '//*[(#id = "div_stats_standard")]') %>%
html_table()
Can anyone tell me where I'm going wrong?
Look for the table.
library(rvest)
url <- "https://fbref.com/en/comps/9/stats/Premier-League-Stats"
page <- read_html(url)
nodes <- html_nodes(page, "table") # you can use Selectorgadget to identify the node
table <- html_table(nodes[[1]]) # each element of the nodes list is one table that can be extracted
head(table)
Result:
head(table)
Playing Time Playing Time Playing Time Performance Performance
1 Squad # Pl MP Starts Min Gls Ast
2 Arsenal 26 27 297 2,430 39 26
3 Aston Villa 28 27 297 2,430 33 27
4 Bournemouth 25 28 308 2,520 27 17
5 Brighton 23 28 308 2,520 28 19
6 Burnley 21 28 308 2,520 32 23
Performance Performance Performance Performance Per 90 Minutes Per 90 Minutes
1 PK PKatt CrdY CrdR Gls Ast
2 2 2 64 3 1.44 0.96
3 1 3 54 1 1.22 1.00
4 1 1 60 3 0.96 0.61
5 1 1 44 2 1.00 0.68
6 2 2 53 0 1.14 0.82
Per 90 Minutes Per 90 Minutes Per 90 Minutes Expected Expected Expected Per 90 Minutes
1 G+A G-PK G+A-PK xG npxG xA xG
2 2.41 1.37 2.33 35.0 33.5 21.3 1.30
3 2.22 1.19 2.19 30.6 28.2 22.0 1.13
4 1.57 0.93 1.54 31.2 30.5 20.8 1.12
5 1.68 0.96 1.64 33.8 33.1 22.4 1.21
6 1.96 1.07 1.89 30.9 29.4 18.9 1.10
Per 90 Minutes Per 90 Minutes Per 90 Minutes Per 90 Minutes
1 xA xG+xA npxG npxG+xA
2 0.79 2.09 1.24 2.03
3 0.81 1.95 1.04 1.86
4 0.74 1.86 1.09 1.83
5 0.80 2.01 1.18 1.98
6 0.68 1.78 1.05 1.73

Error in producing the output

I have problem with my code. I can't trace the error. I have coor data (40 by 2 matrix) as below and a rainfall data (14610 by 40 matrix).
No Longitude Latitude
1 100.69 6.34
2 100.77 6.24
3 100.39 6.11
4 100.43 5.53
5 100.39 5.38
6 101.00 5.71
7 101.06 5.30
8 100.80 4.98
9 101.17 4.48
10 102.26 6.11
11 102.22 5.79
12 102.28 5.31
13 102.02 5.38
14 101.97 4.88
15 102.95 5.53
16 103.13 5.32
17 103.06 4.94
18 103.42 4.76
19 103.42 4.23
20 102.38 4.24
21 101.94 4.23
22 103.04 3.92
23 103.36 3.56
24 102.66 3.03
25 103.19 2.89
26 101.35 3.70
27 101.41 3.37
28 101.75 3.16
29 101.39 2.93
30 102.07 3.09
31 102.51 2.72
32 102.26 2.76
33 101.96 2.74
34 102.19 2.36
35 102.49 2.29
36 103.02 2.38
37 103.74 2.26
38 103.97 1.85
39 103.72 1.76
40 103.75 1.47
rainfall= 14610 by 40 matrix;
coor= 40 by 2 matrix
my_prog=function(rainrain,coordinat,misss,distance)
{
rain3<-rainrain # target station i**
# neighboring stations for target station i
a=coordinat # target station i**
diss=as.matrix(distHaversine(a,coor,r=6371))
mmdis=sort(diss,decreasing=F,index.return=T)
mdis=as.matrix(mmdis$x)
mdis1=as.matrix(mmdis$ix)
dist=cbind(mdis,mdis1)
# NA creation
# create missing values in rainfall data
set.seed(100)
b=sample(1:nrow(rain3),(misss*nrow(rain3)),replace=F)
k=replace(rain3,b,NA)
# pick i closest stations
neig=mdis1[distance] # neighbouring selection distance
# target (with NA) and their neighbors
rainB=rainfal00[,neig]
rainA0=rainB[,2:ncol(rainB)]
rainA<-as.matrix(cbind(k,rainA0))
rain2=na.omit(rainA)
x=as.matrix(rain2[,1]) # used to calculate the correlation
n1=ncol(rainA)-1
#1) normal ratio(nr)
jum=as.matrix(apply(rain2,2,mean))
nr0=(jum[1]/jum)
nr=as.matrix(nr0[2:nrow(nr0),])
m01=as.matrix(rainA[is.na(k),])
m1=m01[,2:ncol(m01)]
out1=as.matrix(sapply(seq_len(nrow(m1)),
function(i) sum(nr*m1[i,],na.rm=T)/n1))
print(out1)
}
impute=my_prog(rainrain=rainfall[,1],coordinat=coor[1,],misss=0.05,distance=mdis<200)
I have run this code and and the output obtained is:
Error in my_prog(rainrain = rainfal00[, 1], misss = 0.05, coordinat = coor[1, :
object 'mdis' not found
I have checked the program, but cannot trace the problem. I would really appreciate if someone could help me.

How to change a column classed as NULL to class integer?

So I'm starting with a dataframe called max.mins that has 153 rows.
day Tx Hx Tn
1 1 10.0 7.83 2.1
2 2 7.7 6.19 2.5
3 3 7.1 4.86 0.0
4 4 9.8 7.37 2.7
5 5 13.4 12.68 0.4
6 6 17.5 17.47 3.5
7 7 16.5 15.58 6.5
8 8 21.5 20.30 6.2
9 9 21.7 21.41 9.7
10 10 24.4 28.18 8.0
I'm applying these statements to the dataframe to look for specific criteria
temp_warnings <- subset(max.mins, Tx >= 32 & Tn >=20)
humidex_warnings <- subset(max.mins, Hx >= 40)
Now when I open up humidex_warnings for example I have this dataframe
row.names day Tx Hx Tn
1 41 10 31.1 40.51 20.7
2 56 25 33.4 42.53 19.6
3 72 11 34.1 40.78 18.1
4 73 12 33.8 40.18 18.8
5 74 13 34.1 41.10 22.4
6 79 18 30.3 41.57 22.5
7 94 2 31.4 40.81 20.3
8 96 4 30.7 40.39 20.2
The next step is to search for 2 or 3 consective numbers in the column row.names and give me a total of how many times this occurs (I asked this in a previous question and have a function that should work once this problem is sorted out). The issue is that row.names is class NULL which is preventing me from applying further functions to this dataframe.
Help? :)
Thanks in advance,
Nick
If you need the row.names as a data as integer:
humidex_warnings$seq <- as.integer(row.names(humidex_warnings))
If you don't need row.names
row.names(humidex_warnings) <- NULL

R program - getting particular values depending on another column

So I have data regarding Id number and time
Id number Time(hr)
1 5
2 6.1
3 7.2
4 8.3
5 9.6
6 10.9
7 13
8 15.1
9 17.2
10 19.3
11 21.4
12 23.5
13 25.6
14 27.1
15 28.6
16 30.1
17 31.8
18 33.5
19 35.2
20 36.9
21 38.6
22 40.3
23 42
24 43.7
25 45.4
I want this output
Time Id number
10 5
20 10
30 16
40 22
So I want the time to be in 10 hour intervals and get the ID that corresponds to that particular hour...I decided to use this code data <- data2[seq(0, nrow(data2), by=5), ] but instead of the Time being in 10 hr intervals...the ID number is at 10 intervals....but I dont want that output..so far I'm getting this output
Id.number Time..s.
10 19.3
20 36.9
You can use %% (mod) operator.
data[data$Time %% 10 == 0, ]
I use cut() and cumsum(table()) but I don't quite get the answer you are expecting. How exactly are you calculating this?
# first load the data
v.txt <- '1 5
2 6.1
3 7.2
4 8.3
5 9.6
6 10.9
7 13
8 15.1
9 17.2
10 19.3
11 21.4
12 23.5
13 25.6
14 27.1
15 28.6
16 30.1
17 31.8
18 33.5
19 35.2
20 36.9
21 38.6
22 40.3
23 42
24 43.7
25 45.4'
# load in the data... awkwardly...
v <- as.data.frame(matrix(as.numeric(unlist(strsplit(strsplit(v.txt, '\n')[[1]], ' +'))), byrow=T, ncol=2))
tens <- seq(from=0, by=10, to=100)
v$cut <- cut(v$Time, tens, labels=tens[-1])
v2 <- as.data.frame(cumsum(table(v$cut)))
names(v2) <- 'Time'
v2$Id <- rownames(v2)
rownames(v2) <- 1:nrow(v2)
v2 <- v2[,c(2,1)]
rm(v, v.txt, tens) # not needed anymore
v2 # the answer... but doesn't quite match your expected answer...
Id Time
1 10 5
2 20 10
3 30 15
4 40 21
5 50 25

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