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I am using R programming language. Suppose I have the following data ("my_data"):
student first_run second_run third_run fourth_run fifth_run sixth_run seventh_run eight_run ninth_run tenth_run
1 student1 19.70847 21.79771 16.49083 19.51691 13.97987 14.60733 13.89703 15.24651 20.75679 18.44020
2 student2 11.22369 15.36253 16.90215 20.20724 15.90227 15.14539 13.74945 18.30090 19.55124 17.24132
3 student3 15.93649 17.03599 14.20214 13.17548 14.70327 15.49697 13.08945 19.94142 22.41674 17.37958
4 student4 16.18733 15.13197 14.79481 16.75177 14.51287 17.71816 13.45054 14.25553 19.89091 18.88981
5 student5 18.71084 18.85453 17.15864 19.38880 15.68862 18.39169 15.26428 16.04526 18.92532 16.62409
6 student6 19.75246 12.74605 18.52214 17.92626 14.48501 17.20780 13.10512 12.46502 20.68583 15.87711
7 student7 14.75144 23.82376 18.51366 20.77424 14.22155 16.08186 12.95981 12.67820 20.12166 15.66006
8 student8 17.06516 15.63075 13.72026 15.02068 14.21098 15.99414 14.64818 16.15603 21.74607 17.07382
9 student9 20.27611 12.44592 12.26502 15.13456 14.61552 18.72192 15.11129 17.60746 18.83831 17.55257
10 student10 17.70736 16.21620 14.10861 17.20014 16.59376 19.50027 13.05073 15.80002 18.09781 18.34313
I want to add 2 columns to this data:
my_mean : the mean of each row
my_median: the median of each row
I tried the following code in R:
my_data$median = apply(my_data, 1, median, na.rm=T)
my_data$mean = apply(my_data, 1, mean, na.rm=T)
But I don't think this code is correct. For instance, when using this code, the median of the second row of data is returned as "16.90215"
But when I manually take the median of this row:
median(11.22369 , 15.36253 , 16.90215 , 20.20724, 15.90227 , 15.14539 , 13.74945 , 18.30090 , 19.55124 , 17.24132)
I get an answer of
11.22
Can someone please show me what I am doing wrong?
Thanks
The calculation is incorrect i.e. the first argument of median is 'x' which can be a vector. The second argument is na.rm, followed by variadic arguments .... So, when write 11.22369, 15.36253, the 'x' is taken as 11.22369 and that is the value returned. Instead, it should be a vector by concatenation c
median(c(11.22369 , 15.36253 , 16.90215 , 20.20724, 15.90227 , 15.14539 , 13.74945 , 18.30090 , 19.55124 , 17.24132))
[1] 16.40221
Also, based on the OP's data, the first column should be dropped which is character or factor
apply(my_data[-1], 1, median, na.rm=TRUE)
1 2 3 4 5 6 7 8 9 10
17.46551 16.40221 15.71673 15.65965 17.77517 16.54246 15.87096 15.81245 16.34356 16.89695
The second row is used in the manual calculation
library(dplyr)
df %>%
rowwise() %>%
mutate(median = median(c_across(where(is.numeric))),
mean = mean(c_across(where(is.numeric))))
c_across and rowwise were created for this type of situation. Most verbs work column-wise. To change this behavior pipe to rowwise first.
c_across will then combine all values in a row that are numeric (hence where(is.numeric) into a numeric vector and then mean or median can be applied.
Note: You will likely want to pipe the output to ungroup since rowwise creates a rowwise grouped data frame.
Here is an alternative using pmap along with passing all the arguments simultaneously thus using ellipsis i.e. .... The output is needed to be unnested with unnest_wider from tidyr:
library(tidyr)
library(dplyr)
library(purrr)
df %>%
mutate(res = pmap(across(where(is.numeric)),
~ list(median = median(c(...)),
avg = mean(c(...))))) %>%
unnest_wider(res)
output:
student first_run second_run third_run fourth_run fifth_run sixth_run seventh_run eight_run ninth_run tenth_run median avg
<chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 student1 19.7 21.8 16.5 19.5 14.0 14.6 13.9 15.2 20.8 18.4 17.5 17.4
2 student2 11.2 15.4 16.9 20.2 15.9 15.1 13.7 18.3 19.6 17.2 16.4 16.4
3 student3 15.9 17.0 14.2 13.2 14.7 15.5 13.1 19.9 22.4 17.4 15.7 16.3
4 student4 16.2 15.1 14.8 16.8 14.5 17.7 13.5 14.3 19.9 18.9 15.7 16.2
5 student5 18.7 18.9 17.2 19.4 15.7 18.4 15.3 16.0 18.9 16.6 17.8 17.5
6 student6 19.8 12.7 18.5 17.9 14.5 17.2 13.1 12.5 20.7 15.9 16.5 16.3
7 student7 14.8 23.8 18.5 20.8 14.2 16.1 13.0 12.7 20.1 15.7 15.9 17.0
8 student8 17.1 15.6 13.7 15.0 14.2 16.0 14.6 16.2 21.7 17.1 15.8 16.1
9 student9 20.3 12.4 12.3 15.1 14.6 18.7 15.1 17.6 18.8 17.6 16.3 16.3
10 student10 17.7 16.2 14.1 17.2 16.6 19.5 13.1 15.8 18.1 18.3 16.9 16.7
You could definitely benefit from the speed of matrixStats library.
matrixStats::rowMedians(as.matrix(d[-1]))
# [1] 17.46551 16.40221 15.71673 15.65965 17.77517 16.54246 15.87096 15.81245 16.34356 16.89695
matrixStats::rowMeans2(as.matrix(d[-1]))
# [1] 17.44417 16.35862 16.33775 16.15837 17.50521 16.27728 16.95862 16.12661 16.25687 16.66180
stopifnot(all.equal(matrixStats::rowMedians(as.matrix(d[-1])),
as.numeric(apply(d[-1], 1, median, na.rm=T))))
stopifnot(all.equal(matrixStats::rowMeans2(as.matrix(d[-1])),
as.numeric(apply(d[-1], 1, mean, na.rm=T))))
Data:
d <- structure(list(student = c("student1", "student2", "student3",
"student4", "student5", "student6", "student7", "student8", "student9",
"student10"), first_run = c(19.70847, 11.22369, 15.93649, 16.18733,
18.71084, 19.75246, 14.75144, 17.06516, 20.27611, 17.70736),
second_run = c(21.79771, 15.36253, 17.03599, 15.13197, 18.85453,
12.74605, 23.82376, 15.63075, 12.44592, 16.2162), third_run = c(16.49083,
16.90215, 14.20214, 14.79481, 17.15864, 18.52214, 18.51366,
13.72026, 12.26502, 14.10861), fourth_run = c(19.51691, 20.20724,
13.17548, 16.75177, 19.3888, 17.92626, 20.77424, 15.02068,
15.13456, 17.20014), fifth_run = c(13.97987, 15.90227, 14.70327,
14.51287, 15.68862, 14.48501, 14.22155, 14.21098, 14.61552,
16.59376), sixth_run = c(14.60733, 15.14539, 15.49697, 17.71816,
18.39169, 17.2078, 16.08186, 15.99414, 18.72192, 19.50027
), seventh_run = c(13.89703, 13.74945, 13.08945, 13.45054,
15.26428, 13.10512, 12.95981, 14.64818, 15.11129, 13.05073
), eight_run = c(15.24651, 18.3009, 19.94142, 14.25553, 16.04526,
12.46502, 12.6782, 16.15603, 17.60746, 15.80002), ninth_run = c(20.75679,
19.55124, 22.41674, 19.89091, 18.92532, 20.68583, 20.12166,
21.74607, 18.83831, 18.09781), tenth_run = c(18.4402, 17.24132,
17.37958, 18.88981, 16.62409, 15.87711, 15.66006, 17.07382,
17.55257, 18.34313)), class = "data.frame", row.names = c("1",
"2", "3", "4", "5", "6", "7", "8", "9", "10"))
I have a dataframe with 6 features like this:
X1 X2 X3 X4 X5 X6
Modern Dog 9.7 21.0 19.4 7.7 32.0 36.5
Golden Jackal 8.1 16.7 18.3 7.0 30.3 32.9
Chinese Wolf 13.5 27.3 26.8 10.6 41.9 48.1
Indian Wolf 11.5 24.3 24.5 9.3 40.0 44.6
Cuon 10.7 23.5 21.4 8.5 28.8 37.6
Dingo 9.6 22.6 21.1 8.3 34.4 43.1
I want to draw a line plot like this:
I'm trying this:
plot(df$X1, type = "o",col = "red", xlab = "Month", ylab = "Rain fall")
lines(c(df$X2, df$X3, df$X4, df$X5, df$X6), type = "o", col = "blue")
But it's only plotting a single variable. I'm sorry if this question is annoying, i'm totally new to R and i just don't know how to get this done. I would really appreciate any help on this.
Thanks in advance
The easiest way would be to convert your dataset to a long format (e.g. by using the gather function in the tidyr package), and then plotting using the group aesthetic in ggplot.
I recreate your dataset, assuming your group variable is named "Group":
df <- read.table(text = "
Group X1 X2 X3 X4 X5 X6
Modern_Dog 9.7 21.0 19.4 7.7 32.0 36.5
Golden_Jackal 8.1 16.7 18.3 7.0 30.3 32.9
Chinese_Wolf 13.5 27.3 26.8 10.6 41.9 48.1
Indian_Wolf 11.5 24.3 24.5 9.3 40.0 44.6
Cuon 10.7 23.5 21.4 8.5 28.8 37.6
Dingo 9.6 22.6 21.1 8.3 34.4 43.1 ",
header = TRUE, stringsAsFactors = FALSE)
Then convert the dataset to long format and plot:
library(tidyr)
library(ggplot2)
df_long <- df %>% gather(X1:X6, key = "Month", value = "Rainfall")
ggplot(df_long, aes(x = Month, y = Rainfall, group = Group, shape = Group)) +
geom_line() +
geom_point() +
theme(legend.position = "bottom")
See also the answers here: Group data and plot multiple lines.
I am an R beginner user and I face the following problem. I have the following data frame:
distance speed
1 61.0 36.4
2 51.4 35.3
3 42.2 34.2
4 33.4 32.8
5 24.9 31.3
6 17.5 28.4
7 11.5 24.1
8 7.1 19.4
9 3.3 16.9
10 0.5 15.5
11 4.4 15.1
12 8.5 15.5
13 13.1 17.3
14 18.8 20.5
15 25.7 24.1
16 33.3 26.3
17 41.0 27.0
18 48.7 27.7
19 56.6 28.4
20 64.8 29.2
21 73.6 31.7
22 83.3 34.2
23 93.4 35.3
The column distance represents the distance of a following object over a specific point and the column speed the object's speed. As you can see the object is getting closer to the point and then it is getting away. I am trying to make its speed profile. I tried the following code but it didn't give me the plot I want (because I want to show how its speed is changing when the moving object moves closer and past the reference point)
ggplot(speedprofile, aes(x = distance, y = speed)) + #speedprofile is the data frame
geom_line(color = "red") +
geom_smooth() +
geom_vline(xintercept = 0) # the vline is the reference line
The plot is the following:
Then, I tried to set the first 10 distances as negative manually which are prior to zero (0). So I get a plot closer to that I want:
But there is a problem. The distance can't be defined as negative.
To sum up, the expected plot is the following (and I am sorry for the quality).
Do you have any ideas on how to solve this?
Thank you in advance!
You can do something like this to auto-compute the change point (to know when the distance should be negative) and then set the axis labels to be positive.
Your data (in case anyone needs it to answer):
read.table(text="distance speed
61.0 36.4
51.4 35.3
42.2 34.2
33.4 32.8
24.9 31.3
17.5 28.4
11.5 24.1
7.1 19.4
3.3 16.9
0.5 15.5
4.4 15.1
8.5 15.5
13.1 17.3
18.8 20.5
25.7 24.1
33.3 26.3
41.0 27.0
48.7 27.7
56.6 28.4
64.8 29.2
73.6 31.7
83.3 34.2
93.4 35.3", stringsAsFactors=FALSE, header=TRUE) -> speed_profile
Now, compute the "real" distance (negative for approaching, positive for receding):
speed_profile$real_distance <- c(-1, sign(diff(speed_profile$distance))) * speed_profile$distance
Now, compute the X axis breaks ahead of time:
breaks <- scales::pretty_breaks(10)(range(speed_profile$real_distance))
ggplot(speed_profile, aes(real_distance, speed)) +
geom_smooth(linetype = "dashed") +
geom_line(color = "#cb181d", size = 1) +
scale_x_continuous(
name = "distance",
breaks = breaks,
labels = abs(breaks) # make all the labels for the axis positive
)
Provided fonts are working well on your system you could even do:
labels <- abs(breaks)
labels[(!breaks == 0)] <- sprintf("%s\n→", labels[(!breaks == 0)])
ggplot(speed_profile, aes(real_distance, speed)) +
geom_smooth(linetype = "dashed") +
geom_line(color = "#cb181d", size = 1) +
scale_x_continuous(
name = "distance",
breaks = breaks,
labels = labels,
)
I am trying to compare colours in 3D CIELuv colourspace, and I want to identify the L, U, and V values for the colour that is closest to my primary colour of interest. I have calculated the Euclidean distance between each source colour (represented by the three coordinates, L, U, and V, for each colour) and the primary colour (for which I also have the LUV coordinates, not shown for space). The distances between each colour and the primary colour are stored in the three DistCol variables. I then found the smallest of these distances using df$Min.Dist <- colnames(df[c(10:12)])[unlist(apply(df[c(10:12)], 1, which.min))]. Example:
Colour1L Colour1U Colour1V Colour2L Colour2U Colour2V Colour3L Colour3U Colour3V DistCol1 DistCol2 DistCol3 Min.Dist
1 25.5 9.0 -54.5 98.8 0.0 -1.6 63.9 55.0 60.2 25.4 82.1 137.8 DistCol1
2 8.7 14.8 5.6 41.7 133.2 27.6 41.7 133.2 27.6 144.2 58.3 133.3 DistCol2
3 83.2 24.7 -42.7 21.6 -0.4 0.8 83.2 24.7 -42.7 12.1 170.6 102.3 DistCol1
4 55.0 -49.8 62.5 99.2 0.1 -1.8 55.0 -49.8 62.5 213.7 103.4 67.7 DistCol3
I want to use the Min.Dist variable (or any other method really, if there's a better way!) to conditionally select all three L, u, and v values for whichever colour is the closest. That is, in the first row, Min.Dist is DistCol1, so the three Source values would all come from the three Colour1 columns. My final output would ideally look like:
Colour1L Colour1U Colour1V Colour2L Colour2U Colour2V Colour3L Colour3U Colour3V DistCol1 DistCol2 DistCol3 Min.Dist SourceL SourceU SourceV
1 25.5 9.0 -54.5 98.8 0.0 -1.6 63.9 55.0 60.2 25.4 82.1 137.8 DistCol1 25.5 9.0 -54.5
2 8.7 14.8 5.6 41.7 133.2 27.6 41.7 133.2 27.6 144.2 58.3 133.3 DistCol2 41.7 133.2 27.6
3 83.2 24.7 -42.7 21.6 -0.4 0.8 83.2 24.7 -42.7 12.1 170.6 102.3 DistCol1 83.2 24.7 -42.7
4 55.0 -49.8 62.5 99.2 0.1 -1.8 55.0 -49.8 62.5 213.7 103.4 67.7 DistCol3 55.0 -49.8 62.5
I have previously obtained a similar result using a long nested ifelse expression for each of the L, U, and V dimensions e.g. df$SourceL <- ifelse(df$Min.Dist =="DistCol1", Colour1L, ifelse(df$Min.Dist == "DistCol2", Colour2L, ifelse(... but I'm dealing with 8-10 colours in my real data and this is extremely tedious and prone to error.
I apologise if this question has already been answered elsewhere, and would very much appreciate any advice or direction to a resource for this. Thank you as well to everyone who answers questions on this forum - your advice has been invaluable for solving many R problems over the past months!
Doing this in a non-vectorized way with base R:
rebuilding your data.frame:
df <- data.frame(c(25.5,8.7,83.2,55),c(9,14.8,24.7,-49.8),c(-54.5,5.6,-42.7,62.5), c(98.8,41.7,21.6,99.2),c(0,133.2,-0.4,0.1),c(-1.6,27.6,0.8,-1.8),c(63.9,41.7,83.2,55),c(55,133.2,24.7,-49.8),c(60.2,27.6,-42.7,62.5),c(25.4,144.2,12.1,213.7),c(82.1,58.3,170.6,103.4),c(137.8,133.3,102.3,67.7),c("DistCol1","DistCol2","DistCol1","DistCol3"))
colnames(df) <- c("Colour1L", "Colour1U", "Colour1V", "Colour2L", "Colour2U", "Colour2V", "Colour3L", "Colour3U", "Colour3V", "DistCol1", "DistCol2", "DistCol3", "Min.Dist")
Looping over rows
for (i in 1:length(df$Colour1L)) {
df$SourceL[i] <- df[[paste0("Colour",substr(df$Min.Dist,8,8)[i],"L")]][i]
df$SourceU[i] <- df[[paste0("Colour",substr(df$Min.Dist,8,8)[i],"U")]][i]
df$SourceV[i] <- df[[paste0("Colour",substr(df$Min.Dist,8,8)[i],"V")]][i]
}
I have the following data:
a=c(1:10)
b=c(16:25)
c=c(24:33)
wa=c(3,7,3,3,3,3,3,3,3,1)
wb=c(3,2,3,3,3,3,3,3,3,8)
wc=c(4,1,4,4,4,4,4,4,4,1)
z=data.frame(a,b,c,wa,wb,wc)
I want to get the weighted mean for each record. So I tried this:
weight=apply(subset(z,select=c(wa,wb,wc)),1,function(x) x)
z$weightMean=apply(subset(z,select=c(a,b,c)),1,function(x) weighted.mean(x,weight))
Which returned the following error message:
Error in weighted.mean.default(x, weight) :
'x' and 'w' must have the same length
So then I tried this:
weight=as.vector(weight)
z$weightMean=apply(subset(z,select=c(a,b,c)),1,function(x) weighted.mean(x,weight))
Which also returned the same error.
What am I doing wrong?
This seems to do the trick:
> apply(z, 1, function(x) weighted.mean(x[1:3], x[4:6]))
[1] 14.7 7.3 16.7 17.7 18.7 19.7 20.7 21.7 22.7 24.3
This will probably be a bit faster, though less clear as to what's going on:
> rowSums(z[,1:3] * z[,4:6]) / rowSums(z[,4:6])
[1] 14.7 7.3 16.7 17.7 18.7 19.7 20.7 21.7 22.7 24.3