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I am trying to use the aggregate function in R to summarise a data using the length function. My data has some NA's and I have tried using 'na.rm = T' or 'na.omit' however none sees to work. I keep getting this error
'Error in FUN(X[[i]], ...) :
2 arguments passed to 'length' which requires 1'
data10 <- structure(list(Group = c(1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 2,
1, 2, 2, 1, 2, 2, 2, 1, 2, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2,
1), SUBJECT = c(1, 1, 2, 3, 3, 4, 5, 5, 6, 7, 8, 8, 9, 10, 10,
11, 12, 14, 14, 15, 16, 16, 17, 18, 19, 19, 20, 21, 21, 22, 23,
23, 24, 25), test = c(1, 2, 1, 1, 2, 2, 1, 2, 2, 1, 1, 2, 2,
1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 2, 1
), trial = c(1, 3, 5, 7, 1, 3, 5, 7, 1, 3, 5, 7, 1, 3, 5, 7,
1, 3, 5, 7, 1, 3, 5, 7, 1, 3, 5, 7, 1, 3, 5, 7, 1, 3), Condition = c(1,
2, 3, 1, 3, 1, 2, 3, 2, 3, 1, 2, 1, 2, 3, 1, 3, 1, 2, 3, 2, 3,
1, 2, 1, 2, 3, 1, 3, 1, 2, 3, 2, 3), Sac2 = c(1, 1, 1, NA, 2,
1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1,
1, 1, 1, 1, 2, 2, 1, 1), Sac = c(1, 1, 1, NA, 3, 1, 1, 1, 1, 3,
1, 1, 1, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 3,
3, 1, 1), Saccade...8 = c(1, 1, 1, NA, 2, 1, 1, 1, 1, 2, 1, 1,
1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 2, 1,
1), T_APPEAR = c(9.236, 17.85, 28.942, 63.724, 9.463, 22.963,
52.068, 57.021, 15.344, 19.783, 37.825, 46.17, 4.339, 21.241,
29.179, 31.823, 12.164, 22.84, 23.954, 73.663, 27.269, 22.131,
30.361, 62.674, 6.928, 16.413, 47.555, 48.893, 7.291, 15.796,
31.788, 54.946, 10.117, 28.83)), row.names = c(NA, -34L), class = c("tbl_df",
"tbl", "data.frame"))
data14 = aggregate(data10,
by = list(data10$SUBJECT,data10$Condition, data10$Group, data10$test),
FUN = length(), na.rm=TRUE)
What is the best way to represent the following trait rating scale? I'd like to label the traits (8 traits) and degrees or each emotion (1 being low feelings, 5 being strong feelings), across the democratic and republican parties? Do I need to aggregate the items? I'm new to R and not sure how to tackle this.
Survey question and scale:
"Below is a list of feelings or moods that could be caused by an object. Please use the list below to describe how the U.S. FEDERAL parties (and its elected officials) make you feel. If the word definitely describes how a party makes you feel, then choose the number 5. If you decide that the word does not at all describe how the party makes you feel, then choose the number 1. Use the intermediate numbers between 1 and 5 to indicate responses between these two extremes."
Survey sample:
dput(df[Book3(1:nrow(df), 30),])
structure(list(TRAITDEM1 = c(3, 4, 3, 3, 3, 3, 3, 1, 2, 2, 2,
3, 3, 2, 2, 1, 1, 3, 1, 5, 1, 1, 3, 1, 4, 4, 3, 1, 2, 4), TRAITDEM2 = c(3,
1, 1, 2, 2, 2, 3, 5, 4, 2, 2, 2, 3, 3, 3, 4, 1, 2, 3, 1, 4, 5,
2, 3, 1, 1, 1, 4, 1, 2), TRAITDEM3 = c(3, 4, 4, 2, 3, 3, 3, 1,
1, 2, 2, 3, 3, 2, 2, 1, 1, 3, 1, 5, 1, 1, 3, 1, 4, 5, 4, 1, 3,
5), TRAITDEM4 = c(3, 2, 1, 2, 2, 2, 4, 5, 4, 5, 2, 3, 2, 3, 3,
4, 3, 4, 3, 1, 5, 4, 1, 4, 3, 4, 2, 4, 2, 1), TRAITDEM5 = c(3,
4, 3, 4, 4, 3, 2, 1, 1, 2, 2, 3, 4, 2, 2, 1, 1, 3, 1, 5, 1, 1,
2, 1, 4, 4, 4, 1, 3, 4), TRAITDEM6 = c(3, 1, 1, 1, 1, 1, 1, 2,
1, 1, 1, 2, 2, 2, 2, 4, 3, 1, 1, 1, 4, 5, 1, 3, 1, 1, 1, 1, 1,
1), TRAITDEM7 = c(3, 1, 3, 3, 2, 2, 1, 1, 1, 2, 3, 4, 3, 2, 2,
1, 1, 2, 2, 5, 1, 1, 1, 3, 3, 4, 2, 1, 5, 5), TRAITDEM8 = c(3,
1, 1, 1, 2, 1, 3, 5, 2, 4, 1, 1, 2, 2, 3, 1, 3, 1, 2, 1, 5, 5,
2, 2, 1, 2, 1, 2, 1, 1), TRAITREP1 = c(1, 1, 1, 1, 1, 1, 1, 1,
1, 4, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1,
1), TRAITREP2 = c(1, 5, 5, 5, 5, 5, 5, 2, 5, 2, 5, 5, 5, 5, 4,
5, 1, 5, 5, 5, 5, 1, 5, 4, 5, 5, 5, 3, 5, 5), TRAITREP3 = c(1,
1, 1, 1, 2, 1, 1, 2, 1, 4, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 3,
1, 1, 1, 1, 1, 1, 1, 2), TRAITREP4 = c(1, 5, 5, 1, 5, 5, 5, 3,
5, 2, 5, 4, 5, 5, 5, 5, 3, 5, 5, 5, 5, 1, 5, 3, 5, 5, 5, 4, 5,
1), TRAITREP5 = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 2,
1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1), TRAITREP6 = c(1,
5, 5, 5, 3, 3, 3, 1, 1, 1, 3, 3, 5, 3, 4, 5, 3, 4, 5, 4, 5, 1,
5, 3, 4, 4, 5, 1, 1, 3), TRAITREP7 = c(1, 1, 1, 1, 2, 2, 1, 1,
1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1,
2), TRAITREP8 = c(1, 5, 5, 5, 4, 5, 5, 2, 5, 2, 5, 4, 5, 5, 4,
1, 3, 5, 5, 5, 5, 3, 4, 4, 5, 5, 5, 3, 5, 5), PARTYID_Strength = c(5,
1, 2, 1, 2, 1, 8, 7, 6, 3, 1, 6, 6, 1, 7, 8, 7, 1, 1, 1, 2, 4,
1, 6, 1, 1, 1, 7, 6, 8)), row.names = c(NA, -30L), class = c("tbl_df",
"tbl", "data.frame"))
"PartyID_Strength" represents 8 measures of political parties:
1 - Strong Democrat
2 - Not very strong Democrat
3 - Strong Republican
4 - Not very strong Republican
5 - Independent
6 - Independent - Democrat
7 - Independent - Republican
8 - Other
I tried it this way (graph below) but it's still not plotting the remaining four traits:
Cleaning the data
In order to solve your problem, we have to transform your data, in order to convert it into tidy format.
Observation
There are few particular problems with your original dataset:
Data are in a wide format, i.e. most of the columns from your data frame, can be represented by 3 variables;
Names of the variables are not self-explanatory. Names are in upper case which, by itself, does not hold any useful information, they are not readable and not good for typing/writing.
There is additional information we can extract from the variable names: Party and Feelings toward the Party. First one is an abbreviation ('dem' or 'rep') second one is the numerically encoded feeling towards the political party. However the order of numbers encoding the feeling does not reflect natural order of emotions from the disgust up to joy;
Variable PARTYID_Strength is numerically encoded Political Party [self-]Identification it also does not reflect natural order from strongest democrats through independent towards strongest republicans;
Plan
Convert data from wide into long format using all variables starting with TRAIT, and leaving PARTYID_Strength variable unchanged;
Extract useful information from the TRAIT... variables (Political Party, Feelings Toward the Party);
Convert all numerically encoded variables into the factors with reasonably ordered levels;
Give all variables meaningful names;
Summarize the data;
Transformations
We need to create several lookup tables, which will simplify the workflow.
Affiliation lookup table:
aff_lookup <- c(
'Strong Democrat',
'Not very strong Democrat',
'Strong Republican',
'Not very strong Republican',
'Independent',
'Independent-Democrat',
'Independent-Republican',
'Other'
)
We can further order aff_lookup by this vector:
aff_order = c(1, 2, 6, 5, 7, 4, 3, 8)
Emotions/Feelings lookup table:
emo_lookup <- c(
'Delighted',
'Angry',
'Happy',
'Annoyed',
'Joy',
'Hateful',
'Relaxed',
'Disgusted'
)
And we can order emo_lookup by this vector:
emo_order <- emo_order <- c(8, 6, 2, 4, 7, 3, 1, 5)
Political party lookup table:
party_lookup <- c(
dem = 'National Democratic Party',
rep = 'National Republican Party'
)
Finally, with all helper variables, we can transform our data into desirable form.
library(tidyverse)
dat %<>%
rename_all(tolower) %>%
pivot_longer(
cols = starts_with('trait'),
names_to = c('party', 'emotion'),
names_pattern = 'trait(dem|rep)(\\d)',
values_to = 'score'
) %>%
mutate(
party = factor(party_lookup[party]),
affiliation = factor(
aff_lookup[partyid_strength],
levels = aff_lookup[aff_order]
),
emotion = factor(
emo_lookup[as.numeric(emotion)],
levels = emo_lookup[emo_order]
)
) %>%
group_by(party, emotion, affiliation) %>%
summarise(score = median(score)) %>%
ungroup()
head(dat)
## A tibble: 6 x 4
# party emotion affiliation score
# <fct> <fct> <fct> <dbl>
#1 National Democratic Party Disgusted Strong Democrat 1
#2 National Democratic Party Disgusted Not very strong Democrat 2
#3 National Democratic Party Disgusted Independent-Democrat 2
#4 National Democratic Party Disgusted Independent 3
#5 National Democratic Party Disgusted Independent-Republican 3
#6 National Democratic Party Disgusted Not very strong Republican 5
Plot the data
Plan
Now we can plot the data, as two separate plots for Democrats and Republicans with Affiliation (Political Party Identification) on X-axis and Emotions (Feelings) on Y-axis.
Each Emotion/Affilation point is going to be represented as a bar with the height of the bar representing the Score.
We can also add color encoding to our plot. From my point of view, encoding Emotions/Feelings with a color gradient from red (Disgust) to green (Joy) could help as to gather the internal structure of our data.
Plot
dat %>%
ggplot(
aes(
x = affiliation,
y = as.numeric(emotion) + (score / max(score) * .95) / 2,
height = (score / max(score) * .95),
width = .95,
fill = emotion,
label = score
)
) +
geom_tile(show.legend = FALSE) +
geom_text(size = 3.5, color = 'gray25', alpha = .75) +
facet_wrap(~ party, scales = 'free') +
scale_fill_brewer(palette = 'RdYlGn') +
scale_y_continuous(breaks = sort(emo_order), labels = emo_lookup[emo_order]) +
labs(x = 'Affiliations', y = 'Emotions') +
ggthemes::theme_tufte() +
theme(
axis.text.x = element_text(angle = 45, hjust = 1),
axis.ticks.x = element_blank(),
axis.text.y = element_text(hjust = 0, vjust = -0.025),
axis.ticks.y = element_blank()
)
Which gives as following figure:
Explanation
There is a trick with this plot: it looks like a series of barplots, bot it is not real barplots (by the fact, not functionally).
What I do:
The core of this solution is the use of geom_tile() for each data point. It is just a rectangle (square by default) with geometrical center of mass determined by the given coordinates (Affilation, Emotion).
Both Affilation and Emotion are factors, not numerics. And it is OK for Affiliation, because we want only to position our tile according to the Affiliation it represents.
It is more complicated with Emotion, because we want to position each tile according to the Emotion it represents, but also we want to encode Score by the height of the tile.
To define the height of the tile we use height parameter within the aes(). We want our tile height to be less or equall to one (with 0.05 offset) so the tiles between let say Angry and Annoyed do not overlap. That's why we use (score / max(score) * .95 for the height parameter.
We also need to give different y-coordinates for each tile, so the center of the tile is placed not on the imaginary line representing each emotion, but half-height up. So when tile is drawn, it's center (on y-axis) is placed half-height up from the "base line" and the tile extends half-height up and down, creating a fake barplot. That's what the following line of code does as.numeric(emotion) + (score / max(score) * .95) / 2.
We also give a tile a fixed width of .95 by width = .95, file the tile with Red-Yellow-Green gradient and lable each tile with the relevant Score.
The rest are just decorations. However, note how we relable the Y-axis. Because, as it defined in aes() it is continuous scale, but we want to make it fake discrete axis we use this row:
scale_y_continuous(breaks = sort(emo_order), labels = emo_lookup[emo_order])
Here we just use our emo_order to say that we want breaks for integers from 1 to 8, and after that we label this breaks with feelings from ordered emo_lookup table.
I have the following three dimensional array:
dput(a)
structure(c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 3, 2, 1, 1, 1, 2, 2,
2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 6, 2, 7, 6, 2, 7, 6, 2, 7, 4, 2, 4, 4, 2, 6, 4, 2, 4, 6, 2,
7, 4, 2, 6, 4, 2, 6, 4, 2, 6, 4, 2, 4, 4, 2, 6, 4, 2, 4, 4, 2,
6, 4, 2, 6, 4, 2, 6, 6, 2, 7, 4, 2, 6, 4, 2, 6, 4, 2, 4, 2, 3,
1, 2, 3, 1, 2, 3, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 3, 7, 2, 3,
7, 2, 3, 7, 2, 3, 7, 2, 3, 7, 2, 3, 7, 2, 3, 7, 2, 3, 7, 2, 3,
7, 2, 3, 7, 1, 2, 5, 2, 3, 7, 1, 2, 4, 2, 3, 7, 2, 3, 7, 2, 3,
7, 2, 3, 7, 2, 3, 7, 2, 3, 7, 2, 3, 7, 2, 6, 3, 2, 6, 3, 2, 6,
3, 2, 6, 3, 2, 6, 3, 2, 6, 3, 2, 6, 3, 2, 6, 3, 2, 6, 3, 2, 6,
3, 1, 1, 1, 2, 6, 3, 1, 5, 5, 2, 6, 3, 2, 6, 3, 2, 6, 3, 2, 6,
3, 2, 6, 3, 2, 6, 3, 2, 6, 3, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 13,
2, 3, 13, 2, 3, 5, 2, 3, 5, 2, 15, 17, 2, 15, 17, 2, 15, 17,
2, 3, 5, 2, 15, 17, 2, 3, 13, 2, 15, 17, 2, 15, 17, 2, 3, 13,
2, 3, 5, 2, 15, 17, 2, 15, 17, 2, 3, 5, 2), .Dim = c(3L, 20L,
6L), .Dimnames = list(c("cl.tmp", "cl.tmp", "cl.tmp"), NULL,
NULL))
The dimension of this array (a) is 3x20x6 (after edits).
I wanted to count the proportion of times that a[,i,] matches a[,j,] element-by-element in the matrix. Basically, I wanted to get mean(a[,i,] == a[,j,]) for all i, j, and I would like to do this fast but in R.
It occurred to me that the outer function might be a possibility but I am not sure how to specify the function. Any suggestions, or any other alternative ways?
The output would be a 20x20 symmetric matrix of nonnegative elements with 1 on the diagonals.
The solution given below works (thanks!) but I have one further question (sorry).
I would like to display the coordinates above in a heatmap. I try the following:
n<-dim(a)[2]
xx <- matrix(apply(a[,rep(1:n,n),]==a[,rep(1:n,each=n),],2,sum),nrow=n)/prod(dim(a)[-2])
image(1:20, 1:20, xx, xlab = "", ylab = "")
This gives me the following heatmap.
However, I would like to display (reorder the coordinate) such that I get all the coordinates that have high-values amongst each other together. However, I would not like to bias the results by deciding on the number of groups myself. I tried
hc <- hclust(as.dist(1-xx), method = "single")
but I can not decide how to cut the resulting tree to decide on bunching the coordinates together. Any suggestions? Bascically, in the figure, I would like the coordinate pairs in the top left (and bottom right off-diagonal blocks) to be as low-valued (in this case as red) as possible.
Looking around on SO, I found that there exists a function heatmap which might do this,
heatmap(xx,Colv=T,Rowv=T, scale='none',symm = T)
and I get the following:
which is all right, but I can not figure out how to get rid of the dendrograms on the sides or the axes labels. It does work if I extract out and do the following:
yy <- heatmap(xx,Colv=T,Rowv=T, scale='none',symm = T,keep.dendro=F)
image(1:20, 1:20, xx[yy$rowInd,yy$colInd], xlab = "", ylab = "")
so I guess that is what I will stick with. Here is the result:
Try this:
n<-dim(a)[2]
matrix(apply(a[,rep(1:n,n),]==a[,rep(1:n,each=n),],2,sum),nrow=n)/prod(dim(a)[-2])
It has to be stressed that the memory usage of this method goes with n^2 so you might have trouble to use it with larger arrays.
I have a data frame with a column "Tag", here with four different levels. I need help to create the "Seq" column, a sequence generated from the "Tag" Column:
df <- data.frame(Tag = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4),
Seq = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3 )
Each "Tag" should be divided into 3 sub-groups defined by "Seq". We need to generate runs of 1, 2, and 3, with a total length of that of each "Tag". Thus, the length of each run of 1, 2, and 3 respectively depends on length of each "Tag".
Note that the length each "Tag" differs. For example, Tag 1 is of length 31, and has a "Seq" 10 times 1, 10 times 2, and 11 times 3.
To begin with, Tag 1 is 31 while tag 2 is 32. Looking at the code below, the first number (1) will always be of lesser length than the next two (2,3). I used a ceiling process to come up with this. There is no clear criteria on what the code should do if the number is eg 31/3.. should it give a length of 10, 10, 11? or even 9, 11,11 will be fine? The code gives a 9, 11, 11 length:
ec=table(Tag)
unlist(mapply(function(x,y)rep(c(1,2,3),c(x,y,y)),ec-2*ceiling(ec/3),ceiling(ec/3)))
To check the outputted results, save the results in a variable.. d=mapply(...
then do sapply(d,table).
Hope this will be of help.
ave(Tag, Tag, FUN = function(x){sort(rep(x = 1:3, length.out = length(x)))})
Explanation: For each level of "Tag" (ave(Tag, Tag, ...): repeat each level of "Seq" (x = 1:3) to the length of the subset of "Tag" (length.out = length(x)). sort the numbers.
I have a data set where number denotes a particular color. Since i have large data set, i am sharing a sample data and work. I am looking forward to create this
Output
d <- matrix(c(
1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1,
1, 2, 2, 2, 2, 1,
1, 2, 3, 3, 2, 1,
1, 2, 3, 3, 2, 1,
1, 2, 3, 3, 2, 1,
1, 2, 3, 3, 2, 1,
1, 2, 2, 2, 2, 1,
1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1), nrow=13, byrow = TRUE)
from this Input :
d_mi <- d
d_mi[ sample(1:length(d), length(d)*0.3) ] <- NA
d_mi
Optional : Output(in color mode)