I have a table with the results of a survey. I mean, I don't have the full survey data, only the frequency tables, Likert scale counts. Is it possible to graph this table in R?
The table looks like this:
Question 1: (response is a factor)
response count
1 5
2 6
3 2
4 2
5 1
It's very easy to do it in Excel, but i can't in R. The only thing I could think of was to repeat the table values based on the count, but there must be a simpler way...
A histogram-like chart from plot
plot(df,type = "h")
Try this ggplot2 approach. You can set your response as x variable and count as y variable and use geom_col() in order to display bars. Here the code:
library(ggplot2)
#Plot
ggplot(df,aes(x=factor(response),y=count))+
geom_col(color='black',fill='cyan3')+
xlab('Response')
Output:
Some data used:
#Data
df <- structure(list(response = 1:5, count = c(5L, 6L, 2L, 2L, 1L)), class = "data.frame", row.names = c(NA,
-5L))
If we wanted to do this without having to install any package, use the base R methods with either a named vector in a single-line
barplot(setNames(df$count, df$response))
Or with the formula method for data.frame
barplot(count ~ response, df)
-output
data
df <- structure(list(response = 1:5, count = c(5L, 6L, 2L, 2L, 1L)),
class = "data.frame", row.names = c(NA,
-5L))
Related
Hi I have been trying for a while to match two large columns of names, several have different spellings etc... so far I have written some code to practice on a smaller dataset
examples%>% mutate(new_ID = case_when(mapply (adist, example_1 , example_2) <= 3 ~ example_1, TRUE ~ example_2))
This manages to create a new column with names the name from example 1 if it is less than an edit distance of 3 away. However, it does not give the name from example 2 if it does not meet this criteria which I need it to do.
This code also only works on the adjacent row of each column, whereas, I need it to work on a dataset which has two columns (one is larger- so cant be put in the same order).
Also needs to not try to match the NAs from the smaller column of names (there to fill it out to equal length to the other one).
Anyone know how to do something like this?
dput(head(examples))
structure(list(. = structure(c(4L, 3L, 2L, 1L, 5L), .Label = c("grarryfieldsred","harroldfrankknight", "sandramaymeres", "sheilaovensnew", "terrifrank"), class = "factor"), example_2 = structure(c(4L, 2L, 3L, 1L,
5L), .Label = c(" grarryfieldsred", "candramymars", "haroldfranrinight",
"sheilowansknew", "terryfrenk"), class = "factor")), row.names = c(NA,
5L), class = "data.frame")
The problem is that your columns have become factors rather than character vectors. When you try to combine two columns together with different factor levels, unexpected results can happen.
First convert your columns to character:
library(dplyr)
examples %>%
mutate(across(contains("example"),as.character)) %>%
mutate(new_ID = case_when(mapply (adist, example_1 , example_2) <= 3 ~ example_1,
TRUE ~ example_2))
# example_1 example_2 new_ID
#1 sheilaovensnew sheilowansknew sheilowansknew
#2 sandramaymeres candramymars candramymars
#3 harroldfrankknight haroldfranrinight harroldfrankknight
#4 grarryfieldsred grarryfieldsred grarryfieldsred
#5 terrifrank terryfrenk terrifrank
In your dput output, somehow the name of example_1 was changed. I ran this first:
names(examples)[1] <- "example_1"
I got a dataframe to work on where I have a bunch of variables as factors in quotation marks like ""x1"".
str(df) gives me something like this:
$ x : Factor w/ 10 Levels "\"\"x1\"\"",..: 1 7 9 ...
I tried to get rid of the quotation marks with the gsub() function but that didn´t work. Probably because I don´t know what to insert as pattern? Would be great if somebody can solve this puzzle and maybe explain to me if the "\"\"x1\"\"" is the solution to this?
An example for the dataframe would look like this:
structure(list(Sent = structure(c(2L, 2L, 2L, 2L, 2L), .Label = c("\"\"Opted out\"\"",
"\"\"Yes\"\""), class = "factor"), Responded = structure(c(2L,
2L, 2L, 2L, 2L), .Label = c("\"\"Complete\"\"", "\"\"No\"\"",
"\"\"Partial\"\""), class = "factor")), row.names = c(NA, -5L
), class = c("tbl_df", "tbl", "data.frame"), .Names = c("Sent",
"Responded"))
Thanks in advance!
vec = c('""x1""', '""x2""', '""x3""')
vec = factor(vec)
levels(vec) <- gsub('["\\]', "", levels(vec))
#> vec
#[1] x1 x2 x3
#Levels: x1 x2 x3
See how I would use ' as wrapper, when I want to use " inside a string.
Another problem it didn't work for you was probably because you didn't use the levels attribute but rather the factor variable itself.
Factor variables are internally stored as 1, 2, 3,... numbers.
As you now have provided data, you can use: (df1 being your data with the factor columns)
df1[] <- lapply(df1, function(vec){ levels(vec) <- gsub('["\\]',"",levels(vec)); vec})
I have a large dataframe and I have a vector to pull out terms of interest. for a previous project I was using:
a=data[data$rn %in% y, "Gene"]
To pull out information into a new vector. Now I have a another job Id like to do.
I have a large dataframe of 15 columns and >100000 rows. I want to search column 3 and 9 for the content in the vector and print this as a new dataframe.
To make this extra annoying the hit could be in v3 and not in v9 and visa versa.
Working example
I have striped the dataframe to 3 cols and few rows.
data <- structure(list(Gene = structure(c(1L, 5L, 3L, 2L, 4L), .Label = c("ibp","leuA", "pLeuDn_02", "repA", "repA1"), class = "factor"), LocusTag = structure(c(1L,2L, 5L, 3L, 4L), .Label = c("pBPS1_01", "pBPS1_02", "pleuBTgp4","pleuBTgp5", "pLeuDn_02"), class = "factor"), hit = structure(c(2L,4L, 3L, 1L, 5L), .Label = c("2-isopropylmalate synthase", "Ibp protein","ORF1", "repA1 protein", "replication-associated protein"), class = "factor")), .Names = c("Gene","LocusTag", "hit"), row.names = c(NA, 5L), class = "data.frame")
y <- c("ibp", "orf1")
First of all R is case sensitive so your example will not collect the third line but I guess you want that extracted. so you would have to change your y to
y <- c("ibp", "ORF1")
Ok from your example I try to see what you want to achieve I am not sure if this is really what you want but R knows the operator | as "or" so you could try something like:
new.data<-data[data$Gene %in% y|data$hit %in% y,]
if you only want to extract certain columns of your data set you can specify them behind the "," e.g.:
new.data<-data[data$Gene %in% y|data$hit %in% y, c("LocusTag","Gene")]
I'm having an issue using apply functions (which I assume is the right way to do the following) across multiple data frames.
Some example data (3 different data frames, but the problem I'm working on has upwards of 50):
biz <- data.frame(
country = c("england","canada","australia","usa"),
businesses = sample(1000:2500,4))
pop <- data.frame(
country = c("england","canada","australia","usa"),
population = sample(10000:20000,4))
restaurants <- data.frame(
country = c("england","canada","australia","usa"),
restaurants = sample(500:1000,4))
Here's what I ultimately want to do:
1) Sort eat data frame from largest to smallest, according to the variable that's included
dataframe <- dataframe[order(dataframe$VARIABLE,)]
2) then create a vector variable that gives me the rank for each
dataframe$rank <- 1:nrow(dataframe)
3) Then create another data frame that has one column of the countries and the rank for each of the variables of interest as other columns. Something that would look like (rankings aren't real here):
country.rankings <- structure(list(country = structure(c(5L, 1L, 6L, 2L, 3L, 4L), .Label = c("brazil",
"canada", "england", "france", "ghana", "usa"), class = "factor"),
restaurants = 1:6, businesses = c(4L, 5L, 6L, 3L, 2L, 1L),
population = c(4L, 6L, 3L, 2L, 5L, 1L)), .Names = c("country",
"restaurants", "businesses", "population"), class = "data.frame", row.names = c(NA,
-6L))
So I'm guessing there's a way to put each of these data frames together into a list, something like:
lib <- c(biz, pop, restaurants)
And then do an lapply across that to 1) sort, 2)create the rank variable and 3) create the matrix or data frame of rankings for each variable (# of businesses, population size, # of restaurants) for each country. Problem I'm running into is that writing the lapply function to sort each data frame runs into issues when I try to order by the variable:
sort <- lapply(lib,
function(x){
x <- x[order(x[,2]),]
})
returns the error message:
Error in `[.default`(x, , 2) : incorrect number of dimensions
because I'm trying to apply column headings to a list. But how else would I tackle this problem when the variable names are different for every data frame (but keeping in mind that the country names are consistent)
(would also love to know how to use this using plyr)
Ideally I'd would recommend data.table for this.
However, here is a quick solution using data.frame
Try this:
Step1: Create a list of all data.frames
varList <- list(biz,pop,restaurants)
Step2: Combine all of them in one data.frame
temp <- varList[[1]]
for(i in 2:length(varList)) temp <- merge(temp,varList[[i]],by = "country")
Step3: Get ranks:
cbind(temp,apply(temp[,-1],2,rank))
You can remove the undesired columns if you want!!
cbind(temp[,1:2],apply(temp[,-1],2,rank))[,-2]
Hope this helps!!
totaldatasets <- c('biz','pop','restaurants')
totaldatasetslist <- vector(mode = "list",length = length(totaldatasets))
for ( i in seq(length(totaldatasets)))
{
totaldatasetslist[[i]] <- get(totaldatasets[i])
}
totaldatasetslist2 <- lapply(
totaldatasetslist,
function(x)
{
temp <- data.frame(
country = totaldatasetslist[[i]][,1],
countryrank = rank(totaldatasetslist[[i]][,2])
)
colnames(temp) <- c('country', colnames(x)[2])
return(temp)
}
)
Reduce(
merge,
totaldatasetslist2
)
Output -
country businesses population restaurants
1 australia 3 3 3
2 canada 2 2 2
3 england 1 1 1
4 usa 4 4 4
I have a dataframe consisting of an ID, that is the same for each element in a group, two datetimes and the time interval between these two. One of the datetime objects is my relevant time marker. Now I like to get a subset of the dataframe that consists of the earliest entry for each group. The entries (especially the time interval) need to stay untouched.
My first approach was to sort the frame according to 1. ID and 2. relevant datetime. However, I wasn't able to return the first entry for each new group.
I then have been looking at the aggregate() as well as ddply() function but I could not find an option in both that just returns the first entry without applying an aggregation function to the time interval value.
Is there an (easy) way to accomplish this?
ADDITION:
Maybe I was unclear by adding my aggregate() and ddply() notes. I do not necessarily need to aggregate. Given the fact that the dataframe is sorted in a way that the first row of each new group is the row I am looking for, it would suffice to just return a subset with each row that has a different ID than the one before (which is the start-row of each new group).
Example data:
structure(list(ID = c(1454L, 1322L, 1454L, 1454L, 1855L, 1669L,
1727L, 1727L, 1488L), Line = structure(c(2L, 1L, 3L, 1L, 1L,
1L, 1L, 1L, 1L), .Label = c("A", "B", "C"), class = "factor"),
Start = structure(c(1357038060, 1357221074, 1357369644, 1357834170,
1357913412, 1358151763, 1358691675, 1358789411, 1359538400
), class = c("POSIXct", "POSIXt"), tzone = ""), End = structure(c(1357110430,
1357365312, 1357564413, 1358230679, 1357978810, 1358674600,
1358853933, 1359531923, 1359568151), class = c("POSIXct",
"POSIXt"), tzone = ""), Interval = c(1206.16666666667, 2403.96666666667,
3246.15, 6608.48333333333, 1089.96666666667, 8713.95, 2704.3,
12375.2, 495.85)), .Names = c("ID", "Line", "Start", "End",
"Interval"), row.names = c(NA, -9L), class = "data.frame")
By reproducing the example data frame and testing it I found a way of getting the needed result:
Order data by relevant columns (ID, Start)
ordered_data <- data[order(data$ID, data$Start),]
Find the first row for each new ID
final <- ordered_data[!duplicated(ordered_data$ID),]
As you don't provide any data, here is an example using base R with a sample data frame :
df <- data.frame(group=c("a", "b"), value=1:8)
## Order the data frame with the variable of interest
df <- df[order(df$value),]
## Aggregate
aggregate(df, list(df$group), FUN=head, 1)
EDIT : As Ananda suggests in his comment, the following call to aggregate is better :
aggregate(.~group, df, FUN=head, 1)
If you prefer to use plyr, you can replace aggregate with ddply :
ddply(df, "group", head, 1)
Using ffirst from collapse
library(collapse)
ffirst(df, g = df$group)
data
df <- data.frame(group=c("a", "b"), value=1:8)
This could also be achieved by dplyr using group_by and slice-family of functions,
data %>%
group_by(ID) %>%
slice_head(n = 1)