I am scrapping data from a website, and in this context, data tidying gets kind of hard.
what I have right now is a string of numbers that go into a sequence, let's say
a<-c(1,2,3,1,2,3,4,5,1,2,3,4)
The first value that I'm looking for is 3, the second one is 5, and the third one will be 4.
So basically, I want to go through the sequence 1:5 and choose the highest value, to have the final output as
a<-c(3,4,5)
I thought about choosing the maximum values, such as
a<-sort(a, decreasing = T)
a<-a[1:3]
But this won't count, cause the final product is:
[1] 5 4 4
where the small values are discriminated. Any ideas if this could be possible?
not entirely sure if this is what you're asking for. i think what you're wanting is to see which of your values you have in your vector.
try this:
a<-c(1,2,3,1,2,3,4,5,1,2,3,4)
search_values = 3:5
# unique values
search_values = a[a %in% search_values]
unique(search_values)
# counts of values
table(search_values)
sort(unlist(lapply(split(a, cumsum(c(1, diff(a)) != 1)), max), use.names = FALSE))
#[1] 3 4 5
Sounds like you want something like this?
a <- c(1,2,3,1,2,3,4,5,1,2,3,4) # Data input
a <- unique(a) # Keep unique numbers
a <- sort(a, dec = F) # Sort ascending
tail(a, 3) # Last three numbers in set
Gives:
[1] 3 4 5
In one line:
tail(sort(unique(a), dec = F), 3)
Related
Generate 1000 random samples of 4 digits with replacement from the population of digits (pop). Plot the distribution of the sample proportion of 1’s using a Histogram.
I've tried this, but don't believe that it is right:
set.seed(123)
pop <- c(0,1,2,3,4,5,6,7,8,9)
y<-replicate(1000,sample(pop, size=4, replace = TRUE), simplify = FALSE)
d<-length(y)
Taking the observation 1 from vector x and storing it in variable b
e<-x[x==1]
The number of ones stored in vector x stored in vector c
f<-length(e)
Finding the proportion of 1s in vector x
prop<-f/d
prop
Additionally, I've tried to make a histogram out of this code, but it won't return anything for some reason. any ideas?
Your code is actually correct:
set.seed(123)
pop <- c(0,1,2,3,4,5,6,7,8,9)
y<-replicate(1000,sample(pop, size=4, replace = TRUE), simplify = FALSE)
This is a list, because you specify, simplify = FALSE, check by doing:
class(y)
[1] "list"
You can see how it's structured:
y[1:3]
[[1]]
[1] 2 2 9 1
[[2]]
[1] 5 4 3 5
[[3]]
[1] 8 9 4 2
If you want to find the proportion etc, you unlist the list you have:
y = unlist(y)
For proportion of 1s, do:
mean(y==1)
[1] 0.09575
To get a histogram, do:
hist(y,col="steelblue",breaks=seq(-0.5,9.5,by=1))
You need to specify breaks in this case because the default is to include the left most, which will give you something really weird. Or in your situation something below is also good:
barplot(table(y))
I am not completely sure if this is what you wish to accomplish, but I think it might be really close.
library("purrr")
#Your population, vector with integers
pop <- 1:40
#get a vector of 1000 repetitions of sampling pop by 4 elements with replacement
df <- unlist(1:1000 %>%
map(function(x) sample(pop,4,replace = T)))
#Get the proportion of ones in the df
prop <- length(df[df == 1]) / length(df)
#For example
# 0.02375
I have a data frame named "mydata" that looks like this this:
A B C D
1. 5 4 4 4
2. 5 4 4 4
3. 5 4 4 4
4. 5 4 4 4
5. 5 4 4 4
6. 5 4 4 4
7. 5 4 4 4
I'd like to delete row 2,4,6. For example, like this:
A B C D
1. 5 4 4 4
3. 5 4 4 4
5. 5 4 4 4
7. 5 4 4 4
The key idea is you form a set of the rows you want to remove, and keep the complement of that set.
In R, the complement of a set is given by the '-' operator.
So, assuming the data.frame is called myData:
myData[-c(2, 4, 6), ] # notice the -
Of course, don't forget to "reassign" myData if you wanted to drop those rows entirely---otherwise, R just prints the results.
myData <- myData[-c(2, 4, 6), ]
You can also work with a so called boolean vector, aka logical:
row_to_keep = c(TRUE, FALSE, TRUE, FALSE, TRUE, FALSE, TRUE)
myData = myData[row_to_keep,]
Note that the ! operator acts as a NOT, i.e. !TRUE == FALSE:
myData = myData[!row_to_keep,]
This seems a bit cumbersome in comparison to #mrwab's answer (+1 btw :)), but a logical vector can be generated on the fly, e.g. where a column value exceeds a certain value:
myData = myData[myData$A > 4,]
myData = myData[!myData$A > 4,] # equal to myData[myData$A <= 4,]
You can transform a boolean vector to a vector of indices:
row_to_keep = which(myData$A > 4)
Finally, a very neat trick is that you can use this kind of subsetting not only for extraction, but also for assignment:
myData$A[myData$A > 4,] <- NA
where column A is assigned NA (not a number) where A exceeds 4.
Problems with deleting by row number
For quick and dirty analyses, you can delete rows of a data.frame by number as per the top answer. I.e.,
newdata <- myData[-c(2, 4, 6), ]
However, if you are trying to write a robust data analysis script, you should generally avoid deleting rows by numeric position. This is because the order of the rows in your data may change in the future. A general principle of a data.frame or database tables is that the order of the rows should not matter. If the order does matter, this should be encoded in an actual variable in the data.frame.
For example, imagine you imported a dataset and deleted rows by numeric position after inspecting the data and identifying the row numbers of the rows that you wanted to delete. However, at some later point, you go into the raw data and have a look around and reorder the data. Your row deletion code will now delete the wrong rows, and worse, you are unlikely to get any errors warning you that this has occurred.
Better strategy
A better strategy is to delete rows based on substantive and stable properties of the row. For example, if you had an id column variable that uniquely identifies each case, you could use that.
newdata <- myData[ !(myData$id %in% c(2,4,6)), ]
Other times, you will have a formal exclusion criteria that could be specified, and you could use one of the many subsetting tools in R to exclude cases based on that rule.
Create id column in your data frame or use any column name to identify the row. Using index is not fair to delete.
Use subset function to create new frame.
updated_myData <- subset(myData, id!= 6)
print (updated_myData)
updated_myData <- subset(myData, id %in% c(1, 3, 5, 7))
print (updated_myData)
By simplified sequence :
mydata[-(1:3 * 2), ]
By sequence :
mydata[seq(1, nrow(mydata), by = 2) , ]
By negative sequence :
mydata[-seq(2, nrow(mydata), by = 2) , ]
Or if you want to subset by selecting odd numbers:
mydata[which(1:nrow(mydata) %% 2 == 1) , ]
Or if you want to subset by selecting odd numbers, version 2:
mydata[which(1:nrow(mydata) %% 2 != 0) , ]
Or if you want to subset by filtering even numbers out:
mydata[!which(1:nrow(mydata) %% 2 == 0) , ]
Or if you want to subset by filtering even numbers out, version 2:
mydata[!which(1:nrow(mydata) %% 2 != 1) , ]
For completeness, I'll add that this can be done with dplyr as well using slice. The advantage of using this is that it can be part of a piped workflow.
df <- df %>%
.
.
slice(-c(2, 4, 6)) %>%
.
.
Of course, you can also use it without pipes.
df <- slice(df, -c(2, 4, 6))
The "not vector" format, -c(2, 4, 6) means to get everything that is not at rows 2, 4 and 6. For an example using a range, let's say you wanted to remove the first 5 rows, you could do slice(df, 6:n()). For more examples, see the docs.
Delete Dan from employee.data - No need to manage a new data.frame.
employee.data <- subset(employee.data, name!="Dan")
Here's a quick and dirty function to remove a row by index.
removeRowByIndex <- function(x, row_index) {
nr <- nrow(x)
if (nr < row_index) {
print('row_index exceeds number of rows')
} else if (row_index == 1)
{
return(x[2:nr, ])
} else if (row_index == nr) {
return(x[1:(nr - 1), ])
} else {
return (x[c(1:(row_index - 1), (row_index + 1):nr), ])
}
}
It's main flaw is it the row_index argument doesn't follow the R pattern of being a vector of values. There may be other problems as I only spent a couple of minutes writing and testing it, and have only started using R in the last few weeks. Any comments and improvements on this would be very welcome!
To identify by a name:
Call out the unique ID and identify the location in your data frame (DF).
Mark to delete. If the unique ID applies to multiple rows, all these rows will be removed.
Code:
Rows<-which(grepl("unique ID", DF$Column))
DF2<-DF[-c(Rows),]
DF2
Another approach when working with Unique IDs is to subset data:
*This came from an actual report where I wanted to remove the chemical standard
Chem.Report<-subset(Chem.Report, Chem_ID!="Standard")
Chem_ID is the column name.
The ! is important for excluding
Another question for me as a beginner. Consider this example here:
n = c(2, 3, 5)
s = c("ABBA", "ABA", "STING")
b = c(TRUE, "STING", "STRING")
df = data.frame(n,s,b)
n s b
1 2 ABBA TRUE
2 3 ABA STING
3 5 STING STRING
How can I search within this dataframe for similar strings, i.e. ABBA and ABA as well as STING and STRING and make them the same (doesn't matter whether ABBA or ABA, either fine) that would not require me knowing any variations? My actual data.frame is very big so that it would not be possible to know all the different variations.
I would want something like this returned:
> n = c(2, 3, 5)
> s = c("ABBA", "ABBA", "STING")
> b = c(TRUE, "STING", "STING")
> df = data.frame(n,s,b)
> print(df)
n s b
1 2 ABBA TRUE
2 3 ABBA STING
3 5 STING STING
I have looked around for agrep, or stringdist, but those refer to two data.frames or are able to name the column which I can't since I have many of those.
Anyone an idea? Many thanks!
Best regards,
Steffi
This worked for me but there might be a better solution
The idea is to use a recursive function, special, that uses agrepl, which is the logical version of approximate grep, https://www.rdocumentation.org/packages/base/versions/3.4.1/topics/agrep. Note that you can specify the 'error tolerance' to group similar strings with agrep. Using agrepl, I split off rows with similar strings into x, mutate the s column to the first-occurring string, and then add a grouping variable grp. The remaining rows that were not included in the ith group are stored in y and recursively passed through the function until y is empty.
You need the dplyr package, install.packages("dplyr")
library(dplyr)
desired <- NULL
grp <- 1
special <- function(x, y, grp) {
if (nrow(y) < 1) { # if y is empty return data
return(x)
} else {
similar <- agrepl(y$s[1], y$s) # find similar occurring strings
x <- rbind(x, y[similar,] %>% mutate(s=head(s,1)) %>% mutate(grp=grp))
y <- setdiff(y, y[similar,])
special(x, y, grp+1)
}
}
desired <- special(desired,df,grp)
To change the stringency of string similarity, change max.distance like agrepl(x,y,max.distance=0.5)
Output
n s b grp
1 2 ABBA TRUE 1
2 3 ABBA STING 1
3 5 STING STRING 2
To remove the grouping variable
withoutgrp <- desired %>% select(-grp)
I have a data frame named "mydata" that looks like this this:
A B C D
1. 5 4 4 4
2. 5 4 4 4
3. 5 4 4 4
4. 5 4 4 4
5. 5 4 4 4
6. 5 4 4 4
7. 5 4 4 4
I'd like to delete row 2,4,6. For example, like this:
A B C D
1. 5 4 4 4
3. 5 4 4 4
5. 5 4 4 4
7. 5 4 4 4
The key idea is you form a set of the rows you want to remove, and keep the complement of that set.
In R, the complement of a set is given by the '-' operator.
So, assuming the data.frame is called myData:
myData[-c(2, 4, 6), ] # notice the -
Of course, don't forget to "reassign" myData if you wanted to drop those rows entirely---otherwise, R just prints the results.
myData <- myData[-c(2, 4, 6), ]
You can also work with a so called boolean vector, aka logical:
row_to_keep = c(TRUE, FALSE, TRUE, FALSE, TRUE, FALSE, TRUE)
myData = myData[row_to_keep,]
Note that the ! operator acts as a NOT, i.e. !TRUE == FALSE:
myData = myData[!row_to_keep,]
This seems a bit cumbersome in comparison to #mrwab's answer (+1 btw :)), but a logical vector can be generated on the fly, e.g. where a column value exceeds a certain value:
myData = myData[myData$A > 4,]
myData = myData[!myData$A > 4,] # equal to myData[myData$A <= 4,]
You can transform a boolean vector to a vector of indices:
row_to_keep = which(myData$A > 4)
Finally, a very neat trick is that you can use this kind of subsetting not only for extraction, but also for assignment:
myData$A[myData$A > 4,] <- NA
where column A is assigned NA (not a number) where A exceeds 4.
Problems with deleting by row number
For quick and dirty analyses, you can delete rows of a data.frame by number as per the top answer. I.e.,
newdata <- myData[-c(2, 4, 6), ]
However, if you are trying to write a robust data analysis script, you should generally avoid deleting rows by numeric position. This is because the order of the rows in your data may change in the future. A general principle of a data.frame or database tables is that the order of the rows should not matter. If the order does matter, this should be encoded in an actual variable in the data.frame.
For example, imagine you imported a dataset and deleted rows by numeric position after inspecting the data and identifying the row numbers of the rows that you wanted to delete. However, at some later point, you go into the raw data and have a look around and reorder the data. Your row deletion code will now delete the wrong rows, and worse, you are unlikely to get any errors warning you that this has occurred.
Better strategy
A better strategy is to delete rows based on substantive and stable properties of the row. For example, if you had an id column variable that uniquely identifies each case, you could use that.
newdata <- myData[ !(myData$id %in% c(2,4,6)), ]
Other times, you will have a formal exclusion criteria that could be specified, and you could use one of the many subsetting tools in R to exclude cases based on that rule.
Create id column in your data frame or use any column name to identify the row. Using index is not fair to delete.
Use subset function to create new frame.
updated_myData <- subset(myData, id!= 6)
print (updated_myData)
updated_myData <- subset(myData, id %in% c(1, 3, 5, 7))
print (updated_myData)
By simplified sequence :
mydata[-(1:3 * 2), ]
By sequence :
mydata[seq(1, nrow(mydata), by = 2) , ]
By negative sequence :
mydata[-seq(2, nrow(mydata), by = 2) , ]
Or if you want to subset by selecting odd numbers:
mydata[which(1:nrow(mydata) %% 2 == 1) , ]
Or if you want to subset by selecting odd numbers, version 2:
mydata[which(1:nrow(mydata) %% 2 != 0) , ]
Or if you want to subset by filtering even numbers out:
mydata[!which(1:nrow(mydata) %% 2 == 0) , ]
Or if you want to subset by filtering even numbers out, version 2:
mydata[!which(1:nrow(mydata) %% 2 != 1) , ]
For completeness, I'll add that this can be done with dplyr as well using slice. The advantage of using this is that it can be part of a piped workflow.
df <- df %>%
.
.
slice(-c(2, 4, 6)) %>%
.
.
Of course, you can also use it without pipes.
df <- slice(df, -c(2, 4, 6))
The "not vector" format, -c(2, 4, 6) means to get everything that is not at rows 2, 4 and 6. For an example using a range, let's say you wanted to remove the first 5 rows, you could do slice(df, 6:n()). For more examples, see the docs.
Delete Dan from employee.data - No need to manage a new data.frame.
employee.data <- subset(employee.data, name!="Dan")
Here's a quick and dirty function to remove a row by index.
removeRowByIndex <- function(x, row_index) {
nr <- nrow(x)
if (nr < row_index) {
print('row_index exceeds number of rows')
} else if (row_index == 1)
{
return(x[2:nr, ])
} else if (row_index == nr) {
return(x[1:(nr - 1), ])
} else {
return (x[c(1:(row_index - 1), (row_index + 1):nr), ])
}
}
It's main flaw is it the row_index argument doesn't follow the R pattern of being a vector of values. There may be other problems as I only spent a couple of minutes writing and testing it, and have only started using R in the last few weeks. Any comments and improvements on this would be very welcome!
To identify by a name:
Call out the unique ID and identify the location in your data frame (DF).
Mark to delete. If the unique ID applies to multiple rows, all these rows will be removed.
Code:
Rows<-which(grepl("unique ID", DF$Column))
DF2<-DF[-c(Rows),]
DF2
Another approach when working with Unique IDs is to subset data:
*This came from an actual report where I wanted to remove the chemical standard
Chem.Report<-subset(Chem.Report, Chem_ID!="Standard")
Chem_ID is the column name.
The ! is important for excluding
I have a data frame named "mydata" that looks like this this:
A B C D
1. 5 4 4 4
2. 5 4 4 4
3. 5 4 4 4
4. 5 4 4 4
5. 5 4 4 4
6. 5 4 4 4
7. 5 4 4 4
I'd like to delete row 2,4,6. For example, like this:
A B C D
1. 5 4 4 4
3. 5 4 4 4
5. 5 4 4 4
7. 5 4 4 4
The key idea is you form a set of the rows you want to remove, and keep the complement of that set.
In R, the complement of a set is given by the '-' operator.
So, assuming the data.frame is called myData:
myData[-c(2, 4, 6), ] # notice the -
Of course, don't forget to "reassign" myData if you wanted to drop those rows entirely---otherwise, R just prints the results.
myData <- myData[-c(2, 4, 6), ]
You can also work with a so called boolean vector, aka logical:
row_to_keep = c(TRUE, FALSE, TRUE, FALSE, TRUE, FALSE, TRUE)
myData = myData[row_to_keep,]
Note that the ! operator acts as a NOT, i.e. !TRUE == FALSE:
myData = myData[!row_to_keep,]
This seems a bit cumbersome in comparison to #mrwab's answer (+1 btw :)), but a logical vector can be generated on the fly, e.g. where a column value exceeds a certain value:
myData = myData[myData$A > 4,]
myData = myData[!myData$A > 4,] # equal to myData[myData$A <= 4,]
You can transform a boolean vector to a vector of indices:
row_to_keep = which(myData$A > 4)
Finally, a very neat trick is that you can use this kind of subsetting not only for extraction, but also for assignment:
myData$A[myData$A > 4,] <- NA
where column A is assigned NA (not a number) where A exceeds 4.
Problems with deleting by row number
For quick and dirty analyses, you can delete rows of a data.frame by number as per the top answer. I.e.,
newdata <- myData[-c(2, 4, 6), ]
However, if you are trying to write a robust data analysis script, you should generally avoid deleting rows by numeric position. This is because the order of the rows in your data may change in the future. A general principle of a data.frame or database tables is that the order of the rows should not matter. If the order does matter, this should be encoded in an actual variable in the data.frame.
For example, imagine you imported a dataset and deleted rows by numeric position after inspecting the data and identifying the row numbers of the rows that you wanted to delete. However, at some later point, you go into the raw data and have a look around and reorder the data. Your row deletion code will now delete the wrong rows, and worse, you are unlikely to get any errors warning you that this has occurred.
Better strategy
A better strategy is to delete rows based on substantive and stable properties of the row. For example, if you had an id column variable that uniquely identifies each case, you could use that.
newdata <- myData[ !(myData$id %in% c(2,4,6)), ]
Other times, you will have a formal exclusion criteria that could be specified, and you could use one of the many subsetting tools in R to exclude cases based on that rule.
Create id column in your data frame or use any column name to identify the row. Using index is not fair to delete.
Use subset function to create new frame.
updated_myData <- subset(myData, id!= 6)
print (updated_myData)
updated_myData <- subset(myData, id %in% c(1, 3, 5, 7))
print (updated_myData)
By simplified sequence :
mydata[-(1:3 * 2), ]
By sequence :
mydata[seq(1, nrow(mydata), by = 2) , ]
By negative sequence :
mydata[-seq(2, nrow(mydata), by = 2) , ]
Or if you want to subset by selecting odd numbers:
mydata[which(1:nrow(mydata) %% 2 == 1) , ]
Or if you want to subset by selecting odd numbers, version 2:
mydata[which(1:nrow(mydata) %% 2 != 0) , ]
Or if you want to subset by filtering even numbers out:
mydata[!which(1:nrow(mydata) %% 2 == 0) , ]
Or if you want to subset by filtering even numbers out, version 2:
mydata[!which(1:nrow(mydata) %% 2 != 1) , ]
For completeness, I'll add that this can be done with dplyr as well using slice. The advantage of using this is that it can be part of a piped workflow.
df <- df %>%
.
.
slice(-c(2, 4, 6)) %>%
.
.
Of course, you can also use it without pipes.
df <- slice(df, -c(2, 4, 6))
The "not vector" format, -c(2, 4, 6) means to get everything that is not at rows 2, 4 and 6. For an example using a range, let's say you wanted to remove the first 5 rows, you could do slice(df, 6:n()). For more examples, see the docs.
Delete Dan from employee.data - No need to manage a new data.frame.
employee.data <- subset(employee.data, name!="Dan")
Here's a quick and dirty function to remove a row by index.
removeRowByIndex <- function(x, row_index) {
nr <- nrow(x)
if (nr < row_index) {
print('row_index exceeds number of rows')
} else if (row_index == 1)
{
return(x[2:nr, ])
} else if (row_index == nr) {
return(x[1:(nr - 1), ])
} else {
return (x[c(1:(row_index - 1), (row_index + 1):nr), ])
}
}
It's main flaw is it the row_index argument doesn't follow the R pattern of being a vector of values. There may be other problems as I only spent a couple of minutes writing and testing it, and have only started using R in the last few weeks. Any comments and improvements on this would be very welcome!
To identify by a name:
Call out the unique ID and identify the location in your data frame (DF).
Mark to delete. If the unique ID applies to multiple rows, all these rows will be removed.
Code:
Rows<-which(grepl("unique ID", DF$Column))
DF2<-DF[-c(Rows),]
DF2
Another approach when working with Unique IDs is to subset data:
*This came from an actual report where I wanted to remove the chemical standard
Chem.Report<-subset(Chem.Report, Chem_ID!="Standard")
Chem_ID is the column name.
The ! is important for excluding