How to get row index number in R? - r

Suppose I have a list or data frame in R, and I would like to get the row index, how do I do that? That is, I would like to know how many rows a certain matrix consists of.

I'm interpreting your question to be about getting row numbers.
You can try as.numeric(rownames(df)) if you haven't set the rownames. Otherwise use a sequence of 1:nrow(df).
The which() function converts a TRUE/FALSE row index into row numbers.

It not quite clear what exactly you are trying to do.
To reference a row in a data frame use df[row,]
To get the first position in a vector of something use match(item,vector), where the vector could be one of the columns of your data frame, eg df$cname if the column name is cname.
Edit:
To combine these you would write:
df[match(item,df$cname),]
Note that the match gives you the first item in the list, so if you are not looking for a unique reference number, you may want to consider something else.

See row in ?base::row. This gives the row indices for any matrix-like object.

rownames(dataframe)
This will give you the index of dataframe

If i understand your question, you just want to be able to access items in a data frame (or list) by row:
x = matrix( ceiling(9*runif(20)), nrow=5 )
colnames(x) = c("col1", "col2", "col3", "col4")
df = data.frame(x) # create a small data frame
df[1,] # get the first row
df[3,] # get the third row
df[nrow(df),] # get the last row
lf = as.list(df)
lf[[1]] # get first row
lf[[3]] # get third row
etc.

Perhaps this complementary example of "match" would be helpful.
Having two datasets:
first_dataset <- data.frame(name = c("John", "Luke", "Simon", "Gregory", "Mary"),
role = c("Audit", "HR", "Accountant", "Mechanic", "Engineer"))
second_dataset <- data.frame(name = c("Mary", "Gregory", "Luke", "Simon"))
If the name column contains only unique across collection values (across whole collection)
then you can access row in other dataset by value of index returned by match
name_mapping <- match(second_dataset$name, first_dataset$name)
match returns proper row indexes of names in first_dataset from given names from second: 5 4 2 1
example here - accesing roles from first dataset by row index (by given name value)
for(i in 1:length(name_mapping)) {
role <- as.character(first_dataset$role[name_mapping[i]])
second_dataset$role[i] = role
}
===
second dataset with new column:
name role
1 Mary Engineer
2 Gregory Mechanic
3 Luke Supervisor
4 Simon Accountant
r

x <- matrix(ceiling(9*runif(20)), nrow=5)
colnames(x) <- c("these", "are", "the", "columnes")
df <- data.frame(x)
Result:
dataframe
which(df == "2") #returns rowIndexes results from the entire dataset, in this case it returns a list of 3 index numb
Result:
5 13 17
length(which(df == "2")) #count numb. of rows that matches a condition
Result:
3
You can also do this column wise, example of:
which(df$columnName == c("2", "7")) #you do the same with strings
length(which(df$columnName == c("2", "7")))

Related

Count number of entries in each column with result in dataframe

I have a dataframe with many columns. I want to count the number of times something is entered into each column.
#Example data
Gender<-c("","Male","Male","","Female","Female")
location<-c("UK","France","USA","","","")
dataset<-data.frame(Gender,location, stringsAsFactors = FALSE)
There are 4 entries in the gender column and 3 entries in the location column.
I want the results to be in a dataframe such as:
result<-data.frame(Results=c("Gender","location"), Totals=c(4,3))
Can anyone suggest an approach to do this?
You can use the namesof datasetas one column for resultand calculate the Totals by counting how often grep matches anything that is a character (as opposed to nothing in an empty cell):
result <- data.frame(
Results = names(dataset),
Totals = sapply(dataset, function(x) length(grep(".", x)))
)
rownames(result) <- NULL
Result:
result
Results Totals
1 Gender 4
2 location 3
A base R option using stack + colSums
setNames(
rev(stack(colSums(dataset != ""))),
c("Results", "Total")
)
gives
Results Total
1 Gender 4
2 location 3
This should work for you:
ngen <- sum(dataset$Gender != "") #sum number entries in column that are not empty
nloc <- sum(dataset$location != "") #sam thing
Totals <- c(ngen,nloc)
result<-data.frame(Results=c("Gender","location"), Totals)
You can simplify some of the steps if you want, but that would be the detailed way.

Substituting or summing based on condition

I have a dataset that looks something like this
df <- data.frame("id" = c("Alpha", "Alpha", "Alpha","Alpha","Beta","Beta","Beta","Beta"),
"Year" = c(1970,1971,1972,1973,1974,1977,1978,1990),
"Group" = c(1,NA,1,NA,NA,2,2,NA),
"Val" = c(2,3,3,5,2,5,3,5))
And I would like to create a cumulative sum of "Val". I know how to do the simple cumulative sum
df <- df %>% group_by(id) %>% mutate(cumval=cumsum(Val))
However, I would like my final data to look like this
final <- data.frame("id" = c("Alpha", "Alpha", "Alpha","Alpha","Beta","Beta","Beta","Beta"),
"Year" = c(1970,1971,1972,1973,1974,1977,1978,1990),
"Group" = c(1,NA,1,NA,NA,2,2,NA),
"Val" = c(2,3,3,5,2,5,3,5),
"cumval" = c(2,5,6,11,2,7,5,10))
The basic idea is that when two "Val"'s are of the same "Group" the one happening later (Year) substitutes the previous one.
For instance, in the sample dataset, observation 3 has a "cumval" of 6 rather than 8 because of the "Val" at time 1972 replaced the "Val" at time 1970. similarly for Beta.
I thank you in advance for your help
In my head, this requires a for loop. First we split the dataframe by the id column into a list of two. Then we create two empty lists. In the og list, we will put the row where the first unique non NA group identifier occurs. For alpha this is the first row and for Beta this is the second row. We will use this to subtract from the cumulative sum when the value gets substituted.
mylist <- split(df, f = df$id)
og <- list()
vals <- list()
df_num <- 1
We shall use a nested loop, the outer loop loops over each object (dataframe in this case) in the list and the inner loop loops over each value in the Group column.
We need to keep track of the row numbers, which we do with the r variable. We initially set it to 0 outside the for loop so we add 1. First we check if we are in the first row of the data frame, in which case the cumulative sum is simply equal to the value in the first row of the Val column. Then within the if test, we use another if test to check if the Group id is an NA. If it isn't then this is the first occurrence of the number that will indicate a substitution of the current value if this number appears again. So we save the number to the temporary variable temp. We also extract and save the row that contains the value to the og list.
After this it, goes to the next iteration. We check if the current Group value is NA. If it is, then we just add the value to the cumulative sum. If it isn't equal to NA, we check if the value is NA and is equal to the value stored in temp. If both are true, then this means we need to substitute. We extract the original value stored in the og list and save it as old. We then subtract the old value from the cumulative sum and add the current value. We also replace the orginal value in og with the current replacement value. This is because if the value needs to replaced again, we will need to subtract the current value and not the original value.
If j is NA but it is not equal to temp, then this is a new instance of Group. So we save the row with the original value to og list, and save the Group. The sum continues as normal as this is not an instance of replacing a value. Note that the variable x that is used to count the elements in the og list is only incremented when a new occurrence is added to the list. Thus, og[[x-1]] will always be the replacement value.
for (my_df in mylist) {
x <- 1
r <- 0
for (j in my_df$Group) {
r <- r + 1
if (r == 1) {
vals[[1]] <- my_df$Val[1]
if (is.na(j)==FALSE) {
og[[x]] <- df[r, c('Group', 'Val'), drop = FALSE]
temp <- j
x <- x + 1
}
next
}
if (is.na(j)==TRUE) {
vals[[r]] <- vals[[r-1]] + my_df$Val[r]
} else if (is.na(j)==FALSE & j==temp) {
old <- og[[x-1]]
old <- old[,2]
vals[[r]] <- vals[[r-1]] - old + df$Val[r]
og[[x-1]] <- df[r, c('Group', 'Val'), drop = FALSE]
} else {
vals[[r]] <- vals[[r-1]] + my_df$Val[r]
og[[x]] <- my_df[r, c('Group', 'Val')]
temp <- j
x <- x + 1
}
}
cumval <- unlist(vals) %>% as.data.frame()
colnames(cumval) <- 'cumval'
my_df <- cbind(my_df, cumval)
mylist[[df_num]] <- my_df
df_num <- df_num + 1
}
Lastly, we combine the two dataframes in the list by binding them on rows with bind_rows from the dplyr package. Then I check if the Final dataframe is identical to your desired output with identical() and it evaluates to TRUE
final_df <- bind_rows(mylist)
identical(final_df, final)
[1] TRUE

How to get all columns with the same column name in R at once?

Let's say I have the following data frame:
> test <- cbind(test=c(1, 2, 3), test=c(1, 2, 3))
> test
test test
[1,] 1 1
[2,] 2 2
[3,] 3 3
Now from such data frame I want to fetch all the columns named "test" to a new data frame:
> new_df <- test[, "test"]
However this last attempt to do so only fetches the first column called "test" in test data frame:
> new_df
[1] 1 2 3
How can I get all of the columns called "test" in this example and put them into a new data frame in a single command? In my real data I have many columns with repeated colnames and I don't know the index of the columns so I can`t get them by number.
It is not advisable to have same column names for practical reasons. But, we can do a comparison (==) to get a logical vector and use that to extract the columns
i1 <- colnames(test) == "test"
new_df <- test[, i1, drop = FALSE]
Note that data.frame doesn't allow duplicate column names and would change it to unique by appending .1 .2 etc at the end with make.unique. With matrix (the OP's dataset), allows to have duplicate column names or row names (not recommended though)
Also, if there are multiple column names that are repeated and want to select them as separate datasets, use split
lst1 <- lapply(split(seq_len(ncol(test)), colnames(test)), function(i)
test[, i, drop = FALSE])
Or loop through the unique column names and do a == by looping through it with lapply
lst2 <- lapply(unique(colnames(test)), function(nm)
test[, colnames(test) == nm, drop = FALSE])

How to extract rows of a data frame between two characters

I've got some poorly structured data I am trying to clean. I have a list of keywords I can use to extract data frames from a CSV file. My raw data is structured roughly as follows:
There are 7 columns with values, the first columns are all string identifiers, like a credit rating or a country symbol (for FX data), while the other 6 columns are either a header like a percentage change string (e.g. +10%) or just a numerical value. Since I have all this data lumped together, I want to be able to extract data for each category. So for instance, I'd like to extract all the rows between my "credit" keyword and my "FX" keyword in my first column. Is there a way to do this in either base R or dplyr easily?
eg.
df %>%
filter(column1 = in_between("credit", "FX"))
Sample dataframe:
row 1: c('random',-1%', '0%', '1%, '2%')
row 2: c('credit', NA, NA, NA, NA)
row 3: c('AAA', 1,2,3,4)
...
row n: c('FX', '-1%', '0%', '1%, '2%')
And I would want the following output:
row 1: c('credit', -1%', '0%', '1%, '2%')
row 2: c('AAA', 1,2,3,4)
...
row n-1: ...
If I understand correctly you could do something like
start <- which(df$column1 == "credit")
end <- which(df$column1 == "FX")
df[start:(end-1), ]
Of course this won't work if "credit" or "FX" is in the column more than once.
Using what Brian suggested:
in_between <- function(df, start, end){
return(df[start:(end-1),])
}
Then loop over the indices in
dividers = which(df$column1 %in% keywords == TRUE)
And save the function outputs however one would like.
lapply(1:(length(dividers)-1), function(x) in_between(df, start = dividers[x], end = dividers[x+1]))
This works. Messy data so I still have the annoying case where I need to keep the offset rows.
I'm still not 100% sure what you are trying to accomplish but does this do what you need it to?
set.seed(1)
df <- data.frame(
x = sample(LETTERS[1:10]),
y = rnorm(10),
z = runif(10)
)
start <- c("C", "E", "F")
df2 <- df %>%
mutate(start = x %in% start,
group = cumsum(start))
split(df2, df2$group)

Store output of sapply into a data frame?

how can I store the output of sapply() to a dataframe where the index value is stored in first column and its value in corresponding 2nd column. For illustration, I have shown only 2 elements here, but there are 110 columns in my data. "loan" is the data frame.
cols <- sapply(loan,function(x) sum(is.na(x)))
cols
id
0
member_id
7
I want output as:
var value
id 0
member_id 7
I know that sapply() returns a vector, but when I print the vector, values are printed along with its some "index" e.g., column name if applied on a data frame. So, now when I want to store it as a data frame with two columns where 1st column contains the index part and the second column contains the value, how can I do it?
I found an answer to my question. For those who actually did understand my problem, this answer might make sense:
cols <- data.frame(sapply(loan ,function(x) sum(is.na(x))))
cols <- cbind(variable = row.names(cols), cols)
I wanted the row.names to be in a column of the same data frame corresponding to the values obtained from sapply.
We can use stack
stack(mylist)[2:1]
data
mylist <- list(df = 1, rf = 2)
Is this what you want?
Your original list:
L <- c("df",1,"rf",2)
L
[1] "df" "1" "rf" "2"
As a data frame:
N <- length(L)
df <- data.frame( var = L[seq(1,N,2)], value = L[seq(2,N,2)] )
df
var value
1 df 1
2 rf 2

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