I am trying to get column entries as a list that match a list of entries from data frame
Showing what I am trying to do:
Dataframe named Tepo
| | name | shortcut |
| -------- | -------------- | ----------|
| 1 | Apples | A |
| 2 | Bannans | B |
| 3 | oranges | O |
| 4 | Carrots | C |
| 5 | Mangos | M |
| 6 | Strawberies | S |
I have a list FruitList as chr
>FruitList
>[1] "Bannas" "Carrots" "Mangos"
And I would like to get a list, shortcutList, of the corresponding columns:
>shortcutList
>[1] "B" "C" "M"
My attempt:
shortcutList <- tepo$shorcut[tepo$name == FruiteList[]]
However, I don't get the desired list output.
Thanks for the help
Use %in% :
shortcutList <- tepo$shortcut[tepo$name %in% FruitList]
Related
This question already has answers here:
Aggregating all unique values of each column of data frame
(2 answers)
Collapse / concatenate / aggregate a column to a single comma separated string within each group
(6 answers)
Closed 1 year ago.
I would like to create a column in my R data frame based on the order in which multiple values occur in one column.
For example, my data frame has an id column and an item type column, and the values of the order column is what I would like to add. Is there a way to tell R to look at the order of values in the item column so that it can spit out "ABCD" or "ADCB" (any other order) as the cell value under the 3rd column?
| id | item | order |
| 11 | A | ABCD |
| 11 | A | ABCD |
| 11 | B | ABCD |
| 11 | B | ABCD |
| 11 | C | ABCD |
| 11 | C | ABCD |
| 11 | D | ABCD |
| 11 | D | ABCD |
| 12 | A | ADCB |
| 12 | A | ADCB |
| 12 | D | ADCB |
| 12 | D | ADCB |
| 12 | C | ADCB |
| 12 | C | ADCB |
| 12 | B | ADCB |
| 12 | B | ADCB |
...
When I bind multiple dataframes together using Out2 = do.call(rbind.data.frame, Out), I obtain the following output. How do I relabel the first column such that it only contains the numbers within the square brackets, i.e. 1 to 5 for each trial number? Is there a way to add a column name to the first column too?
| V1 | V2 | Trial |
+--------+--------------+--------------+-------+
| [1,] | 0.130880519 | 0.02085533 | 1 |
| [2,] | 0.197243133 | -0.000502744 | 1 |
| [3,] | -0.045241653 | 0.106888902 | 1 |
| [4,] | 0.328759949 | -0.106559163 | 1 |
| [5,] | 0.040894969 | 0.114073454 | 1 |
| [1,]1 | 0.103130056 | 0.013655756 | 2 |
| [2,]1 | 0.133080106 | 0.038049071 | 2 |
| [3,]1 | 0.067975054 | 0.03036033 | 2 |
| [4,]1 | 0.132437217 | 0.022887103 | 2 |
| [5,]1 | 0.124950463 | 0.007144698 | 2 |
| [1,]2 | 0.202996317 | 0.004181205 | 3 |
| [2,]2 | 0.025401354 | 0.045672932 | 3 |
| [3,]2 | 0.169469266 | 0.002551237 | 3 |
| [4,]2 | 0.2303046 | 0.004936579 | 3 |
| [5,]2 | 0.085702254 | 0.020814191 | 3 |
+--------+--------------+--------------+-------+
We can use parse_number to extract the first occurence of numbers
library(dplyr)
df1 %>%
mutate(newcol = readr::parse_number(row.names(df1)))
Or in base R, use sub to capture the digits after the [ in the row names
df1$newcol <- sub("^\\[(\\d+).*", "\\1", row.names(df1))
If I have a list similar to x <- c("Name,Age,Gender", "Rob,21,M", "Matt,30,M"), how can I convert to a dataframe where Name, Age, and Gender become the column headers.
Currently my approach is,
dataframe <- data.frame(matrix(unlist(x), nrow=3, byrow=T))
which gives me
matrix.unlist.user_data...nrow...num_rows..byrow...T.
1 Name,Age,Gender
2 Rob,21,M
3 Matt,30,M
and doesn't help me at all.
How can I get something which resembles the following from the list mentioned above?
+---------------------------------------------+
| name | age | gender |
| | | |
+---------------------------------------------+
| | | |
| | | |
| ... | ... | ... |
| | | |
| | | ++
+---------------------------------------------+
| | | |
| ... | ... | ... |
| | | |
| | | |
+---------------------------------------------+
We paste the strings into a single string with \n and use either read.csv or read.table from base R
read.table(text=paste(x, collapse='\n'), header = TRUE, stringsAsFactors = FALSE, sep=',')
Alternatively,
data.table::fread(paste(x, collapse = "\n"))
Name Age Gender
1: Rob 21 M
2: Matt 30 M
Everything is in the title, I got from a database many columns, paired two-by-two containing codes and labels for some variables, I want an easy way to create half as many factors, with, for each factor levels/codes matching to the original two variables.
Here is an exemple of original data for two factors
| customer_type | customer_type_name | customer_status | customer_status_name |
|----------------------|----------------------|----------------------|----------------------|
| 1 | A | 2 | Beta |
| 2 | B | 2 | Beta |
| 3 | C | 1 | Alpha |
| 2 | B | 3 | Gamma |
| 1 | A | 4 | Delta |
| 3 | C | 2 | Beta |
i.e. a simpler way (simpler to call in a function for lots of variables) to do from dataframe "accounts"
a<-accounts[,c("customertypecode","customertypecodename")]
a<-a[!duplicated(a),]
a<-a[order(a$customertypecode),]
accounts$customertypecode<-factor(accounts$customertypecode,labels=a$customertypecodename[!is.na(a$customertypecodename)])
I load a text file (tree.txt) to R, with the below content (copy pasted from JWEKA - J48 command).
I use the following command to load the text file:
data3 <-read.table (file.choose(), header = FALSE,sep = ",")
I would like to insert each column into a separate variables named like the following format COL1, COL2 ... COL8 (in this example since we have 8 columns). If you load it to EXCEL with delimited separation each row will be separated in one column (this is the required result).
Each COLn will contain the relevant characters of the tree in this example.
How can separate and insert the text file into these columns automatically while ignoring the header and footer content of the file?
Here is the text file content:
[[1]]
J48 pruned tree
------------------
MSTV <= 0.4
| MLTV <= 4.1: 3 -2
| MLTV > 4.1
| | ASTV <= 79
| | | b <= 1383:00:00 2 -18
| | | b > 1383
| | | | UC <= 05:00 1 -2
| | | | UC > 05:00 2 -2
| | ASTV > 79:00:00 3 -2
MSTV > 0.4
| DP <= 0
| | ALTV <= 09:00 1 (170.0/2.0)
| | ALTV > 9
| | | FM <= 7
| | | | LBE <= 142:00:00 1 (27.0/1.0)
| | | | LBE > 142
| | | | | AC <= 2
| | | | | | e <= 1058:00:00 1 -5
| | | | | | e > 1058
| | | | | | | DL <= 04:00 2 (9.0/1.0)
| | | | | | | DL > 04:00 1 -2
| | | | | AC > 02:00 1 -3
| | | FM > 07:00 2 -2
| DP > 0
| | DP <= 1
| | | UC <= 03:00 2 (4.0/1.0)
| | | UC > 3
| | | | MLTV <= 0.4: 3 -2
| | | | MLTV > 0.4: 1 -8
| | DP > 01:00 3 -8
Number of Leaves : 16
Size of the tree : 31
An example of the COL1 content will be:
MSTV
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MSTV
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COL2 content will be:
MLTV
MLTV
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>
DP
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DP
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Try this:
cleaned.txt <- capture.output(cat(paste0(tail(head(readLines("FILE_LOCATION"), -4), -4), collapse = '\n'), sep = '\n'))
cleaned.df <- read.fwf(file = textConnection(cleaned.txt),
header = FALSE,
widths = rep.int(4, max(nchar(cleaned.txt)/4)),
strip.white= TRUE
)
cleaned.df <- cleaned.df[,colSums(is.na(cleaned.df))<nrow(cleaned.df)]
For the cleaning process, I end up using a combination of head and tail to remove the 4 spaces on the top and the bottom. There's probably a more efficient way to do this outside of R, but this isn't so bad. Generally, I'm just making the file readable to R.
Your file looks like a fixed-width file so I use read.fwf, and use textConnection() to point the function to the cleaned output.
Finally, I'm not sure how your data is actually structured, but when I copied it from stackoverflow, it pasted with a bunch of whitespace at the end of each line. I'm using some tricks to guess at how long the file is, and removing extraneous columns over here
widths = rep.int(4, max(nchar(cleaned.txt)/4))
cleaned.df <- cleaned.df[,colSums(is.na(cleaned.df))<nrow(cleaned.df)]
Next, I'm creating the data in the way you would like it structured.
for (i in colnames(cleaned.df)) {
assign(i, subset(cleaned.df, select=i))
assign(i, capture.output(cat(paste0(unlist(get(i)[get(i)!=""])),sep = ' ', fill = FALSE)))
}
rm(i)
rm(cleaned.df)
rm(cleaned.txt)
What this does is it creates a loop for each column header in your data frame.
From there it uses assign() to put all the data in each column into its' own data frame. In your case, they are named V1 through V15.
Next, it uses a combination of cat() and paste() with unlist() an capture.output() to concatenate your list into a single character vectors, for each of the data frames, so they are now character vectors, instead of data frames.
Keep in mind that because you wanted a space at each new character, I'm using a space as a separator. But because this is a fixed-width file, some columns are completely blank, which I'm removing using
get(i)[get(i)!=""]
(Your question said you wanted COL2 to be: MLTV MLTV | | | | | | > DP | | | | | | | | | | | | DP | | | | | |).
If we just use get(i), there will be a leading whitespace in the output.