We have brands data in a column/variable which is delimited by semicolon(;). Our task is to split these column data to multiple columns which we were able to do with the following syntax.
Attached the data as Screen shot.
Data set
Here is the R code:
x<-dataset$Pref_All
point<-df %>% separate(x, c("Pref_01","Pref_02","Pref_03","Pref_04","Pref_05"), ";")
point[is.na(point)] <- ""
However our question is: We have this type of brands data in more than 10 to 15 columns and if we use the above syntax the maximum number of columns to be split is to be decided on the number of brands each column holds (which we manually calculated and taken as 5 columns).
We would like to know is there any way where we can write the code in a dynamic way such that it should calculate the maximum number of brands each column holds and accordingly it should create those many new columns in a data frame. for e.g.
Pref_01,Pref_02,Pref_03,Pref_04,Pref_05.
the preferred output is given as a screen shot.
Output
Thanks for the help in advance.
x <- c("Swift;Baleno;Ciaz;Scross;Brezza", "Baleno;swift;celerio;ignis", "Scross;Baleno;celerio;brezza", "", "Ciaz;Scross;Brezza")
strsplit(x,";")
library(dplyr)
library(tidyr)
x <- data.frame(ID = c(1,2,3,4,5),
Pref_All = c("S;B;C;S;B",
"B;S;C;I",
"S;B;C;B",
" ",
"C;S;B"))
x$Pref_All <- as.character(levels(x$Pref_All))[x$Pref_All]
final_df <- x %>%
tidyr::separate(Pref_All, c(paste0("Pref_0", 1:b[[which.max(b)]])), ";")
final_df$ID <- x$Pref_All
final_df <- rename(final_df, Pref_All = ID)
final_df[is.na(final_df)] <- ""
Pref_All Pref_01 Pref_02 Pref_03 Pref_04 Pref_05
1 S;B;C;S;B S B C S B
2 B;S;C;I B S C I
3 S;B;C;B S B C B
4
5 C;S;B C S B
The trick for the column names is given by paste0 going from 1 to the maximum number of brands in your data!
I would use str_split() which returns a list of character vectors. From that, we can work out the max number of preferences in the dataframe and then apply over it a function to add the missing elements.
df=data.frame("id"=1:5,
"Pref_All"=c("brand1", "brand1;brand2;brand3", "", "brand2;brand4", "brand5"))
spl = str_split(df$Pref_All, ";")
# Find the max number of preferences
maxl = max(unlist(lapply(spl, length)))
# Add missing values to each element of the list
spl = lapply(spl, function(x){c(x, rep("", maxl-length(x)))})
# Bind each element of the list in a data.frame
dfr = data.frame(do.call(rbind, spl))
# Rename the columns
names(dfr) = paste0("Pref_", 1:maxl)
print(dfr)
# Pref_1 Pref_2 Pref_3
#1 brand1
#2 brand1 brand2 brand3
#3
#4 brand2 brand4
#5 brand5
Related
Just a quick question: how can I replace some values with others if these values are present in all the dataframe's column? Functions like mapvalues and recode work only if the column is specified, but in my case the dataframe has 89 columns so that would be time-consuming.
For the sake of clarity, take in consideration the following example. I want to replace [NULL] with another value.
Example:
a <- c("NULL",2,"NULL")
b <- c(3, "NULL", 1)
df <- data.frame(a, b)
df
a b
0 NULL 3
1 2 NULL
2 NULL 1
The difference between the example and my case is that the dataset is [35383 x 89], and the values I want to replace are more than one.
Thank you in advance for your time.
An extension to the comment by Ronak Shah. You can add 0 if you want like that. Or you can replace it with desired values, if you like that.
For example, replace the NULLs with mean of the respective columns:
#Run a loop to convert the characters into numbers because for your case it is all characters
#This will change the NULL to NAs.
for (i in colnames(df)){
df[,i] <- as.numeric(df[,i])
}
#Now replace the NAs with the mean of the column
for (i in colnames(df)){
df[,i][is.na(df[,i])] <- mean(df[,i], na.rm=TRUE)
}
You can similarly do this for median also. Let me know in the comment if you have any doubts.
For starters, I have added a few more rows to your example to better show how the code works
df
# a b
#1 NULL 3
#2 2 NULL
#3 NULL 1
#4 a 14
#5 1 a
#6 14 5
First, create two vectors: one with whe values you want to replace (pattern) and one with replacements in the same order. To make sure you have done it right, put them together in a data frame and take a look at the rows (this will also help in next step)
In this case, I want NULL to be 0, "a" to be "alpha", and so on, as shown below
pattern <- c("NULL", "a", 14, 1)
replacement <- c(0, "alpha", "fourteen", "one")
subs <- data.frame(pattern, replacement)
subs
# pattern replacement
#1 NULL 0
#2 a alpha
#3 14 fourteen
#4 1 one
To finish it, we will make a for tthat each time we will pick a pattern and its replacement from the subs data frame we created, and with these values execute a map_df(). This function iterates over the columns from our original data frame (df) and apply the gsub() function with the pattern and replacement
for (i in 1:nrow(subs)) {
df <- map_df(df, gsub, pattern = subs$pattern[i], replacement = subs$replacement[i])
}
df
# a b
#1 0 3
#2 2 0
#3 0 one
#4 alpha fourteen
#5 one alpha
#6 fourteen 5
I hope this was clear. Let me know if you have any doubts
I have a df containing 3 variables, and I want to create an extra variable for each possible product combination.
test <- data.frame(a = rnorm(10,0,1)
, b = rnorm(10,0,1)
, c = rnorm(10,0,1))
I want to create a new df (output) containing the result of a*b, a*c, b*c.
output <- data.frame(d = test$a * test$b
, e = test$a * test$c
, f = test$b * test$c)
This is easily doable (manually) with a small number of columns, but even above 5 columns, this activity can get very lengthy - and error-prone, when column names contain prefix, suffix or codes inside.
It would be extra if I could also control the maximum number of columns to consider at the same time (in the example above, I only considered 2 columns, but it would be great to select that parameter too, so to add an extra variable a*b*c - if needed)
My initial idea was to use expand.grid() with column names and then somehow do a lookup to select the whole columns values for the product - but I hope there's an easier way to do it that I am not aware of.
You can use combn to create combination of column names taken 2 at a time and multiply them to create new columns.
cbind(test, do.call(cbind, combn(names(test), 2, function(x) {
setNames(data.frame(do.call(`*`, test[x])), paste0(x, collapse = '-'))
}, simplify = FALSE)))
#. a b c a-b a-c b-c
#1 0.4098568 -0.3514020 2.5508854 -0.1440245 1.045498 -0.8963863
#2 1.4066395 0.6693990 0.1858557 0.9416031 0.261432 0.1244116
#3 0.7150305 -1.1247699 2.8347166 -0.8042448 2.026909 -3.1884040
#4 0.8932950 1.6330398 0.3731903 1.4587864 0.333369 0.6094346
#5 -1.4895243 1.4124826 1.0092224 -2.1039271 -1.503261 1.4255091
#6 0.8239685 0.1347528 1.4274288 0.1110321 1.176156 0.1923501
#7 0.7803712 0.8685688 -0.5676055 0.6778060 -0.442943 -0.4930044
#8 -1.5760181 2.0014636 1.1844449 -3.1543428 -1.866707 2.3706233
#9 1.4414434 1.1134435 -1.4500410 1.6049658 -2.090152 -1.6145388
#10 0.3526583 -0.1238261 0.8949428 -0.0436683 0.315609 -0.1108172
Could this one also be a solution. Ronak's solution is more elegant!
library(dplyr)
# your data
test <- data.frame(a = rnorm(10,0,1)
, b = rnorm(10,0,1)
, c = rnorm(10,0,1))
# new dataframe output
output <- test %>%
mutate(a_b= prod(a,b),
a_c= prod(a,c),
b_c= prod(b,c)
) %>%
select(-a,-b,-c)
I want to cbind a column to the data frame with the column name dynamically assigned from a string
y_attribute = "Survived"
cbind(test_data, y_attribute = NA)
this results in a new column added as y_attribute instead of the required Survived attribute which in provided as a string to the y_attribute variable. What needs to be done to get a column in the data frame with the column name provided from a variable?
You don't actually need cbind to add a new column. Any of these will work:
test_data[, y_attribute] = NA # data frame row,column syntax
test_data[y_attribute] = NA # list syntax (would work for multiple columns at once)
test_data[[y_attribute]] = NA # list single item syntax (single column only)
New columns are added after the existing columns, just like cbind.
We can use tidyverse to do this
library(dplyr)
test_data %>%
mutate(!! y_attribute := NA)
# col1 Survived
#1 1 NA
#2 2 NA
#3 3 NA
#4 4 NA
#5 5 NA
data
test_data <- data.frame(col1 = 1:5)
Not proud of this but I usually will do somethingl like this:
dyn.col <- "XYZ"
cbind(test.data, UNIQUE_NAMEXXX=NA)
colnames(test.data)[colnames(test.data == 'UNIQUE_NAMEXXX')] <- dyn.col
We can also do it with data.table
library(data.table)
setDT(test_data)[, (y_attribute) := NA]
I am trying to train a data that's converted from a document term matrix to a dataframe. There are separate fields for the positive and negative comments, so I wanted to add a string to the column names to serve as a "tag", to differentiate the same word coming from the different fields - for example, the word hello can appear both in the positive and negative comment fields (and thus, represented as a column in my dataframe), so in my model, I want to differentiate these by making the column names positive_hello and negative_hello.
I am looking for a way to rename columns in such a way that a specific string will be appended to all columns in the dataframe. Say, for mtcars, I want to rename all of the columns to have "_sample" at the end, so that the column names would become mpg_sample, cyl_sample, disp_sample and so on, which were originally mpg, cyl, and disp.
I'm considering using sapplyor lapply, but I haven't had any progress on it. Any help would be greatly appreciated.
Use colnames and paste0 functions:
df = data.frame(x = 1:2, y = 2:1)
colnames(df)
[1] "x" "y"
colnames(df) <- paste0('tag_', colnames(df))
colnames(df)
[1] "tag_x" "tag_y"
If you want to prefix each item in a column with a string, you can use paste():
# Generate sample data
df <- data.frame(good=letters, bad=LETTERS)
# Use the paste() function to append the same word to each item in a column
df$good2 <- paste('positive', df$good, sep='_')
df$bad2 <- paste('negative', df$bad, sep='_')
# Look at the results
head(df)
good bad good2 bad2
1 a A positive_a negative_A
2 b B positive_b negative_B
3 c C positive_c negative_C
4 d D positive_d negative_D
5 e E positive_e negative_E
6 f F positive_f negative_F
Edit:
Looks like I misunderstood the question. But you can rename columns in a similar way:
colnames(df) <- paste(colnames(df), 'sample', sep='_')
colnames(df)
[1] "good_sample" "bad_sample" "good2_sample" "bad2_sample"
Or to rename one specific column (column one, in this case):
colnames(df)[1] <- paste('prefix', colnames(df)[1], sep='_')
colnames(df)
[1] "prefix_good_sample" "bad_sample" "good2_sample" "bad2_sample"
You can use setnames from the data.table package, it doesn't create any copy of your data.
library(data.table)
df <- data.frame(a=c(1,2),b=c(3,4))
# a b
# 1 1 3
# 2 2 4
setnames(df,paste0(names(df),"_tag"))
print(df)
# a_tag b_tag
# 1 1 3
# 2 2 4
I think I'm missing something super simple, but I seem to be unable to find a solution directly relating to what I need: I've got a data frame that has a letter as the row name and a two columns of numerical values. As part of a loop I'm running I create a new vector (from an index) that has both a letter and number (e.g. "f2") which I then need to be the name of a new row, then add two numbers next to it (based on some other section of code, but I'm fine with that). What I get instead is the name of the vector/index as the title of the row name, and I'm not sure if I'm missing a function of rbind or something else to make it easy.
Example code:
#Data frame and vector creation
row.names <- letters[1:5]
vector.1 <- c(1:5)
vector.2 <- c(2:6)
vector.3 <- letters[6:10]
data.frame <- data.frame(vector.1,vector.2)
rownames(data.frame) <- row.names
data.frame
index.vector <- "f2"
#what I want the data frame to look like with the new row
data.frame <- rbind(data.frame, "f2" = c(6,11))
data.frame
#what the data frame looks like when I attempt to use a vector as a row name
data.frame <- rbind(data.frame, index.vector = c(6,11))
data.frame
#"why" I can't just type "f" every time
index.vector2 = paste(index.vector, "2", sep="")
data.frame <- rbind(data.frame, index.vector2 = c(6,11))
data.frame
In my loop the "index.vector" is a random sample, hence where I can't just write the letter/number in as a row name, so need to be able to create the row name from a vector or from the index of the sample.
The loop runs and a random number of new rows will be created, so I can't specify what number the row is that needs a new name - unless there's a way to just do it for the newest or bottom row every time.
Any help would be appreciated!
Not elegant, but works:
new_row <- data.frame(setNames(list(6, 11), colnames(data.frame)), row.names = paste(index.vector, "2", sep=""))
data.frame <- rbind(data.frame, new_row)
data.frame
# vector.1 vector.2
# a 1 2
# b 2 3
# c 3 4
# d 4 5
# e 5 6
# f22 6 11
I Understood the problem , but not able to resolve the issue. Hence, suggesting an alternative way to achieve the same
Alternate solution: append your row labels after the data binding in your loop and then assign the row names to your dataframe at the end .
#Data frame and vector creation
row.names <- letters[1:5]
vector.1 <- c(1:5)
vector.2 <- c(2:6)
vector.3 <- letters[6:10]
data.frame <- data.frame(vector.1,vector.2)
#loop starts
index.vector <- "f2"
data.frame <- rbind(data.frame,c(6,11))
row.names<-append(row.names,index.vector)
#loop ends
rownames(data.frame) <- row.names
data.frame
output:
vector.1 vector.2
a 1 2
b 2 3
c 3 4
d 4 5
e 5 6
f2 6 11
Hope this would be helpful.
If you manipulate the data frame with rbind, then the newest elements will always be at the "bottom" of your data frame. Hence you could also set a single row name by
rownnames(data.frame)[nrow(data.frame)] = "new_name"