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I am using this dataset: http://www.openintro.org/stat/data/cdc.R
to create a table from a subset that only contains the means and standard deviations of male participants. The table should look like this:
Mean Standard Deviation
Age: 44.27 16.715
Height: 70.25 3.009219
Weight: 189.3 36.55036
Desired Weight: 178.6 26.25121
I created a subset for males and females with this code:
mdata <- subset(cdc, cdc$gender == ("m"))
fdata <- subset(cdc, cdc$gender == ("f"))
How should I create a table that only contains means and SDs of age, height, weight, and desired weight using these subsets?
The data frame you provided sucked up all the memory on my laptop, and it's not needed to provide that much data to solve your problem. Here's a dplyr/tidyr solution to create a summary table grouped by categories, using the starwars dataset available with dplyr:
library(dplyr)
library(tidyr)
starwars |>
group_by(sex) |>
summarise(across(
where(is.numeric),
.fns = list(Mean = mean, SD = sd), na.rm = TRUE,
.names = "{col}__{fn}"
)) |>
pivot_longer(-sex, names_to = c("var", ".value"), names_sep = "__")
# A tibble: 15 × 4
sex var Mean SD
<chr> <chr> <dbl> <dbl>
1 female height 169. 15.3
2 female mass 54.7 8.59
3 female birth_year 47.2 15.0
4 hermaphroditic height 175 NA
5 hermaphroditic mass 1358 NA
6 hermaphroditic birth_year 600 NA
7 male height 179. 36.0
8 male mass 81.0 28.2
9 male birth_year 85.5 157.
10 none height 131. 49.1
11 none mass 69.8 51.0
12 none birth_year 53.3 51.6
13 NA height 181. 2.89
14 NA mass 48 NA
15 NA birth_year 62 NA
Just make a data frame of colMeans and column sd. Note, that you may also select columns.
fdata <- subset(cdc, gender == "f", select=c("age", "height", "weight", "wtdesire"))
data.frame(mean=colMeans(fdata), sd=apply(fdata, 2, sd))
# mean sd
# age 45.79772 17.584420
# height 64.36775 2.787304
# weight 151.66619 34.297519
# wtdesire 133.51500 18.963014
You can also use by to do it simultaneously for both groups, it's basically a combination of split and lapply. (To avoid apply when calculating column SDs, you could also use sd=matrixStats::colSds(as.matrix(fdata)) which is considerably faster.)
res <- by(cdc[c("age", "height", "weight", "wtdesire")], cdc$gender, \(x) {
data.frame(mean=colMeans(x), sd=matrixStats::colSds(as.matrix(x)))
})
res
# cdc$gender: m
# mean sd
# age 44.27307 16.719940
# height 70.25165 3.009219
# weight 189.32271 36.550355
# wtdesire 178.61657 26.251215
# ------------------------------------------------------------------------------------------
# cdc$gender: f
# mean sd
# age 45.79772 17.584420
# height 64.36775 2.787304
# weight 151.66619 34.297519
# wtdesire 133.51500 18.963014
To extract only one of the data frames in the list-like object use e.g. res$m.
Usually we use aggregate for this, which you also might consider:
aggregate(cbind(age, height, weight, wtdesire) ~ gender, cdc, \(x) c(mean=mean(x), sd=sd(x))) |>
do.call(what=data.frame)
# gender age.mean age.sd height.mean height.sd weight.mean weight.sd wtdesire.mean wtdesire.sd
# 1 m 44.27307 16.71994 70.251646 3.009219 189.32271 36.55036 178.61657 26.25121
# 2 f 45.79772 17.58442 64.367750 2.787304 151.66619 34.29752 133.51500 18.96301
The pipe |> call(what=data.frame) is just needed to get rid of matrix columns, which is useful in case you aim to further process the data.
Note: R >= 4.1 used.
Data:
source('https://www.openintro.org/stat/data/cdc.R')
or
cdc <- structure(list(genhlth = structure(c(3L, 3L, 1L, 5L, 3L, 3L), levels = c("excellent",
"very good", "good", "fair", "poor"), class = "factor"), exerany = c(0,
1, 0, 0, 1, 1), hlthplan = c(1, 1, 1, 1, 1, 1), smoke100 = c(1,
0, 0, 0, 0, 1), height = c(69, 66, 73, 65, 67, 69), weight = c(224L,
215L, 200L, 216L, 165L, 170L), wtdesire = c(224L, 140L, 185L,
150L, 165L, 165L), age = c(73L, 23L, 35L, 57L, 81L, 83L), gender = structure(c(1L,
2L, 1L, 2L, 2L, 1L), levels = c("m", "f"), class = "factor")), row.names = c("19995",
"19996", "19997", "19998", "19999", "20000"), class = "data.frame")
I have multiple data.frames with an equal number of columns. I want to combine these into a single pivot table that I can write to excel.
Example data.frames:
> net_imports[,1:5]
1979 1980 1981 1982 1983
beginning_stocks NA -53 -83 -110 -60.000
production NA -390 -585 -510 -434.996
consumption NA 370 380 390 410.000
ending_stocks 53 83 110 60 46.000
predicted NA 10 -178 -170 -38.996
> area_harvested_output[,1:5]
1979 1980 1981 1982 1983
area_harvested_lag 51.22632 51.2263243 41.6213885 57.6296148 54.4279695
area_harvested_trend 0.00000 0.1007849 0.2015699 0.3023548 0.4031397
import_price_cpi NA 20.4610740 18.7566970 16.8987151 15.2273790
predicted NA 71.7881832 60.5796553 74.8306847 70.0584883
error NA 58.2118168 119.4203447 95.1693153 99.9415117
pred_err NA 130.0000000 180.0000000 170.0000000 170.0000000
I want the resulting table in excel to look something like this
Basically, I just want to maintain the variable names like "net_imports" and "area_harvested_output" as grouped data.
I'd pivot_longer both data.frames to long, so that year becomes one instead of (in your example) five columns, rbind or bind_rows them and export the accumulated long table to Excel (in which I'd then build the interactive Excel pivot table).
your example data:
net_imports <- structure(list(parameter = c("beginning_stocks", "production",
"consumption", "ending_stocks", "predicted"), X1979 = c(NA, NA,
NA, 53L, NA), X1980 = c(-53L, -390L, 370L, 83L, 10L), X1981 = c(-83L,
-585L, 380L, 110L, -178L), X1982 = c(-110L, -510L, 390L, 60L,
-170L), X1983 = c(-60, -434.996, 410, 46, -38.996)), class = "data.frame", row.names = c(NA,
5L))
area_harvested_output <- structure(list(parameter = c("area_harvested_lag", "area_harvested_trend", "import_price_cpi", "predicted", "error", "pred_err"), X1979 = c(51.22632,
0, NA, NA, NA, NA), X1980 = c(51.2263243, 0.1007849, 20.461074,
71.7881832, 58.2118168, 130), X1981 = c(41.6213885, 0.2015699,
18.756697, 60.5796553, 119.4203447, 180), X1982 = c(57.6296148,
0.3023548, 16.8987151, 74.8306847, 95.1693153, 170), X1983 = c(54.4279695,
0.4031397, 15.227379, 70.0584883, 99.9415117, 170)), class = "data.frame", row.names = c(NA,
6L))
the code:
library(dplyr)
library(tidyr)
library(rio) ## convenience package for imports/exports
long_table <-
net_imports %>%
pivot_longer(cols = -parameter,
names_to = 'year') %>%
bind_rows(
area_harvested_output %>%
pivot_longer(cols = -parameter,
names_to = 'year')
)
long_table %>% export('long_table.xlsx')
In R markdown through R Studio (R v. 4.0.3), I'm looking for a better solution to combining similarly structured dataframes while keeping all rows and matching entries on a key. Piping full_join() into a filter into a bind_rows() directly wasn't working, possibly because of the error message:
Error: Can't combine ..1$term_code 'character> and ..2$term_code '<integer.
I have 23 dataframes (let's call these "semester data") of data I'm looking to combine into a single dataframe (intended to be a single dataset of individuals outcomes from semester-to-semester).
Each semester dataframe is roughly 3000-4000 observations (individuals) with 45-47 variables of relevant data. A simplified example of a semester (or term) dataframe is shown below.
Simplified example of a "semester" dataframe:
id
ACT_math
course_code
section_code
term_code
grade
term_GPA
0001
23
101
001
FA12
3.45
3.8
0002
28
201
003
FA12
3.2
3.4
Individuals will show up in multiple semester dataframes as they progress through the program (taking course 101 in the fall and course 102 in the spring).
I want to use the dplyr full_join() to match these individuals on an ID key.
Using the suffix argument, I hope to keep track of which semester and course a set of data (grade, term_GPA, etc) for an individual comes from.
There's some data (ACT score, gender, state residency, etc) that is the stable for an individual across semester dataframes. Ideally I could take the first input and drop the rest, but if I had to clean this afterwards, that's fine.
I started by defining an object programatic_database using the first semester of data SP11. To cut down on the duplication of stable data for an individual, I selected the relevant columns that I wanted to join.
programmatic_database <- programmatic_database %>%
full_join(select(fa12, id, course_code, section_code, grade, term_gpa), by = "id", copy = TRUE, suffix = c(".sp11", ".fa12"), keep = FALSE, name = "id")
However, every semester new students join the program. I would like to add these entries to the bottom of the growing programmatic_database.
I'm also looking to use rbind() or bind_rows() to add these individuals to the bottom of the programmatic_database, along with their relevant data.
After full_join(), I'm filtering out the entries that have already been added horizontally to the dataframe, then piping the remaining entries into bind_rows()
programmatic_database <- fa12[!which(fa12$id %in% programmatic_database),] %>% dplyr::bind_rows(programmatic_database, fa12)
Concatenated example of what my code is producing after several iterations:
id
ACT_math
course_code
section_code
section_code.db
section_code.db.db
term_code
grade.sp11
grade.fa12
grade.sp13
grade.sp15
term_GPA.sp11
term_GPA.fa12
term_GPA.sp15
0001
23
102
001
001
001
FA12
3.45
3.8
3.0
-
3.8
3.7
-
0002
28
201
003
003
003
FA12
3.2
3.4
3.0
-
3.8
3.7
-
1020
28
201
003
003
003
FA12
3.2
3.4
-
-
3.8
3.7
-
6783
30
101
-
-
-
SP15
-
-
-
3.8
-
-
4.0
where I have successfully added horizontally for students 0001 and 0002 for outcomes in subsequent courses in subsequent semesters. I have also managed to add vertically, like with student 6783, leaving blanks for previous semesters before they enrolled but still adding the relevant columns.
Questions:
Is there a way to pipe full_join() into a filter() into a bind_rows() without running into these errors?
rbind number of columns do not match
OR
Error: Can't combine ..1$term_code 'character> and ..2$term_code '<integer.
Is there a easy way to keep certain columns and only add the suffix ".fa12" to certain columns? As you can see, the .db is piling up.
Is there any way to automate this? Loops aren't my strong suit, but I'm sure there's a better-looking code than doing each of the 23 joins/binds by hand.
Thank you for assistance!
Current code for simplicity:
#reproducible example
fa11 <- structure(list(id = c("1001", "1002", "1003",
"1013"), act6_05_composite = c(33L, 26L, 27L, 25L), course_code = c("101",
"101", "101", "101"), term_code = c("FA11", "FA11", "FA11", "FA11"
), section_code = c(1L, 1L, 1L, 1L), grade = c(4, 0, 0, 2.5
), repeat_status_flag = c(NA, "PR", NA, NA), class_code = c(1L,
1L, 1L, 1L), cum_atmpt_credits_prior = c(16, 0, 0, 0), cum_completed_credits_prior = c(0L,
0L, 0L, 0L), cum_passed_credits_prior = c(16, 0, 0, 0), cum_gpa_prior = c(0,
0, 0, 0), cum_atmpt_credits_during = c(29, 15, 18, 15), cum_completed_credits_during = c(13L,
1L, 10L, 15L), cum_passed_credits_during = c(29, 1, 14, 15),
term_gpa = c(3.9615, 0.2333, 2.3214, 2.9666), row.names = c(NA, 4L
), class = "data.frame")
sp12 <- structure(list(id = c("1007", "1013", "1355",
"2779", "2302"), act6_05_composite = c(24L, 26L, 25L, 24L,
24L), course_code = c(101L, 102L, 101L, 101L, 101L
), term_code = c(NA_integer_, NA_integer_, NA_integer_, NA_integer_,
NA_integer_), section_code = c(1L, 1L, 1L, 1L, 1L), grade = c(2,
2.5, 2, 1.5, 3.5), repeat_status_flag = c(NA_character_,
NA_character_, NA_character_, NA_character_, NA_character_
), class_code = c(2L, 2L, 1L, 2L, 2L), cum_atmpt_credits_prior = c(44,
43, 12, 43, 30), cum_completed_credits_prior = c(41L, 43L,
12L, 43L, 12L), cum_passed_credits_prior = c(41, 43, 12,
43, 30), cum_gpa_prior = c(3.3125, 3.186, 3.5416, 3.1785,
3.8636), cum_atmpt_credits_during = c(56, 59, 25, 64, 43),
cum_completed_credits_during = c(53L, 56L, 25L, 56L, 25L),
cum_passed_credits_during = c(53, 59, 25, 64, 43), term_gpa = c(2.8333,
3.423, 3.1153, 2.1923, 3.6153), row.names = c(NA,
5L), class = "data.frame")
# make object from fall 2011 semester dataframe
programmatic_database <- fa11
# join the spring 2012 semester dataframe by id using select variables and attaching relevant suffix
programmatic_database <- programmatic_database %>%
full_join(select(sp12, id, course_code, section_code, grade, term_gpa), by = "id", copy = TRUE, suffix = c(".fa11", ".sp12"), keep = FALSE, name = "id")
#view results of join, force integer type on certain variables as needed (see error above)
#filter the joined entries from fall 2012 database, then bind the remaining entries to the bottom of the growing dataset
programmatic_database <- sp12[!which(sp12$id %in% programmatic_database),] %>% dplyr::bind_rows(programmatic_database, sp12)
It would be possible to use bind_rows here if you make the column types consistent between tables. For instance, you could make a function to re-type any particular columns that aren't consistent in your original data. (That might also be something you could fix upstream as you read it in.)
library(dplyr)
set_column_types <- function(df) {
df %>%
mutate(term_code = as.character(term_code),
course_code = as.character(course_code))
}
bind_rows(
fa11 %>% set_column_types(),
sp12 %>% set_column_types() %>% mutate(term_code = "SP12")
)
This will stack your data into a relatively "long" format, like below. You may want to then reshape it depending on what kind of subsequent calculations you want to do.
id act6_05_composite course_code term_code section_code grade repeat_status_flag class_code cum_atmpt_credits_prior cum_completed_credits_prior cum_passed_credits_prior cum_gpa_prior cum_atmpt_credits_during cum_completed_credits_during cum_passed_credits_during term_gpa
1 1001 33 101 FA11 1 4.0 <NA> 1 16 0 16 0.0000 29 13 29 3.9615
2 1002 26 101 FA11 1 0.0 PR 1 0 0 0 0.0000 15 1 1 0.2333
3 1003 27 101 FA11 1 0.0 <NA> 1 0 0 0 0.0000 18 10 14 2.3214
4 1013 25 101 FA11 1 2.5 <NA> 1 0 0 0 0.0000 15 15 15 2.9666
5 1007 24 101 SP12 1 2.0 <NA> 2 44 41 41 3.3125 56 53 53 2.8333
6 1013 26 102 SP12 1 2.5 <NA> 2 43 43 43 3.1860 59 56 59 3.4230
7 1355 25 101 SP12 1 2.0 <NA> 1 12 12 12 3.5416 25 25 25 3.1153
8 2779 24 101 SP12 1 1.5 <NA> 2 43 43 43 3.1785 64 56 64 2.1923
9 2302 24 101 SP12 1 3.5 <NA> 2 30 12 30 3.8636 43 25 43 3.6153
I was wondering if there was a more elegant, less clunky and faster way to do this. I have millions of rows with ICD coding for clinical data. A short example provided below. I was to subset the dataset based on either of the columns meeting a specific set of diagnosis codes. The code below works but takes ages in R and was wondering if there is a faster way.
structure(list(eid = 1:10, mc1 = structure(c(4L, 3L, 5L, 2L,
1L, 1L, 1L, 1L, 1L, 1L), .Label = c("345", "410", "413.9", "I20.1",
"I23.4"), class = "factor"), oc1 = c(350, 323, 12, 35, 413.1,
345, 345, 345, 345, 345), oc2 = structure(c(5L, 6L, 4L, 1L, 1L,
2L, 2L, 2L, 3L, 2L), .Label = c("", "345", "I20.3", "J23.6",
"K50.1", "K51.4"), class = "factor")), .Names = c("eid", "mc1",
"oc1", "oc2"), class = c("data.table", "data.frame"), row.names = c(NA,
-10L), .internal.selfref = <pointer: 0x102812578>)
The code below subsets all rows that meet the code of either "I20" or "413" (this would include all codes that have for example been coded as "I20.4" or "413.9" etc.
dat2 <- dat [substr(dat$mc1,1,3)== "413"|
substr(dat$oc1,1,3)== "413"|
substr(dat$oc2,1,3)== "413"|
substr(dat$mc1,1,3)== "I20"|
substr(dat$oc1,1,3)== "I20"|
substr(dat$oc2,1,3)== "I20"]
Is there a faster way to do this? For example can i loop through each of the columns looking for the specific codes "I20" or "413" and subset those rows?
We can specify the columns of interest in .SDcols, loop through the Subset of Data.table (.SD), get the first 3 characters with substr, check whether it is %in% a vector of values and Reduce it to a single logical vector for subsetting the rows
dat[dat[,Reduce(`|`, lapply(.SD, function(x)
substr(x, 1, 3) %chin% c('413', 'I20'))), .SDcols = 2:4]]
# eid mc1 oc1 oc2
#1: 1 I20.1 350.0 K50.1
#2: 2 413.9 323.0 K51.4
#3: 5 345 413.1
#4: 9 345 345.0 I20.3
For larger data it could help if we dont chech all rows:
minem <- function(dt, colsID = 2:4) {
cols <- colnames(dt)[colsID]
x <- c('413', 'I20')
set(dt, j = "inn", value = F)
for (i in cols) {
dt[inn == F, inn := substr(get(i), 1, 3) %chin% x]
}
dt[inn == T][, inn := NULL][]
}
n <- 1e7
set.seed(13)
dt <- dts[sample(.N, n, replace = T)]
dt <- cbind(dt, dts[sample(.N, n, replace = T), 2:4])
setnames(dt, make.names(colnames(dt), unique = T))
dt
# eid mc1 oc1 oc2 mc1.1 oc1.1 oc2.1
# 1: 8 345 345.0 345 345 345 345
# 2: 3 I23.4 12.0 J23.6 413.9 323 K51.4
# 3: 4 410 35.0 413.9 323 K51.4
# 4: 1 I20.1 350.0 K50.1 I23.4 12 J23.6
# 5: 10 345 345.0 345 345 345 345
# ---
# 9999996: 3 I23.4 12.0 J23.6 I20.1 350 K50.1
# 9999997: 5 345 413.1 I20.1 350 K50.1
# 9999998: 4 410 35.0 345 345 345
# 9999999: 4 410 35.0 410 35
# 10000000: 10 345 345.0 345 345 345 I20.3
system.time(r1 <- akrun(dt, 2:ncol(dt))) # 22.88 sek
system.time(r2 <- minem(dt, 2:ncol(dt))) # 17.72 sek
all.equal(r1, r2)
# [1] TRUE
I would like to know how to divide the column values in to three different ranges based on scores.
Here's the following data I have
Name V1.1 V1.2 V2.1 V2.2 V3.1 V3.2
John French 86 Math 78 English 56
Sam Math 97 French 86 English 79
Viru English 93 Math 44 French 34
If I consider three ranges. First rangewith 0-60, Second rangewith 61-90 and third range with 91-100.
I would like to the subject name across all the skills.
Expected result would be
Name Level1 Level2 Level3
John English Math,French Null
Sam Null French,Eng Math
Viru Math,Fren Null English
First you need to convert the data to long form, one row per observation (where an observation is a single score. You need to do a melt, but it is complicated because your wide form consists of not only observations but observation classes. One way to do it is to use melt.data.table twice, but you may be more comfortable with dplyr, which has more accessible syntax.
# first you need to convert to long form
library("data.table")
setDT(df)
lhs <- melt.data.table(df, id = "Name", measure = patterns("\\.2"),
variable.name = "obs", value.name = "score")
lhs[, obs := gsub("(V\\d+)\\.\\d+","\\1",obs)]
lhs
rhs <- melt.data.table(df, id = "Name", measure = patterns("V\\d\\.1"),
variable.name = "obs", value.name = "subject")
rhs[, obs := gsub("(V\\d+)\\.\\d+","\\1",obs)]
rhs
df2 <- merge(lhs, rhs, by = c("Name","obs"))
# Name obs score1 subject1
# 1: John V1 86 French
# 2: John V2 78 Math
# 3: John V3 56 English
# 4: Sam V1 97 Math
# 5: Sam V2 86 French
# 6: Sam V3 79 English
# 7: Viru V1 93 English
# 8: Viru V2 44 Math
# 9: Viru V3 34 French
Then you need to use cut or some other function to create the three levels based on score1.
Then you should group by these levels and apply concatenation to the subjects, such as paste(..., collapse = ",").
Then you need to use cast or spread to return it to wide form.
Do give it some effort, and edit your question with what you've tried, and try to come up with a more specific question, not just "please do this for me".
Another option using splitstackshape and nested ifelse
library(splitstackshape)
library(tidyr)
# prepare data to convert in long format
data$subjects = do.call(paste, c(data[,grep("\\.1", colnames(data))], sep = ','))
data$marks = do.call(paste, c(data[,grep("\\.2", colnames(data))], sep = ','))
data = data[,-grep("V", colnames(data))]
# use cSplit to convert wide to long
out = cSplit(setDT(data), sep = ",", c("subjects", "marks"), "long")
# nested ifelse to assign level based on the score range
out[, level := ifelse(marks <= 60, "level1",
ifelse(marks <= 90, "level2", "level3"))]
req = out[, toString(subjects), by= c("Name","level")]
this will give
#> req
# Name level V1
#1: John level2 French, Math
#2: John level1 English
#3: Sam level3 Math
#4: Sam level2 French, English
#5: Viru level3 English
#6: Viru level1 Math, French
you can reshape either using dcast or spread from tidyr
spread(req, level, V1)
# Name level1 level2 level3
#1: John English French, Math NA
#2: Sam NA French, English Math
#3: Viru Math, French NA English
data
data = structure(list(Name = structure(1:3, .Label = c("John", "Sam",
"Viru"), class = "factor"), V1.1 = structure(c(2L, 3L, 1L), .Label = c("English",
"French", "Math"), class = "factor"), V1.2 = c(86L, 97L, 93L),
V2.1 = structure(c(2L, 1L, 2L), .Label = c("French", "Math"
), class = "factor"), V2.2 = c(78L, 86L, 44L), V3.1 = structure(c(1L,
1L, 2L), .Label = c("English", "French"), class = "factor"),
V3.2 = c(56L, 79L, 34L)), .Names = c("Name", "V1.1", "V1.2",
"V2.1", "V2.2", "V3.1", "V3.2"), class = "data.frame", row.names = c(NA,
-3L))
Not very intuitive, but leads to the requested output. Requires the package sjmisc!
library(sjmisc)
mydat <- data.frame(Name = c("John", "Sam", "Viru"),
V1.1 = c("French", "Math", "English"),
V1.2 = c(86, 97, 93),
V2.1 = c("Math", "French", "Math"),
V2.2 = c(78, 86, 44),
V3.1 = c("English", "English", "French"),
V3.2 = c(56, 79, 34))
# recode into groups
rec(mydat[, c(3,5,7)]) <- "min:60=1; 61:90=2; 91:max=3"
# convert to long format
newdf <- to_long(mydat, "no_use",
c("subject", "score"),
c("V1.1", "V2.1", "V3.1"),
c("V1.2", "V2.2", "V3.2")) %>%
select(-no_use) %>%
arrange(Name, score)
# at this point we are at a similar stage as described in the
# other answers, so we have our data in a long format
newdf
fdf <- list()
# iterate all unique names
for (i in unique(newdf$Name)) {
dummy <- c()
# iterare all three scores
for (s in 1:3) {
# find subjects related to the score
dat <- newdf$subject[newdf$Name == i & newdf$score == s]
if (length(dat) == 0) dat <- ""
dat <- paste0(dat, collapse = ",")
dummy <- c(dummy, dat)
}
# add character vector with sorted subjects to list
fdf[[length(fdf) + 1]] <- dummy
}
# list to data frame
finaldf <- as.data.frame(t(as.data.frame(fdf)))
finaldf <- cbind(unique(newdf$Name), finaldf)
# proper row/col names
colnames(finaldf) <- c("Names", "Level1", "Level2", "Level3")
rownames(finaldf) <- 1:nrow(finaldf)
finaldf