Iteratively create global environment objects from tibble - r

I'm trying to make objects directly from information listed in a tibble that can be called on by later functions/tibbles in my environment. I can make the objects manually but I'm working to do this iteratively.
library(tidyverse)
##determine mean from 2x OD Negatives in experimental plates, then save summary for use in appending table
ELISA_negatives = "my_file.csv"
neg_tibble <- as_tibble(read_csv(ELISA_negatives, col_names = TRUE)) %>%
group_by(Species_ab, Antibody, Protein) %>%
filter(str_detect(Animal_ID, "2x.*")) %>%
summarize(ave_neg_U_mL = mean(U_mL, na.rm = TRUE), n=sum(!is.na(U_mL)))
neg_tibble
# A tibble: 4 x 5
# Groups: Species_ab, Antibody [2]
Species_ab Antibody Protein ave_neg_U_mL n
<chr> <chr> <chr> <dbl> <int>
1 Mouse IgG GP 28.2 6
2 Mouse IgG NP 45.9 6
3 Rat IgG GP 5.24 4
4 Rat IgG NP 1.41 1
I can write the object manually based off the above tibble:
Mouse_IgG_GP_cutoff <- as.numeric(neg_tibble[1,4])
Mouse_IgG_GP_cutoff
[1] 28.20336
In my attempt to do this iteratively, I can make a new tibble neg_tibble_string with the information I need. All I would need to do now is make a global object from the Name in the first column Test_Name, and assign it to the numeric value in the second column ave_neg_U_mL (which is where I'm getting stuck).
neg_tibble_string <- neg_tibble %>%
select(Species_ab:Protein) %>%
unite(col='Test_Name', c('Species_ab', 'Antibody', 'Protein'), sep = "_") %>%
mutate(Test_Name = str_c(Test_Name, "_cutoff")) %>%
bind_cols(neg_tibble[4])
neg_tibble_string
# A tibble: 4 x 2
Test_Name ave_neg_U_mL
<chr> <dbl>
1 Mouse_IgG_GP_cutoff 28.2
2 Mouse_IgG_NP_cutoff 45.9
3 Rat_IgG_GP_cutoff 5.24
4 Rat_IgG_NP_cutoff 1.41
I feel like there has to be a way to do this to get this from the above tibble neg_tibble_string, and make this for all four of the rows. I've tried a variant of this and this, but can't get anywhere.
> list_df <- mget(ls(pattern = "neg_tibble_string"))
> list_output <- map(list_df, ~neg_tibble_string$ave_neg_U_mL)
Warning message:
Unknown or uninitialised column: `ave_neg_U_mL`.
> list_output
$neg_tibble_string
NULL
As always, any insight is appreciated! I'm making progress on my R journey but I know I am missing large gaps in knowledge.

As we already returned the object value in a list, we need only to specify the lambda function i.e. .x returns the value of the list element which is a tibble and extract the column
library(purrr)
list_output <- map(list_df, ~.x$ave_neg_U_ml)
If the intention is to create global objects, deframe, convert to a list and then use list2env
library(tibble)
list2env(as.list(deframe(neg_tibble_string)), .GlobalEnv)

Related

R for loop to extract info from a file and add it into tibble?

I am not great with tidyverse so forgive me if this is a simple question. I have a bunch of files with data that I need to extract and add into distinct columns in a tibble I created.
I want the the row names to start with the file IDs which I did manage to create:
filelist <- list.fileS(pattern=".txt") # Gives me the filenames in current directory.
# The filenames are something like AA1230.report.txt for example
file_ID <- trimws(filelist, whitespace="\\..*") # Gives me the ID which is before the "report.txt"
metadata <- as_tibble(file_ID[1:181]) # create dataframe with IDs as row names for 180 files.
Now in these report files are information on species and abundance (kraken report files for those familiar with kraken) and all I need is to extract the number of reads for each domain. I can easily search up in each file the domains and number of reads that fall into that domain using something like:
sample_data <- as_tibble(read.table("AA1230.report.txt", sep="\t", header=FALSE, strip.white=TRUE))
sample_data <- rename(sample_data, Percentage=V1, Num_reads_root=V2, Num_reads_taxon=V3, Rank=V4, NCBI_ID=V5, Name=V6) # Just renaming the column headers for clarity
sample_data %>% filter(Rank=="D") # D for domain
This gives me a clear output such as:
Percentage Num_Reads_Root Num_Reads_Taxon Rank NCBI_ID Name
<dbl> <int> <int> <fct> <int> <fct>
1 75.9 60533 28 D 2 Bacteria
2 0.48 386 0 D 2759 Eukaryota
3 0.01 4 0 D 2157 Archaea
4 0.02 19 0 D 10239 Viruses
Now, I want to just grab the info in the second column and final column and save this info into my tibble so that I can get something like:
> metadata
value Bacteria_Counts Eukaryota_Counts Viruses_Counts Archaea_Counts
<chr> <int> <int> <int> <int>
1 AA1230 60533 386 19 4
2 AB0566
3 AA1231
4 AB0567
5 BC1148
6 AW0001
7 AW0002
8 BB1121
9 BC0001
10 BC0002
....with 171 more rows
I'm just having trouble coming up with a for loop to create these sample_data outputs, then from that, extract the info and place into a tibble. I guess my first loop should create these sample_data outputs so something like:
for (files in file.list()) {
>> get_domains <<
}
Then another loop to extract that info from the above loop and insert it into my metadata tibble.
Any suggestions? Thank you so much!
PS: If regular dataframes in R is better for this let me know, I have just recently learned that tidyverse is a better way to organize dataframes in R but I have to learn more about it.
You could also do:
library(tidyverse)
filelist <- list.files(pattern=".txt")
nms <- c("Percentage", "Num_reads_root", "Num_reads_taxon", "Rank", "NCBI_ID", "Name")
set_names(filelist,filelist) %>%
map_dfr(read_table, col_names = nms, .id = 'file_ID') %>%
filter(Rank == 'D') %>%
select(file_ID, Name, Num_reads_root) %>%
pivot_wider(id_cols = file_ID, names_from = Name, values_from = Num_reads_root) %>%
mutate(file_ID = str_remove(file_ID, '.txt'))
I've found that using a for loop is nice sometimes because saves all the progress along the way in case you hit an error. Then you can find the problem file and debug it or use try() but throw a warning().
library(tidyverse)
filelist <- list.files(pattern=".txt") #list files
tmp_list <- list()
for (i in seq_along(filelist)) {
my_table <- read_tsv(filelist[i]) %>% # It looks like your files are all .tsv's
rename(Percentage=V1, Num_reads_root=V2, Num_reads_taxon=V3, Rank=V4, NCBI_ID=V5, Name=V6) %>%
filter(Rank=="D") %>%
mutate(file_ID <- trimws(filelist[i], whitespace="\\..*")) %>%
select(file_ID, everything())
tmp_list[[i]] <- my_table
}
out <- bind_rows(tmp_list)
out

Using dynamic names with dplyr

I have a data frame in which I would I would like to compute some extra column as a function of the existing columns, but want to specify both each new column name and the function dynamically. I have a vector of column names that are already in the dataframe df_daily:
DAILY_QUESTIONS <- c("Q1_Daily", "Q2_Daily", "Q3_Daily", "Q4_Daily", "Q5_Daily")
The rows of the dataframe have responses to each question from each user each time they answer the questionnaire, as well as a column with the number of days since the user first answered the questionnaire (i.e. Days_From_First_Use = 0 on the very first use, = 1 if it is used the next day etc.). I want to average the responses to these questions by Days_From_First_Use . I start by by grouping my dataframe by Days_From_First_Use:
df_test <- df_daily %>%
group_by(Days_From_First_Use)
and then try averaging the responses in a loop as follows:
for(i in 1:5){
df_test <- df_test %>%
mutate(!! paste0('Avg_Score_', DAILY_QUESTIONS[i]) :=
paste0('mean(', DAILY_QUESTIONS[i], ')'))
}
Unfortunately, while my new variable names are correct ("Avg_Score_Q1_Daily", "Avg_Score_Q2_Daily", "Avg_Score_Q3_Daily", "Avg_Score_Q4_Daily", "Avg_Score_Q5_Daily"), my answers are not: every row in my data frame has a string such as "mean(Q1_Daily)" in the relevant column .
So I'm clearly doing something wrong - what do I need to do fix this and get the average score across all users on each day?
Sincerely and with many thanks in advance
Thomas Philips
I took a somewhat different approach, using summarize(across(...)) after group_by(Days_From_First_Use) I achieve the dynamic names by using rename_with and a custom function that replaces (starts with)"Q" with "Avg_Score_Q"
library(dplyr, warn.conflicts = FALSE)
# fake data -- 30 normalized "responses" from 0 to 2 days from first use to 5 questions
DAILY_QUESTIONS <- c("Q1_Daily", "Q2_Daily", "Q3_Daily", "Q4_Daily", "Q5_Daily")
df_daily <- as.data.frame(do.call('cbind', lapply(1:5, function(i) rnorm(30, i))))
colnames(df_daily) <- DAILY_QUESTIONS
df_daily$Days_From_First_Use <- floor(runif(30, 0, 3))
df_test <- df_daily %>%
group_by(Days_From_First_Use) %>%
summarize(across(.fns = mean)) %>%
rename_with(.fn = function(x) gsub("^Q","Avg_Score_Q",x))
#> `summarise()` ungrouping output (override with `.groups` argument)
df_test
#> # A tibble: 3 x 6
#> Days_From_First… Avg_Score_Q1_Da… Avg_Score_Q2_Da… Avg_Score_Q3_Da…
#> <dbl> <dbl> <dbl> <dbl>
#> 1 0 1.26 1.75 3.02
#> 2 1 0.966 2.14 3.48
#> 3 2 1.08 2.45 3.01
#> # … with 2 more variables: Avg_Score_Q4_Daily <dbl>, Avg_Score_Q5_Daily <dbl>
Created on 2020-12-06 by the reprex package (v0.3.0)

R Beginner struggling with extremely messy XLSX

I got an XLSX with data from a questionnaire for my master thesis.
The questions and answers for an interviewee are in one row in the second column. The first column contains the date.
The data of the second column comes in a form like this:
"age":"52","height":"170","Gender":"Female",...and so on
I started with:
test12 <- read_xlsx("Testdaten.xlsx")
library(splitstackshape)
test13 <- concat.split(data = test12, split.col= "age", sep =",")
Then I got the questions and the answers as a column divided by a ":".
For e.g. column 1: "age":"52" and column2:"height":"170".
But the data is so messy that sometimes in the column of the age question and answer there is a height question and answer and for some questionnaires questions and answers double.
I would need the questions as variables and the answers as observations. But I have no clue how to get there. I could clean the data in excel first, but with the fact that columns are not constant and there are for e.g. some height questions in the age column I see no chance to do it as I will get new data regularly, formated the same way.
Here is an example of the data:
A tibble: 5 x 2
partner.createdAt partner.wphg.info
<chr> <chr>
1 2019-11-09T12:13:11.099Z "{\"age_years\":\"50\",\"job_des\":\"unemployed\",\"height_cm\":\"170\",\"Gender\":\"female\",\"born_in\":\"Italy\",\"Alcoholic\":\"false\",\"knowledge_selfass\":\"5\",\"total_wealth\":\"200000\""
2 2019-11-01T06:43:22.581Z "{\"age_years\":\"34\",\"job_des\":\"self-employed\",\"height_cm\":\"158\",\"Gender\":\"male\",\"born_in\":\"Germany\",\"Alcoholic\":\"true\",\"knowledge_selfass\":\"3\",\"total_wealth\":\"10000\""
3 2019-11-10T07:59:46.136Z "{\"age_years\":\"24\",\"height_cm\":\"187\",\"Gender\":\"male\",\"born_in\":\"England\",\"Alcoholic\":\"false\",\"knowledge_selfass\":\"3\",\"total_wealth\":\"150000\""
4 2019-11-11T13:01:48.488Z "{\"age_years\":\"59\",\"job_des\":\"employed\",\"height_cm\":\"167\",\"Gender\":\"female\",\"born_in\":\"United States\",\"Alcoholic\":\"false\",\"knowledge_selfass\":\"2\",\"total_wealth\":\"1000000~
5 2019-11-08T14:54:26.654Z "{\"age_years\":\"36\",\"height_cm\":\"180\",\"born_in\":\"Germany\",\"Alcoholic\":\"false\",\"knowledge_selfass\":\"5\",\"total_wealth\":\"170000\",\"job_des\":\"employed\",\"Gender\":\"male\""
Thank you so much for your time!
You can loop through each entry, splitting at , as you did. Then you can loop through them all again, splitting at :.
The result will be a bunch of variable/value pairings. This can be all done stacked. Then you just want to pivot back into columns.
data
Updated the data based on your edit.
data <- tribble(~partner.createdAt, ~partner.wphg.info,
'2019-11-09T12:13:11.099Z', '{\"age_years\":\"50\",\"job_des\":\"unemployed\",\"height_cm\":\"170\",\"Gender\":\"female\",\"born_in\":\"Italy\",\"Alcoholic\":\"false\",\"knowledge_selfass\":\"5\",\"total_wealth\":\"200000\"',
'2019-11-01T06:43:22.581Z', '{\"age_years\":\"34\",\"job_des\":\"self-employed\",\"height_cm\":\"158\",\"Gender\":\"male\",\"born_in\":\"Germany\",\"Alcoholic\":\"true\",\"knowledge_selfass\":\"3\",\"total_wealth\":\"10000\"',
'2019-11-10T07:59:46.136Z', '{\"age_years\":\"24\",\"height_cm\":\"187\",\"Gender\":\"male\",\"born_in\":\"England\",\"Alcoholic\":\"false\",\"knowledge_selfass\":\"3\",\"total_wealth\":\"150000\"',
'2019-11-11T13:01:48.488Z', '{\"age_years\":\"59\",\"job_des\":\"employed\",\"height_cm\":\"167\",\"Gender\":\"female\",\"born_in\":\"United States\",\"Alcoholic\":\"false\",\"knowledge_selfass\":\"2\",\"total_wealth\":\"1000000\"',
'2019-11-08T14:54:26.654Z', '{\"age_years\":\"36\",\"height_cm\":\"180\",\"born_in\":\"Germany\",\"Alcoholic\":\"false\",\"knowledge_selfass\":\"5\",\"total_wealth\":\"170000\",\"job_des\":\"employed\",\"Gender\":\"male\"')
libraries
We need a few here. Or you can just call tidyverse.
library(stringr)
library(purrr)
library(dplyr)
library(tibble)
library(tidyr)
function
This function will create a data frame (or tibble) for each question. The first column is the date, the second is the variable, the third is the value.
clean_record <- function(date, text) {
clean_records <- str_split(text, pattern = ",", simplify = TRUE) %>%
str_remove_all(pattern = "\\\"") %>% # remove double quote
str_remove_all(pattern = "\\{|\\}") %>% # remove curly brackets
str_split(pattern = ":", simplify = TRUE)
tibble(date = as.Date(date), variable = clean_records[,1], value = clean_records[,2])
}
iteration
Now we use pmap_dfr from purrr to loop over the rows, outputting each row with an id variable named record.
This will stack the data as described in the function. The mutate() line converts all variable names to lowercase. The distinct() line will filter out rows that are exact duplicates.
What we do then is just pivot on the variable column. Of course, replace data with whatever you name your data frame.
data_clean <- pmap_dfr(data, ~ clean_record(..1, ..2), .id = "record") %>%
mutate(variable = tolower(variable)) %>%
distinct() %>%
pivot_wider(names_from = variable, values_from = value)
result
The result is something like this. Note how I had reordered some of the columns, but it still works. You are probably not done just yet. All columns are now of type character. You need to figure out the desired type for each and convert.
# A tibble: 5 x 10
record date age_years job_des height_cm gender born_in alcoholic knowledge_selfass total_wealth
<chr> <date> <chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr>
1 1 2019-11-09 50 unemployed 170 female Italy false 5 200000
2 2 2019-11-01 34 self-employed 158 male Germany true 3 10000
3 3 2019-11-10 24 NA 187 male England false 3 150000
4 4 2019-11-11 59 employed 167 female United States false 2 1000000
5 5 2019-11-08 36 employed 180 male Germany false 5 170000
For example, convert age_years to numeric.
data_clean %>%
mutate(age_years = as.numeric(age_years))
I am sure you may run into other things, but this should be a start.

Calling specific cells in the same column (using dplyr?)

I have a dataframe with character and numeric data. I would like to use dplyr to create a summary grouped by time points and trials generating the following:
averages
standard deviations
variation
ratio between time points
(etc etc)
I feel like all of this could be done in the dplyr pipe, but I am struggling to make a ratio of averages between time points within trials.
I fully admit that I may be carrying around a hammer looking for nails, so please feel free to recommend solutions that utilize other packages or functions, but ideally I'd like simple/straight forward code for ease of use by multiple collaborators.
library(dplyr)
# creating an example DF
num <- runif(100, 50, 3200)
smpl <- 1:100
df <- data.frame( num, smpl)
df$time <- "time1"
df$time[seq(2,100,2)] <- "time2"
df$trial <- "a"
df$trial[26:50] <- "b"
df$trial[51:75] <- "c"
df$trial[75:100] <- "d"
# using the magic of pipelines to calculate useful things
df1 <- df %>%
group_by(time, trial) %>%
summarise(avg = mean(num),
var = var(num),
stdev = sd(num))
I'd love to get [the ratio time2/time1 of the avg for each trial] included in this block above, but I don't know how to call "avg" specifically by "time1" vs "time2" within the pipe.
From here on, nothing does quite what I'm hoping for...
df1 <- df1[with(df1,order(trial,time)),]
# this better ressembles my actual DF structure,
# so reordering it will make some of my next attempts to solve this make more sense
I tried to use the fact that 'every other line' is different (this is not ideal because each df will have a different number of rows, so I will either introduce NAs or it will require constantly change these #'s (or writing a function to constantly change them))
tm2 <- data.frame(x=df1$avg[seq(2,4,2)])
tm1 <- data.frame(x=df1$avg[seq(1,3,2)])
so minimally, this is the ratio I'd like included in the df, but tied to the avg & trial columns:
tm2/tm1
It doesn't matter to me 'which' time row this ratio ends up in, so long as it is consistent across all the trials (so if a column of ratios has "blank" for every "time1" and "value" for every "time2", that's fine).
# I added in a separate column to allow 'match' later
tm1$time <- "time1"
tm2$time <- "time1" # to keep them all 'in row'
df1$avg_tm1 <- tm1$x[match(df1$time, tm1$time)]
df1$avg_tm2 <- tm2$x[match(df1$time, tm2$time)]
but this fails to match by 'trial' also, since that info is lost in this new tm1 df ; this really makes me think it should all be done in dplry the first time...
Then I tried to create a new column in the tm1 df with the ratio
tm2$ratio <-tm2$x/tm1$x
and add in the ratio values only if the avg matches
df1$ratio <- tm2$ratio[match(tm2$x, df1$avg)]
This might work, but when I extract the avg values, it rounds, so the numbers do not match exactly. I'm also cautious about this because if I process ridiculous amounts of data, there's a higher and higher chance that two random averages will be similar enough to misplace these ratios.
I tried several other things that completely failed, so let's pretend that something worked and entered the ratio into the df1 as separate columns
Then any further calculations or annotations are straight forward:
df2 <- df1 %>%
mutate(ratio = avg_tm2/avg_tm1,
lost = 1- ratio,
word = paste0(round(lost*100),"%"))
But I am still stuck on 'how' to call specific cells inside the pipe or which other tools/packages to use to calculate deltas or ratios between cells in the same column.
Thanks in advance
We could group by 'trial' and mutate to create the 'ratio' column
df1 %>%
group_by(trial) %>%
mutate(ratio = last(avg)/first(avg))
# A tibble: 8 x 6
# Groups: trial [4]
# time trial avg var stdev ratio
# <chr> <chr> <dbl> <dbl> <dbl> <dbl>
#1 time1 a 1815. 715630. 846. 0.795
#2 time1 b 2012. 1299823. 1140. 0.686
#3 time1 c 1505. 878168. 937. 1.09
#4 time1 d 1387. 902364. 950. 1.17
#5 time2 a 1444. 998943. 999. 0.795
#6 time2 b 1380. 720135. 849. 0.686
#7 time2 c 1641. 1205778. 1098. 1.09
#8 time2 d 1619. 582418. 763. 1.17
NOTE: We used set.seed(2) for creating the dataset
Work out a separate data.frame:
set.seed(2)
# your code above to generate df1
df2 <- select(df1, time, trial, avg) %>%
spread(time, avg) %>%
mutate(ratio = time2/time1)
df2
# # A tibble: 4 × 4
# trial time1 time2 ratio
# <chr> <dbl> <dbl> <dbl>
# 1 a 1815.203 1443.731 0.7953555
# 2 b 2012.436 1379.981 0.6857266
# 3 c 1505.474 1641.439 1.0903135
# 4 d 1386.876 1619.341 1.1676176
and now you can merge the relevant column onto the original frame:
left_join(df1, select(df2, trial, ratio), by="trial")
# Source: local data frame [8 x 6]
# Groups: time [?]
# time trial avg var stdev ratio
# <chr> <chr> <dbl> <dbl> <dbl> <dbl>
# 1 time1 a 1815.203 715630.4 845.9494 0.7953555
# 2 time1 b 2012.436 1299823.3 1140.0979 0.6857266
# 3 time1 c 1505.474 878168.3 937.1063 1.0903135
# 4 time1 d 1386.876 902363.7 949.9282 1.1676176
# 5 time2 a 1443.731 998943.3 999.4715 0.7953555
# 6 time2 b 1379.981 720134.6 848.6074 0.6857266
# 7 time2 c 1641.439 1205778.0 1098.0792 1.0903135
# 8 time2 d 1619.341 582417.5 763.1629 1.1676176

Programmatically create new variables using purrr?

Intro
After recently taking Hadley Wickham's functional programming class I decided I'd try applying some of the lessons to my projects at work. Naturally, the first project I tried has proven to be more complicated than the examples worked demonstrated in the class. Does anyone have recommendations for a way to use the purrr package to make the task described below more efficient?
Project Background
I need to assign quintile groups to records in a spatial polygon dataframe. In addition to the record identifier there are several other variables and I need to calculate the quintile group for each.
Here's the crux of the problem: I have been asked to identify outliers in one particular variable and to omit those records from the entire analysis as long as it doesn't change the quintile composition of the first quintile group for any of the other variables.
Question
I have put together a dplyr pipeline (see the example below) that performs this checking process for a single variable, but how might I rewrite this process so that I can efficiently check each variable?
EDIT: While it is certainly possible to change the shape of the data from wide to long as an intermediary step, in the end it needs to return to its wide format so that it matches up with the #polygons slot of the spatial polygons dataframe.
Reproducible Example
You can find the complete script here: https://gist.github.com/tiernanmartin/6cd3e2946a77b7c9daecb51aa11e0c94
Libraries and Settings
library(grDevices) # boxplot.stats()
library(operator.tools) # %!in% logical operator
library(tmap) # 'metro' data set
library(magrittr) # piping
library(dplyr) # exploratory data analysis verbs
library(purrr) # recursive mapping of functions
library(tibble) # improved version of a data.frame
library(ggplot2) # dot plot
library(ggrepel) # avoid label overlap
options(scipen=999)
set.seed(888)
Load the example data and take a small sample of it
data("metro")
m_spdf <- metro
# Take a sample
m <-
metro#data %>%
as_tibble %>%
select(-name_long,-iso_a3) %>%
sample_n(50)
> m
# A tibble: 50 x 10
name pop1950 pop1960 pop1970 pop1980 pop1990
<chr> <dbl> <dbl> <dbl> <dbl> <dbl>
1 Sydney 1689935 2134673 2892477 3252111 3631940
2 Havana 1141959 1435511 1779491 1913377 2108381
3 Campinas 151977 293174 540430 1108903 1693359
4 Kano 123073 229203 541992 1349646 2095384
5 Omsk 444326 608363 829860 1032150 1143813
6 Ouagadougou 33035 59126 115374 265200 537441
7 Marseille 755805 928768 1182048 1372495 1418279
8 Taiyuan 196510 349535 621625 1105695 1636599
9 La Paz 319247 437687 600016 809218 1061850
10 Baltimore 1167656 1422067 1554538 1748983 1848834
# ... with 40 more rows, and 4 more variables:
# pop2000 <dbl>, pop2010 <dbl>, pop2020 <dbl>,
# pop2030 <dbl>
Calculate quintile groups with and without outlier records
# Calculate the quintile groups for one variable (e.g., `pop1990`)
m_all <-
m %>%
mutate(qnt_1990_all = dplyr::ntile(pop1990,5))
# Find the outliers for a different variable (e.g., 'pop1950')
# and subset the df to exlcude these outlier records
m_out <- boxplot.stats(m$pop1950) %>% .[["out"]]
m_trim <-
m %>%
filter(pop1950 %!in% m_out) %>%
mutate(qnt_1990_trim = dplyr::ntile(pop1990,5))
# Assess whether the outlier trimming impacted the first quintile group
m_comp <-
m_trim %>%
select(name,dplyr::contains("qnt")) %>%
left_join(m_all,.,"name") %>%
select(name,dplyr::contains("qnt"),everything()) %>%
mutate(qnt_1990_chng_lgl = !is.na(qnt_1990_trim) & qnt_1990_trim != qnt_1990_all,
qnt_1990_chng_dir = if_else(qnt_1990_chng_lgl,
paste0(qnt_1990_all," to ",qnt_1990_trim),
"No change"))
With a little help from ggplot2, I can see that in this example six outliers were identified and that their omission did not affect the first quintile group for pop1990.
Importantly, this information is tracked in two new variables: qnt_1990_chng_lgl and qnt_1990_chng_dir.
> m_comp %>% select(name,qnt_1990_chng_lgl,qnt_1990_chng_dir,everything())
# A tibble: 50 x 14
name qnt_1990_chng_lgl qnt_1990_chng_dir qnt_1990_all qnt_1990_trim
<chr> <lgl> <chr> <dbl> <dbl>
1 Sydney FALSE No change 5 NA
2 Havana TRUE 4 to 5 4 5
3 Campinas TRUE 3 to 4 3 4
4 Kano FALSE No change 4 4
5 Omsk FALSE No change 3 3
6 Ouagadougou FALSE No change 1 1
7 Marseille FALSE No change 3 3
8 Taiyuan TRUE 3 to 4 3 4
9 La Paz FALSE No change 2 2
10 Baltimore FALSE No change 4 4
# ... with 40 more rows, and 9 more variables: pop1950 <dbl>, pop1960 <dbl>,
# pop1970 <dbl>, pop1980 <dbl>, pop1990 <dbl>, pop2000 <dbl>, pop2010 <dbl>,
# pop2020 <dbl>, pop2030 <dbl>
I now need to find a way to repeat this process for every variable in the dataframe (i.e., pop1960 - pop2030). Ideally, two new variables would be created for each existing pop* variable and their names would be preceded by qnt_ and followed by either _chng_dir or _chng_lgl.
Is purrr the right tool to use for this? dplyr::mutate_? data.table?
It turns out this problem is solvable using tidyr::gather + dplyr::group_by + tidyr::spread functions. While #shayaa and #Gregor didn't provide the solution I was looking for, their advice helped me course-correct away from the functional programming methods I was researching.
I ended up using #shayaa's gather and group_by combination, followed by mutate to create the variable names (qnt_*_chng_lgl and qnt_*_chng_dir) and then using spread to make it wide again. An anonymous function passed to summarize_all removed all the extra NA's that the wide-long-wide transformations created.
m_comp <-
m %>%
mutate(qnt = dplyr::ntile(pop1950,5)) %>%
filter(pop1950 %!in% m_out) %>%
gather(year,pop,-name,-qnt) %>%
group_by(year) %>%
mutate(qntTrim = dplyr::ntile(pop,5),
qnt_chng_lgl = !is.na(qnt) & qnt != qntTrim,
qnt_chng_dir = ifelse(qnt_chng_lgl,
paste0(qnt," to ",qntTrim),
"No change"),
year_lgl = paste0("qnt_chng_",year,"_lgl"),
year_dir = paste0("qnt_chng_",year,"_dir")) %>%
spread(year_lgl,qnt_chng_lgl) %>%
spread(year_dir,qnt_chng_dir) %>%
spread(year,pop) %>%
select(-qnt,-qntTrim) %>%
group_by(name) %>%
summarize_all(function(.){subset(.,!is.na(.)) %>% first})
Nothing wrong with your analysis it seems to me,
After this part
m <- metro#data %>%
as_tibble %>%
select(-name_long,-iso_a3) %>%
sample_n(50)
Just melt your data and continue your analysis but with group_by(year)
library(reshape2)
library(stringr)
mm <- melt(m)
mm[,2] <- as.factor(str_sub(mm[,2],-4))
names(mm)[2:3] <- c("year", "population")
e.g.,
mm %>% group_by(year) %>%
+ mutate(qnt_all = dplyr::ntile(population,5))

Resources