R programming_ Subsetting rows on logic conditions - r

Sample data:
sampleData
Ozone Solar.R Wind Temp Month Day sampleData.Ozone
1 41 190 7.4 67 5 1 41
2 36 118 8.0 72 5 2 36
3 12 149 12.6 74 5 3 12
.........
Want to extract records on the condition $ozone > 31
Here is the code:
data <- sampleData[sampleData$ozone > 31]
And get the error below:
Error in if (inherits(X[[j]], "data.frame") && ncol(xj) > 1L) X[[j]] <- as.matrix(X[[j]]) :
missing value where TRUE/FALSE needed
How should I correct it? Thanks!

R is case sensitive, so your ozone has to match the name in your data.frame. Also to subset a data.frame, you need two indices (row and column) separated by a comma. If there is nothing after the comma, it means that you are selecting all the columns:
sampleData[sampleData$Ozone > 31,]
Other methods to subset a data.frame:
subset(sampleData, Ozone > 31)
or with dplyr:
library(dplyr)
sampleData %>%
filter(Ozone > 31)
Result:
Ozone Solar.R Wind Temp Month Day sampleData.Ozone
1 41 190 7.4 67 5 1 41
2 36 118 8.0 72 5 2 36
Data:
sampleData = structure(list(Ozone = c(41L, 36L, 12L), Solar.R = c(190L, 118L,
149L), Wind = c(7.4, 8, 12.6), Temp = c(67L, 72L, 74L), Month = c(5L,
5L, 5L), Day = 1:3, sampleData.Ozone = c(41L, 36L, 12L)), .Names = c("Ozone",
"Solar.R", "Wind", "Temp", "Month", "Day", "sampleData.Ozone"
), class = "data.frame", row.names = c("1", "2", "3"))

Related

Calculating Percent Change in R for Multiple Variables

I'm trying to calculate percent change in R with each of the time points included in the column label (table below). I have dplyr loaded and my dataset was loaded in R and I named it data. Below is the code I'm using but it's not calculating correctly. I want to create a new dataframe called data_per_chg which contains the percent change from "v1" each variable from. For instance, for wbc variable, I would like to calculate percent change of wbc.v1 from wbc.v1, wbc.v2 from wbc.v1, wbc.v3 from wbc.v1, etc, and do that for all the remaining variables in my dataset. I'm assuming I can probably use a loop to easily do this but I'm pretty new to R so I'm not quite sure how proceed. Any guidance will be greatly appreciated.
id
wbc.v1
wbc.v2
wbc.v3
rbc.v1
rbc.v2
rbc.v3
hct.v1
hct.v2
hct.v3
a1
23
63
30
23
56
90
13
89
47
a2
81
45
46
N/A
18
78
14
45
22
a3
NA
27
14
29
67
46
37
34
33
data_per_chg<-data%>%
group_by(id%>%
arrange(id)%>%
mutate(change=(wbc.v2-wbc.v1)/(wbc.v1))
data_per_chg
Assuming the NA values are all NA and no N/A
library(dplyr)
library(stringr)
data <- data %>%
na_if("N/A") %>%
type.convert(as.is = TRUE) %>%
mutate(across(-c(id, matches("\\.v1$")), ~ {
v1 <- get(str_replace(cur_column(), "v\\d+$", "v1"))
(.x - v1)/v1}, .names = "{.col}_change"))
-output
data
id wbc.v1 wbc.v2 wbc.v3 rbc.v1 rbc.v2 rbc.v3 hct.v1 hct.v2 hct.v3 wbc.v2_change wbc.v3_change rbc.v2_change rbc.v3_change hct.v2_change hct.v3_change
1 a1 23 63 30 23 56 90 13 89 47 1.7391304 0.3043478 1.434783 2.9130435 5.84615385 2.6153846
2 a2 81 45 46 NA 18 78 14 45 22 -0.4444444 -0.4320988 NA NA 2.21428571 0.5714286
3 a3 NA 27 14 29 67 46 37 34 33 NA NA 1.310345 0.5862069 -0.08108108 -0.1081081
If we want to keep the 'v1' columns as well
data %>%
na_if("N/A") %>%
type.convert(as.is = TRUE) %>%
mutate(across(ends_with('.v1'), ~ .x - .x,
.names = "{str_replace(.col, 'v1', 'v1change')}")) %>%
transmute(id, across(ends_with('change')),
across(-c(id, matches("\\.v1$"), ends_with('change')),
~ {
v1 <- get(str_replace(cur_column(), "v\\d+$", "v1"))
(.x - v1)/v1}, .names = "{.col}_change")) %>%
select(id, starts_with('wbc'), starts_with('rbc'), starts_with('hct'))
-output
id wbc.v1change wbc.v2_change wbc.v3_change rbc.v1change rbc.v2_change rbc.v3_change hct.v1change hct.v2_change hct.v3_change
1 a1 0 1.7391304 0.3043478 0 1.434783 2.9130435 0 5.84615385 2.6153846
2 a2 0 -0.4444444 -0.4320988 NA NA NA 0 2.21428571 0.5714286
3 a3 NA NA NA 0 1.310345 0.5862069 0 -0.08108108 -0.1081081
data
data <- structure(list(id = c("a1", "a2", "a3"), wbc.v1 = c(23L, 81L,
NA), wbc.v2 = c(63L, 45L, 27L), wbc.v3 = c(30L, 46L, 14L), rbc.v1 = c("23",
"N/A", "29"), rbc.v2 = c(56L, 18L, 67L), rbc.v3 = c(90L, 78L,
46L), hct.v1 = c(13L, 14L, 37L), hct.v2 = c(89L, 45L, 34L), hct.v3 = c(47L,
22L, 33L)), class = "data.frame", row.names = c(NA, -3L))

How to join and bind similar dataframes in R, iterative solutions welcome

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

Correlation of similar variables in R

I have slightly edited the data table.
I would like to correlate variable with similar name in my dataset:
A_y B_y C_y A_p B_p C_p
1 15 52 32 30 98 56
2 30 99 60 56 46 25
3 10 25 31 20 22 30
..........
n 55 23 85 12 34 52
I would like to obtain correlation of
A_y-A_p: 0.78
B_y-B_p: 0.88
C_y-C_p: 0.93
How can I do it in R? Is it possible?
This is really dangerous. Behavior of data.frames with invalid column names is undefined by the language definition. Duplicated column names are invalid.
You should restructure your input data. Anyway, here is an approach with your input data.
DF <- read.table(text = " A B C A B C
1 15 52 32 30 98 56
2 30 99 60 56 46 25
3 10 25 31 20 22 30", header = TRUE, check.names = FALSE)
sapply(unique(names(DF)), function(s) do.call(cor, unname(DF[, names(DF) == s])))
# A B C
#0.9995544 0.1585501 -0.6004010
#compare:
cor(c(15, 30, 10), c(30, 56, 20))
#[1] 0.9995544
Here is another base R option
within(
rev(
stack(
Map(
function(x) do.call(cor, unname(x)),
split.default(df, unique(gsub("_.*", "", names(df))))
)
)
),
ind <- sapply(
ind,
function(x) {
paste0(grep(paste0("^", x), names(df), value = TRUE),
collapse = "-"
)
}
)
)
which gives
ind values
1 A_y-A_p 0.9995544
2 B_y-B_p 0.1585501
3 C_y-C_p -0.6004010
Data
df <- structure(list(A_y = c(15L, 30L, 10L), B_y = c(52L, 99L, 25L),
C_y = c(32L, 60L, 31L), A_p = c(30L, 56L, 20L), B_p = c(98L,
46L, 22L), C_p = c(56L, 25L, 30L)), class = "data.frame", row.names = c("1",
"2", "3"))

Adding the Century to 2-Digit Year

I currently have a df that looks like
STA YR MO DA MAX date
58716 33013 43 3 11 60 0043-03-11
58717 33013 43 3 12 55 0043-03-12
58718 33013 43 3 13 63 0043-03-13
58719 33013 43 3 14 50 0043-03-14
58720 33013 43 3 15 58 0043-03-15
58721 33013 43 3 16 63 0043-03-16
I did df$date <- as.Date(with(df, paste(YR, MO, DA,sep="-")), "%Y-%m-%d")as you can see to get the date column, but clearly because there's no '19' in front of the year column, the year in the date comes out wacky. These are all 19xx dates. What would be a good way to fix this?
Try
df$date <- as.Date(with(df, paste(1900+YR, MO, DA,sep="-")), "%Y-%m-%d")
You should use %y since you have two digit year.
df$date <- as.Date(with(df, paste(YR, MO, DA,sep="-")), "%y-%m-%d")
However, this doesn't solve your problem since anything less than 69 is prefixed with 20 in 2 digit-years so 43 becomes 2043.
If you know that all your years are in the form of 19XX, you can do
df$date <- as.Date(with(df, sprintf('19%d-%d-%d', YR, MO, DA)))
If your years contain a mixture of 2-digit years from more than one century, then this code converts them all into valid dates in the past (no future dates).
dates_y2Y <- function(y,m,d) {
library(stringr)
y <- stringr::str_pad(y, width=2, pad="0")
m <- stringr::str_pad(m, width=2, pad="0")
d <- stringr::str_pad(d, width=2, pad="0")
toyear <- format(Sys.Date(), "%y")
tomnth <- format(Sys.Date(), "%m")
today <- format(Sys.Date(), "%d")
as.Date(
ifelse(y<toyear | y==toyear & m<tomnth | y==toyear & m==tomnth & d<=today,
as.Date(paste(y,m,d,sep="-"), format="%y-%m-%d"),
as.Date(paste(paste0("19",y),m,d,sep="-"), format="%Y-%m-%d"))
, origin="1970-01-01")
}
df$date <- dates_y2Y(df$YR, df$MO, df$DA)
df
STA YR MO DA date
1 33013 23 1 31 1923-01-31
2 33013 43 2 30 <NA>
3 33013 63 5 5 1963-05-05
4 33013 83 7 27 1983-07-27
5 33013 3 12 9 2003-12-09
6 33013 20 4 21 2020-04-21
7 33013 20 4 22 1920-04-22
Data:
df <- structure(list(STA = c(33013L, 33013L, 33013L, 33013L, 33013L,
33013L, 33013L), YR = c(23L, 43L, 63L, 83L, 3L, 20L, 20L), MO = c(1L,
2L, 5L, 7L, 12L, 4L, 4L), DA = c(31L, 30L, 5L, 27L, 9L, 21L,
22L), date = structure(c(-17137, NA, -2433, 4955, 12395, 18373,
-18151), class = "Date")), row.names = c(NA, -7L), class = "data.frame")
another solution
library(lubridate)
df %>%
mutate(date = make_date(year = 1900 + YR, month = MO, day = DA))
Another option with sprintf
df$date <- as.Date(do.call(sprintf, c(f = '19%d-%d-%d', df[2:4])))
Or with unite
library(dplyr)
library(tidyr)
library(stringr)
df %>%
mutate(YR = str_c('19', YR)) %>%
unite(date, YR, MO, DA, sep="-", remove = FALSE) %>%
mutate(date = as.Date(date))

Identifying Duplicate/Unique Teams (and Restructuring Data) in R

I have a data set that looks like this:
Person Team
1 30
2 30
3 30
4 30
11 40
22 40
1 50
2 50
3 50
4 50
15 60
16 60
17 60
1 70
2 70
3 70
4 70
11 80
22 80
My overall goal is to organize that team identification codes so that it is easy to see which teams are duplicates of one another and which teams are unique. I want to summarize the data so that it looks like this:
Team Duplicate1 Duplicate2
30 50 70
40 80
60
As you can see, teams 30, 50, and 70 have identical members, so they share a row. Similarly, teams 40 and 80 have identical members, so they share a row. Only team 60 (in this example) is unique.
In situations where teams are duplicated, I don't care which team id goes in which column. Also, there may be more than 2 duplicates of a team. Teams range in size from 2 members to 8 members.
This answer gives the output data format you asked for. I left the duplicate teams in a single variable because I think it's a better way to handle an arbitrary number of duplicates.
require(dplyr)
df %>%
arrange(Team, Person) %>% # this line is necessary in case the rest of your data isn't sorted
group_by(Team) %>%
summarize(players = paste0(Person, collapse = ",")) %>%
group_by(players) %>%
summarize(teams = paste0(Team, collapse = ",")) %>%
mutate(
original_team = ifelse(grepl(",", teams), substr(teams, 1, gregexpr(",", teams)[[1]][1]-1), teams),
dup_teams = ifelse(grepl(",", teams), substr(teams, gregexpr(",", teams)[[1]][1]+1, nchar(teams)), NA)
)
The result:
Source: local data frame [3 x 4]
players teams original_team dup_teams
1 1,2,3,4 30,50,70 30 50,70
2 11,22 40,80 40 80
3 15,16,17 60 60 NA
Not exactly the format you're wanting, but pretty useful:
# using MrFlick's data
library(dplyr)
dd %>% group_by(Team) %>%
arrange(Person) %>%
summarize(team.char = paste(Person, collapse = "_")) %>%
group_by(team.char) %>%
arrange(team.char, Team) %>%
mutate(duplicate = 1:n())
Source: local data frame [6 x 3]
Groups: team.char
Team team.char duplicate
1 40 11_22 1
2 80 11_22 2
3 60 15_16_17 1
4 30 1_2_3_4 1
5 50 1_2_3_4 2
6 70 1_2_3_4 3
(Edited in the arrange(Person) line in case the data isn't already sorted, got the idea from #Reed's answer.)
Using this for your sample data
dd<-structure(list(Person = c(1L, 2L, 3L, 4L, 11L, 22L, 1L, 2L, 3L,
4L, 15L, 16L, 17L, 1L, 2L, 3L, 4L, 11L, 22L), Team = c(30L, 30L,
30L, 30L, 40L, 40L, 50L, 50L, 50L, 50L, 60L, 60L, 60L, 70L, 70L,
70L, 70L, 80L, 80L)), .Names = c("Person", "Team"),
class = "data.frame", row.names = c(NA, -19L))
You could try a table()/interaction() to find duplicate groups. For example
tt <- with(dd, table(Team, Person))
grp <- do.call("interaction", c(data.frame(unclass(tt)), drop=TRUE))
split(rownames(tt), grp)
this returns
$`1.1.1.1.0.0.0.0.0`
[1] "30" "50" "70"
$`0.0.0.0.0.1.1.1.0`
[1] "60"
$`0.0.0.0.1.0.0.0.1`
[1] "40" "80"
so the group "names" are really just indicators for membership for each person. You could easily rename them if you like with setNames(). But here it collapse the appropriate teams.
Two more base R options (though not exactly the desired output):
DF2 <- aggregate(Person ~ Team, DF, toString)
> split(DF2$Team, DF2$Person)
$`1, 2, 3, 4`
[1] 30 50 70
$`11, 22`
[1] 40 80
$`15, 16, 17`
[1] 60
Or
( DF2$DupeGroup <- as.integer(factor(DF2$Person)) )
Team Person DupeGroup
1 30 1, 2, 3, 4 1
2 40 11, 22 2
3 50 1, 2, 3, 4 1
4 60 15, 16, 17 3
5 70 1, 2, 3, 4 1
6 80 11, 22 2
Note that the expected output as shown in the question would either require to add NAs or empty strings in some of the columns entries because in a data.frame, all columns must have the same number of rows. That is different for lists in, as you can see in some of the answers.
The second option, but using data.table, since aggregate tends to be slow for large data:
library(data.table)
setDT(DF)[, toString(Person), by=Team][,DupeGroup := .GRP, by=V1][]
Team V1 DupeGroup
1: 30 1, 2, 3, 4 1
2: 40 11, 22 2
3: 50 1, 2, 3, 4 1
4: 60 15, 16, 17 3
5: 70 1, 2, 3, 4 1
6: 80 11, 22 2
Using uniquecombs from the mgcv package:
library(mgcv)
library(magrittr) # for the pipe %>%
# Using MrFlick's data
team_names <- sort(unique(dd$Team))
unique_teams <- with(dd, table(Team, Person)) %>% uniquecombs %>% attr("index")
printout <- unstack(data.frame(team_names, unique_teams))
> printout
$`1`
[1] 60
$`2`
[1] 40 80
$`3`
[1] 30 50 70
Now you could use something like this answer to print it in tabular form (note that the groups are column-wise, not row-wise as in your question):
attributes(printout) <- list(names = names(printout)
, row.names = 1:max(sapply(printout, length))
, class = "data.frame")
> printout
1 2 3
1 60 40 30
2 <NA> 80 50
3 <NA> <NA> 70
Warning message:
In format.data.frame(x, digits = digits, na.encode = FALSE) :
corrupt data frame: columns will be truncated or padded with NAs

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