# Data Management Libraries

In recent years, RStudio has spearheaded development of a series of libraries that make data refactoring, selecting, and management simple and fast for large data sets. Many of these tools are equiva lent to what you can do using selection, sorting, aggregate, and tapply of normal data frames. Some of them offer very useful capabilities that are otherwise very difficult to manage. Most of these are developed by Hadley Wickham, who also created ggplot2. Part of the reason for the proliferation of libraries is the philosophy to not break what people rely on, and so when improved functionality is made, a new library is created so that compatibility can be broken without harming anyone relying on certain functionality.

Some relevant libraries include:

## plyr and dplyr

These libraries are sets of tools for splitting, applying, and combining data. The goal is to have a coherent set of tools for breaking down data into smaller pieces, operating on each chunk of data and reassembling them–an idiom called “split-apply-combine”.

dplyr is a successor to plyr, written to be much faster, to integrae with remote databases, but it works only with data frames. The dplyr library seems to be better supported, and tests show it can be more than a hundred times faster than plyr.

## reshape, reshape2 and tidyr

The reshape2 library is a ‘reboot’ of reshape, that is faster and better. These libraries allow easily transforming a data set from ‘long’ to ‘wide’ format and back again. That is, you can take a data set with multiple columns you are treating as distinct DVs, and reframe the data set so they are both in a single DV column, with a separate column specifiying which level of IV a row belongs to. The tidyr library is the newest entry into data management libraries, also by Wickham, and is described as an “evolution” of reshape2.

# Overview of dplyr

The following creates a couple data sets for use in these examples:

dat0 <- data.frame(sub= c(1,1,1,2,2,2,3,3,3,4,4,4),
question = c("a","b","c","a","b","c","a","b","c","a","b","c"),
dv = c(5,3,1,2,3,6,4,2,3,1,3,5))

dat <- data.frame(sub = sample(letters,100,replace=T),
cond = sample(c("A","B","C"),100,replace=T),
group = sample(1:10,100,replace=T),
dv1 = runif(100)*5)

The dplyr library implements a number of functions that are available in one form or another within R, but may be difficult to use, inconsistent, or slow.

The dplyr library does not create side-effects. That is, it always makes a copy of your original data and returns it, rather than altering the form of your original data. Consequently, you need to usually assign the outcome to a new variable. Sometimes, it is acceptable to assign it to its old name, as in the following:

library(dplyr)
data <- dat
filter(data,sub=="b")
  sub cond group       dv1
1   b    C     7 0.9104921
2   b    C    10 4.8048437
3   b    B     9 3.5090899
4   b    C     4 1.4837364
data <-filter(data,sub=="b")
head(data)
  sub cond group       dv1
1   b    C     7 0.9104921
2   b    C    10 4.8048437
3   b    B     9 3.5090899
4   b    C     4 1.4837364

However, this is often not the best practice, because it means that the data variable depends on whether you have run some code or not.

# slice and filter

The following use dplyr to rearrange and filter rows of a data frame. filter picks out rows based on a boolean vector of the same size (number of rows)

head((dat$sub=="b")) ##shows the first 6 elements of the boolean [1] FALSE FALSE FALSE FALSE FALSE FALSE filter(dat,sub=="b") ##use filter to pick out just tho subject B rows  sub cond group dv1 1 b C 7 0.9104921 2 b C 10 4.8048437 3 b B 9 3.5090899 4 b C 4 1.4837364 Similarly, slice allows you to do this based on the row index (number) slice(dat,1) ##first row  sub cond group dv1 1 v C 1 1.390465 slice(dat,2:10) ##9 rows after the first  sub cond group dv1 1 g C 1 4.6790032 2 q C 6 1.1066056 3 h C 7 2.6110340 4 v C 10 1.8685713 5 o C 10 3.7582408 6 c B 6 0.7141249 7 l B 5 4.8820045 8 h C 3 1.7563706 9 q B 4 1.4864425 slice(dat,1:20*2) ##even rows 2..40  sub cond group dv1 1 g C 1 4.6790032 2 h C 7 2.6110340 3 o C 10 3.7582408 4 l B 5 4.8820045 5 q B 4 1.4864425 6 g A 6 0.2497757 7 x B 6 0.6285890 8 w B 1 2.6781465 9 b C 7 0.9104921 10 v C 2 4.3696741 11 a C 5 3.7754780 12 j B 8 0.5748146 13 d C 1 4.3373926 14 q C 1 0.7422173 15 a B 3 3.4336640 16 i C 3 4.0976893 17 t C 2 0.7880525 18 e C 5 1.6160224 [ reached getOption("max.print") -- omitted 2 rows ] slice(dat,-1)  sub cond group dv1 1 g C 1 4.67900324 2 q C 6 1.10660561 3 h C 7 2.61103398 4 v C 10 1.86857133 5 o C 10 3.75824076 6 c B 6 0.71412492 7 l B 5 4.88200452 8 h C 3 1.75637062 9 q B 4 1.48644253 10 x A 7 0.96795371 11 g A 6 0.24977571 12 f A 4 3.21570340 13 x B 6 0.62858898 14 e C 3 2.53825601 15 w B 1 2.67814646 16 q A 1 1.91787686 17 b C 7 0.91049211 18 w C 3 3.41260566 [ reached getOption("max.print") -- omitted 81 rows ] # arrange() The arrange function reorders the rows by the levels of a specific factor arrange(dat,sub)  sub cond group dv1 1 a C 5 3.77547801 2 a B 3 3.43366398 3 a B 2 1.35059336 4 a C 10 0.79434535 5 b C 7 0.91049211 6 b C 10 4.80484367 7 b B 9 3.50908989 8 b C 4 1.48373643 9 c B 6 0.71412492 10 c C 5 1.42298564 11 c B 6 4.12378558 12 c B 10 0.25254923 13 d C 1 4.33739260 14 d B 6 4.28931767 15 d A 7 3.72928259 16 d C 8 3.95967892 17 d C 8 2.56633554 18 d B 2 1.15588444 [ reached getOption("max.print") -- omitted 82 rows ] arrange(dat,sub,group)  sub cond group dv1 1 a B 2 1.35059336 2 a B 3 3.43366398 3 a C 5 3.77547801 4 a C 10 0.79434535 5 b C 4 1.48373643 6 b C 7 0.91049211 7 b B 9 3.50908989 8 b C 10 4.80484367 9 c C 5 1.42298564 10 c B 6 0.71412492 11 c B 6 4.12378558 12 c B 10 0.25254923 13 d C 1 4.33739260 14 d B 2 1.15588444 15 d B 6 4.28931767 16 d A 7 3.72928259 17 d C 8 3.95967892 18 d C 8 2.56633554 [ reached getOption("max.print") -- omitted 82 rows ] # select() • The select function picks out columns by name select(dat0,sub,dv)  sub dv 1 1 5 2 1 3 3 1 1 4 2 2 5 2 3 6 2 6 7 3 4 8 3 2 9 3 3 10 4 1 11 4 3 12 4 5 select(dat0,sub:dv)  sub question dv 1 1 a 5 2 1 b 3 3 1 c 1 4 2 a 2 5 2 b 3 6 2 c 6 7 3 a 4 8 3 b 2 9 3 c 3 10 4 a 1 11 4 b 3 12 4 c 5 select(dat0,-question)  sub dv 1 1 5 2 1 3 3 1 1 4 2 2 5 2 3 6 2 6 7 3 4 8 3 2 9 3 3 10 4 1 11 4 3 12 4 5 There are a lot of matching functions: select(dat0,starts_with("s"))  sub 1 1 2 1 3 1 4 2 5 2 6 2 7 3 8 3 9 3 10 4 11 4 12 4 This function can be very handy for situations like survey data where you have dozens or hundreds of columns/variables. You may be interested in just a few of these, and select will pick these out. # rename() • The rename function renames columns. rename(dat0,participant=sub)  participant question dv 1 1 a 5 2 1 b 3 3 1 c 1 4 2 a 2 5 2 b 3 6 2 c 6 7 3 a 4 8 3 b 2 9 3 c 3 10 4 a 1 11 4 b 3 12 4 c 5 # distinct() • The distinct function finds distinct combinations of values (typically IVs). This is similar to doing a table, or identifying the levels of a factor. dat2 <- data.frame(a=sample(1:10,20,replace=T), b=sample(c(100,200,300),20,replace=T)) distinct(dat2)  a b 1 1 100 2 3 100 3 5 200 4 1 200 5 8 100 6 8 300 7 6 100 8 4 300 9 7 300 10 6 200 11 9 200 12 2 300 13 9 100 You can also specify specific variables you wish to use: distinct(dat,sub)  sub 1 v 2 g 3 q 4 h 5 o 6 c 7 l 8 x 9 f 10 e 11 w 12 b 13 a 14 m 15 j 16 n 17 d 18 i 19 s 20 t 21 k 22 r 23 z 24 p 25 u Retain all columns of distinct data: distinct(dat,sub,.keep_all=T)  sub cond group dv1 1 v C 1 1.3904651 2 g C 1 4.6790032 3 q C 6 1.1066056 4 h C 7 2.6110340 5 o C 10 3.7582408 6 c B 6 0.7141249 7 l B 5 4.8820045 8 x A 7 0.9679537 9 f A 4 3.2157034 10 e C 3 2.5382560 11 w B 1 2.6781465 12 b C 7 0.9104921 13 a C 5 3.7754780 14 m C 2 2.1973824 15 j B 8 0.5748146 16 n A 10 2.0001063 17 d C 1 4.3373926 18 i C 3 4.0976893 [ reached getOption("max.print") -- omitted 7 rows ] # mutate() and transmute() • The mutate function adds a column that is a function of other columns. Transmute does the same thing, but returns only the new variable. This can be really useful for creating summarized data, composite values of ratings scales, and the like. ##reverse code a scale dat1 <- mutate(dat0,newdv=6-dv) More complex mutations are possible: mutate(dat1,newdv2 = dv*newdv)  sub question dv newdv newdv2 1 1 a 5 1 5 2 1 b 3 3 9 3 1 c 1 5 5 4 2 a 2 4 8 5 2 b 3 3 9 6 2 c 6 0 0 7 3 a 4 2 8 8 3 b 2 4 8 9 3 c 3 3 9 10 4 a 1 5 5 11 4 b 3 3 9 12 4 c 5 1 5 # merging and joining dplyr has a lot of functions to merge data frames, and these are especially useful when you may not have an exact match between the levels (so you cant just do a cbind) A <- data.frame(sub=c("A","B","C","E"),data1=1:4) B <- data.frame(sub=c("A","B","D","F"),data2=11:14) • left_join(A,B) Joins everything into A that is in B left_join(A,B, by="sub")  sub data1 data2 1 A 1 11 2 B 2 12 3 C 3 NA 4 E 4 NA • right_join(A,B) right_join(A,B, by="sub")  sub data1 data2 1 A 1 11 2 B 2 12 3 D NA 13 4 F NA 14 *inner_join(A,B) inner_join(A,B, by="sub")  sub data1 data2 1 A 1 11 2 B 2 12 *full_join(A,B) adds all data, incorporating NAs when one or the other are missing. full_join(A,B, by="sub")  sub data1 data2 1 A 1 11 2 B 2 12 3 C 3 NA 4 E 4 NA 5 D NA 13 6 F NA 14 *semi_join picks out just the first argument for variables where both exist; anti_join picks out the first argument for those where the second doesn’t exist. These can be useful for imputing data and the like–you can choose the values for which the other value is missing. semi_join(A,B, by="sub")  sub data1 1 A 1 2 B 2 anti_join(A,B,by="sub")  sub data1 1 C 3 2 E 4 #Combining data frames row-wise The bind_rows acts like rbind, stacking two data frames on top o fone another. ##This doesn't make any sense, but it works: bind_rows(left_join(A,B,by="sub"), right_join(A,B,by="sub"))  sub data1 data2 1 A 1 11 2 B 2 12 3 C 3 NA 4 E 4 NA 5 A 1 11 6 B 2 12 7 D NA 13 8 F NA 14 # Advanced exercises suppose every other item was reverse coded dat0$coding <- rep(c(-1,1),6)

Recode using mutate and filter:

d1<-mutate(filter(dat0,coding==1),newdv=dv)
d2<-mutate(filter(dat0,coding==-1),newdv=6-dv)
dat0b <- bind_rows(d1,d2)
arrange(dat0b,sub,question)
   sub question dv coding newdv
1    1        a  5     -1     1
2    1        b  3      1     3
3    1        c  1     -1     5
4    2        a  2      1     2
5    2        b  3     -1     3
6    2        c  6      1     6
7    3        a  4     -1     2
8    3        b  2      1     2
9    3        c  3     -1     3
10   4        a  1      1     1
11   4        b  3     -1     3
12   4        c  5      1     5

# Big five coding

Load the data set using the big five personality questionnaire.

• The Q1..Q44 are the personality questions. Some are reverse coded, so that the proper coding is 6-X instead of X.
• The questions alternate between 5 factors, but at the end they are a bit off.
• Some of them are reverse coded.
big5 <- read.csv("bigfive.csv")
qtype <- c("E","A","C","N","O", "E","A","C","N","O",
"E","A","C","N","O", "E","A","C","N","O",
"E","A","C","N","O", "E","A","C","N","O",
"E","A","C","N","O", "E","A","C","N","O",
"O","A","C","O")
valence <- c(1,-1,1,1,1,  -1,1,-1,-1,1,
1,-1,1,1,1,  1,1,-1,1,1,
-1,1,-1,-1,1,  1,-1,1,1,1,
-1,1,1,-1,-1,  1,-1,1,1,1,
-1,1,-1,1)

## Exercise:

Use the above data and dplyr to recode the responses by valence, and then select out each of five personality variables as sums of the proper dimension.

# The reshape2 library

The following gives instructions for using the (older) reshape2 library. The tidyr library is its successor, and can also be used (diffenet function names, different arguments) for doing much of the same thing, but instructions for using that will not be covered here.

Load the library and a survey for examples:

library(reshape2)
head(dat1)
  subcode question                timestamp  type  time  answer
1     207        1 Fri Oct 24 14:27:59 2014  inst 88803
2     207        2 Fri Oct 24 14:28:04 2014 short  5172      20
3     207        3 Fri Oct 24 14:28:11 2014 short  6582 english
4     207        4 Fri Oct 24 14:28:29 2014 short 18461      na
5     207        5 Fri Oct 24 14:28:49 2014 multi 19452       1
6     201        1 Mon Oct 20 17:55:59 2014  inst 29450        

Notice that here, we have five questions of different types in a survey, across a bunch of respondents. This is ‘long’ format (what Wickham calls ‘tidy’). What if we want “wide”? We can use dcast to reorganize into a data frame (d= data frame):

dat2 <-dcast(dat1,subcode~question,value.var="answer")
dat2
   subcode 1  2       3    4 5
1      101   20 english   na 1
2      102   19 english <NA> 1
3      103   20 English <NA> 1
4      104   18 English <NA> 1
5      201   19 english <NA> 1
6      202   19 english   na 1
7      203   19 english   na 1
8      204   20 English <NA> 1
9      206   19 english <NA> 1
10     207   20 english   na 1
11     209   16 english   na 3
12     210   22 english <NA> 1
[ reached getOption("max.print") -- omitted 12 rows ]

This is good, but the variable names are a bit inconvenient.

colnames(dat2) <- c("subcode","q1","q2","q3","q4","q5")

or, use acast for a vector/matrix. This is not appropriate in this case:

dat3 <-acast(dat1,subcode~question,value.var="answer")
dat3[1:5,]
    1 2  3       4    5
101   20 english na   1
102   19 english <NA> 1
103   20 English <NA> 1
104   18 English <NA> 1
201   19 english <NA> 1
Levels:  1 16 18 19 20 22 3 english English na

What if we want a table of timestamps for each question–maybe to look at how long each one took? Specify this as value.var.

dat4 <-dcast(dat1,subcode~question,value.var="timestamp")
dat4
   subcode                        1                        2
1      101 Fri Oct 24 11:28:24 2014 Fri Oct 24 11:28:33 2014
2      102 Fri Oct 24 13:03:34 2014 Fri Oct 24 13:03:41 2014
3      103 Fri Nov 07 09:53:40 2014 Fri Nov 07 09:54:06 2014
4      104 Fri Nov 07 12:59:11 2014 Fri Nov 07 12:59:23 2014
5      201 Mon Oct 20 17:55:59 2014 Mon Oct 20 17:56:05 2014
6      202 Thu Oct 23 15:58:06 2014 Thu Oct 23 15:58:13 2014
7      203 Fri Oct 24 09:57:43 2014 Fri Oct 24 09:57:51 2014
8      204 Fri Oct 24 11:36:44 2014 Fri Oct 24 11:37:07 2014
9      206 Fri Oct 24 13:04:24 2014 Fri Oct 24 13:04:28 2014
10     207 Fri Oct 24 14:27:59 2014 Fri Oct 24 14:28:04 2014
11     209 Fri Nov 07 09:55:46 2014 Fri Nov 07 09:55:49 2014
12     210 Fri Nov 07 11:31:02 2014 Fri Nov 07 11:31:13 2014
3                        4
1  Fri Oct 24 11:28:40 2014 Fri Oct 24 11:28:54 2014
2  Fri Oct 24 13:03:45 2014 Fri Oct 24 13:03:54 2014
3  Fri Nov 07 09:54:18 2014 Fri Nov 07 09:54:26 2014
4  Fri Nov 07 12:59:31 2014 Fri Nov 07 12:59:37 2014
5  Mon Oct 20 17:56:12 2014 Mon Oct 20 17:56:19 2014
6  Thu Oct 23 15:58:19 2014 Thu Oct 23 15:58:26 2014
7  Fri Oct 24 09:58:02 2014 Fri Oct 24 09:58:13 2014
8  Fri Oct 24 11:37:11 2014 Fri Oct 24 11:37:17 2014
9  Fri Oct 24 13:04:31 2014 Fri Oct 24 13:04:37 2014
10 Fri Oct 24 14:28:11 2014 Fri Oct 24 14:28:29 2014
11 Fri Nov 07 09:56:03 2014 Fri Nov 07 09:56:11 2014
12 Fri Nov 07 11:31:16 2014 Fri Nov 07 11:31:22 2014
5
1  Fri Oct 24 11:28:57 2014
2  Fri Oct 24 13:03:57 2014
3  Fri Nov 07 09:54:30 2014
4  Fri Nov 07 12:59:41 2014
5  Mon Oct 20 17:56:22 2014
6  Thu Oct 23 15:58:32 2014
7  Fri Oct 24 09:58:18 2014
8  Fri Oct 24 11:37:22 2014
9  Fri Oct 24 13:04:40 2014
10 Fri Oct 24 14:28:49 2014
11 Fri Nov 07 09:56:17 2014
12 Fri Nov 07 11:31:27 2014
[ reached getOption("max.print") -- omitted 12 rows ]

Now, do the same for time:

dat4 <-dcast(dat1,subcode~question,value.var="time")
dat4
   subcode     1     2     3     4     5
1      101 32764  9226  6762 13743  3104
2      102 20689  7266  4396  8204  2891
3      103 38236 25939 12205  7573  4403
4      104 45862 12164  7875  5612  4136
5      201 29450  5183  7235  6557  3187
6      202 74307  6757  6266  7033  5502
7      203 34879  7859 11528 10525  5120
8      204 37176 22599  4510  5656  5098
9      206 31742  3629  3055  5933  3415
10     207 88803  5172  6582 18461 19452
11     209 30038  3523 13551  7792  6457
12     210 42821 10280  3643  5601  4601
[ reached getOption("max.print") -- omitted 12 rows ]

# Using melt to re-form wide data frames

The *cast function take long (tidy) format and make data frames based on a category label. We can do the opposite too, a process referred to as ‘melting’ (in tidyr, you can use ‘gather’). Before, question was used as the label.

this doesn’t work right. It uses q1..q5 as id varaibles, because they are non-numeric.

melt(dat2)
   q1 q2      q3   q4 q5 variable value
1     20 english   na  1  subcode   101
2     19 english <NA>  1  subcode   102
3     20 English <NA>  1  subcode   103
4     18 English <NA>  1  subcode   104
5     19 english <NA>  1  subcode   201
6     19 english   na  1  subcode   202
7     19 english   na  1  subcode   203
8     20 English <NA>  1  subcode   204
9     19 english <NA>  1  subcode   206
10    20 english   na  1  subcode   207
[ reached getOption("max.print") -- omitted 14 rows ]

Instead, we can specify id.vars, which gets us closer

melt(dat2,id.vars = c("subcode"))
    subcode variable   value
1       101       q1
2       102       q1
3       103       q1
4       104       q1
5       201       q1
6       202       q1
7       203       q1
8       204       q1
9       206       q1
10      207       q1
11      209       q1
12      210       q1
13      211       q1
14      212       q1
15      301       q1
16      302       q1
17      303       q1
18      304       q1
19      305       q1
20      306       q1
21      307       q1
22      308       q1
23      309       q1
24      310       q1
25      101       q2      20
[ reached getOption("max.print") -- omitted 95 rows ]

It is a bit puzzling why this works. It uses only subcode as the id variable. Any variable we wanting tagging each row we can move out of the variable set and into the id set, for example, language:

melt(dat2,id.vars = c("subcode","q3"))
   subcode      q3 variable value
1      101 english       q1
2      102 english       q1
3      103 English       q1
4      104 English       q1
5      201 english       q1
6      202 english       q1
7      203 english       q1
8      204 English       q1
9      206 english       q1
10     207 english       q1
11     209 english       q1
12     210 english       q1
13     211 English       q1
14     212 English       q1
15     301 english       q1
16     302 English       q1
17     303 English       q1
18     304 english       q1
[ reached getOption("max.print") -- omitted 78 rows ]

id.vars specify the variables you want to keep and not split on. These appear several times in the new data . Notice that value.name names the value that the matrix is being unfolded to.

we can name the response like this:

melt(dat2,id.vars = c("subcode","q3"),value.name="response",variable.name="Question")
   subcode      q3 Question response
1      101 english       q1
2      102 english       q1
3      103 English       q1
4      104 English       q1
5      201 english       q1
6      202 english       q1
7      203 english       q1
8      204 English       q1
9      206 english       q1
10     207 english       q1
11     209 english       q1
12     210 english       q1
13     211 English       q1
14     212 English       q1
15     301 english       q1
16     302 English       q1
17     303 English       q1
18     304 english       q1
[ reached getOption("max.print") -- omitted 78 rows ]

Notice that q1 was empty, so we can specify just the measure variables we care about:

melt(dat2,id.vars = c("subcode","q3"),
measure.vars=c("q2","q4","q5"),
value.name="response",variable.name="Question")
   subcode      q3 Question response
1      101 english       q2       20
2      102 english       q2       19
3      103 English       q2       20
4      104 English       q2       18
5      201 english       q2       19
6      202 english       q2       19
7      203 english       q2       19
8      204 English       q2       20
9      206 english       q2       19
10     207 english       q2       20
11     209 english       q2       16
12     210 english       q2       22
13     211 English       q2       20
14     212 English       q2       20
15     301 english       q2       20
16     302 English       q2       20
17     303 English       q2       19
18     304 english       q2       19
[ reached getOption("max.print") -- omitted 54 rows ]

# Exercises

• Using the big5 data set, add a unique subject code to each row. Then, use melt’’ to create a data frame that has the following columns: subject code, gender, question and answer.
big5 <- read.csv("bigfive.csv")
qtype <- c("E","A","C","N","O",   "E","A","C","N","O",
"E","A","C","N","O",   "E","A","C","N","O",
"E","A","C","N","O",   "E","A","C","N","O",
"E","A","C","N","O",   "E","A","C","N","O",
"O","A","C","O")
valence <- c(1,-1,1,1,1,  -1,1,-1,-1,1,
1,-1,1,1,1,  1,1,-1,1,1,
-1,1,-1,-1,1,  1,-1,1,1,1,
-1,1,1,-1,-1,  1,-1,1,1,1,
-1,1,-1,1)
varnames <- colnames(big5)[2:45]

##first, recode the negative codings.

##mutate the columns with -1 valence:

melted <- melt(mutate(recoded,sub=1:nrow(recoded)),
id.vars = c("sub")
)

arrange(melted,sub,variable)
     sub variable value
1      1       Q1     3
2      1       Q2     2
3      1       Q3     4
4      1       Q4     2
5      1       Q5     3
6      1       Q6     2
7      1       Q7     5
8      1       Q8     2
9      1       Q9     1
10     1      Q10     5
11     1      Q11     3
12     1      Q12     4
13     1      Q13     2
14     1      Q14     4
15     1      Q15     4
16     1      Q16     2
17     1      Q17     5
18     1      Q18     2
19     1      Q19     1
20     1      Q20     4
21     1      Q21     4
22     1      Q22     5
23     1      Q23     3
24     1      Q24     1
25     1      Q25     4
[ reached getOption("max.print") -- omitted 5563 rows ]

# Solution to Exercise 1.

big5 <- read.csv("bigfive.csv")
qtype <- c("E","A","C","N","O",   "E","A","C","N","O",
"E","A","C","N","O",   "E","A","C","N","O",
"E","A","C","N","O",   "E","A","C","N","O",
"E","A","C","N","O",   "E","A","C","N","O",
"O","A","C","O")
valence <- c(1,-1,1,1,1,  -1,1,-1,-1,1,
1,-1,1,1,1,  1,1,-1,1,1,
-1,1,-1,-1,1,  1,-1,1,1,1,
-1,1,1,-1,-1,  1,-1,1,1,1,
-1,1,-1,1)
varnames <- colnames(big5)[2:45]

##first, recode the negative codings.

##mutate the columns with -1 valence:

##check this. For negative valence, 2 becomes 4 etc.
bind_rows(recoded[1,],answers[1,])
  Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q15 Q16 Q17 Q18 Q19 Q20
1  3  2  4  2  3  2  5  2  1   5   3   4   2   4   4   2   5   2   1   4
Q21 Q22 Q23 Q24 Q25 Q26 Q27 Q28 Q29 Q30 Q31 Q32 Q33 Q34 Q35 Q36 Q37 Q38
1   4   5   3   1   4   4   2   3   4   2   1   3   5   2   1   3   3   2
Q39 Q40 Q41 Q42 Q43 Q44
1   3   1   2   2   2   4
[ reached getOption("max.print") -- omitted 1 row ]
##create composite subsets
b5.e <- select(recoded, one_of(varnames[qtype=="E"]))
b5.a <- select(recoded, one_of(varnames[qtype=="A"]))
b5.c <- select(recoded, one_of(varnames[qtype=="C"]))
b5.n <- select(recoded, one_of(varnames[qtype=="N"]))
b5.o <- select(recoded, one_of(varnames[qtype=="O"]))

composites1 <- data.frame(e=rowMeans(b5.e,na.rm=T),
a=rowMeans(b5.a,na.rm=T),
c=rowMeans(b5.c,na.rm=T),
n=rowMeans(b5.n,na.rm=T),
o=rowMeans(b5.o,na.rm=T)
)

2019-01-16