607 Project 2 a

Date cleaned

It looks like all of the dates follow the European format (day/month/year), so that makes things easier than if they were mixed formats, especially with lubridate, my favorite function ever.

df <- mydata

df$date_cleaned <- dmy(df$date_cleaned)  # dmy is a lubridate function

kable(df[,1])

Knowing what day of the week each of these days corresponds to will also be useful

df$dow <- weekdays(as.Date(df$date_cleaned)) # add column that reflects DoW

df$dow <- as.factor(df$dow)

df <- df[,c(1,7,2:6)] # Move DoW column next to Date

colnames(df)

## [1] "date_cleaned"    "dow"             "name"            "sides_and_sink" 
## [5] "dishwasher_full" "no_of_mugs"      "notes"

Sides and Sink

Creating columns for sides and sink that check whether that column has the words “side”, “sink”, or “both”. These are logical vectors and depending on what was going to be done with the data I might transform these to binary (also might do the same with dishwasher full).

df$sides <- str_detect(df$sides_and_sink,("side|both"))

df$sink <- str_detect(df$sides_and_sink,"sink|both")

kable(df[,8:9])

Number of Mugs

The easiest thing to do here is to assume that either the largest or smallest number written is accurate enough. A little harder would be to take the average of the high and low (if present). Why not have some fun?

df$mugslow <- as.numeric(str_extract(df$no_of_mugs,"[:digit:]+"))     # extract the first set of uninterupted digits
df$mugshigh <- as.numeric(str_extract(df$no_of_mugs,"(?<=\\D)[:digit:]+"))   # extract uninterupted digits following non-digits

n <- which(is.na(df$mugshigh))  # replace NA in mugshigh with the value in mugslow
df[n, "mugshigh"] <- df[n, "mugslow"]

df$no_of_mugs <- rowMeans(df[c('mugshigh', 'mugslow')])     # calculate the average of the mugs

mugcount <- c(df$mugslow-df$mugshigh) # creating a vector that will be used later to determine if the column 
                                      # for number of columns is being filled out correctly

kable(df[,6:11])

Final cleaning

Reduce and reorder columns

df <- df %>% select("date_cleaned", "dow", "name", "no_of_mugs", "dishwasher_full", "sides", "sink", "notes")

kable(df)

Analysis

What day are there the most mugs?

df_day <- df %>% 
  group_by(dow) %>% 
  summarize(mugs = sum(wday(no_of_mugs))) # group by day of week

df_day$dow <- factor(df_day$dow, levels= c("Monday", "Tuesday", "Wednesday", "Thursday", "Friday")) #set day of week factors in order

df_day <- df_day[order(df_day$dow), ] # Order days chronologically

df_day

## # A tibble: 5 x 2
##   dow        mugs
##   <fct>     <dbl>
## 1 Monday       16
## 2 Tuesday      13
## 3 Wednesday    18
## 4 Thursday     24
## 5 Friday       13

Thursdays are clearly the biggest days for mug usage with Wednesdays close behind.

Lets graph it

p<-ggplot(data=df_day, aes(x=dow, y=mugs), main="Dirty Mugs by Day of Week") +
  geom_bar(stat="identity", fill = "darksalmon") + geom_text(aes(label=mugs), vjust=1.6, size=5)
p

I guess on Monday folks are rested from the weekend, then they need some help getting through the rest of the week until Friday. As these data were collected in the summer, maybe the company had summer Fridays.

Are men or women more likely to wash either the sides or sink?

Gender wasn’t collected with this information but can be determined. There are resources available to determine likely gender if it were needed for a larger data set.

dfga <- df  # create new dataframe

dfga$gender <- as.factor(c('F', 'M', 'F', 'M', 'M', 'F', 'M', 'M', 'M' ))  # create a vector with genders

gcount <- data.frame("gender" = c("F", "M"), "count" =c(sum(dfga$gender == 'F'),sum(dfga$gender == 'M'))) #create a df of gender and counts of gender

dfga$anywash <- (dfga$sides + dfga$sink) # create a vector that sums the sink and sides columns

dfga$anytrue <- dfga$anywash>0  # create a logical vector where TRUE is any cell that is over 0

dfga$anytrue <- as.integer(as.logical(dfga$anytrue)) # coerce the anywtrue vector to integer

dfgb <- dfga[,c(9:11)]  # subset to just have the gender count and anywtrue columns

dfgb <- na.omit(dfgb) # remove NA from analysis

dfgwash <- dfgb %>% 
  group_by(gender) %>% 
  summarize("washed" = sum(anytrue)) # group by gender for anywtrue

dfgwash <- data.frame(dfgwash) # make df of prior group

dfw <- join(gcount, dfgwash) # join the washed and gender count dataframes

## Joining by: gender

dfw$pct <- dfw$washed / dfw$count # create vector for % of each gender to have washed

dfw

##   gender count washed       pct
## 1      F     3      2 0.6666667
## 2      M     6      4 0.6666667

Are men or women more likely to fill out the form completely?

dffa <- df  # create new dataframe

dffa$gender <- as.factor(c('F', 'M', 'F', 'M', 'M', 'F', 'M', 'M', 'M' ))  # create a vector with genders

# gcount <- data.frame("gender" = c("F", "M"), "count" =c(sum(dfga$gender == 'F'),sum(dfga$gender == 'M'))) #create a df of gender and counts of gender

dffa$mugcount <- mugcount # create a vector that reflects whether there was a range put into the mug count instead of a number

dffa$mugcount <- dffa$mugcount!=0  # create a logical vector where TRUE is any cell does not equal 0 (where there was a diff b/n high and low count)

dffa$na_count <- apply(dffa, 1, function(x) sum(is.na(x)))

dffa$mugcount <- as.integer(as.logical(dffa$mugcount)) # coerce the anywtrue vector to integer

dffb <- dffa[,c(10:11)] # subset to just have the mugcount and na_count columns

dffb$complete <- rowSums (dffb) # create vector to add mugcount and na_count

dffb$gender <- dffa[,c(9)] # bring in gender vector

dffb$complete <- as.integer(as.logical(dffb$complete)) # coerce the complete vector to integer

#dffb$complete <- dffb$complete==0  # create a logical vector where TRUE is any cell equals 0

dffb

##   mugcount na_count complete gender
## 1        0        0        0      F
## 2        1        4        1      M
## 3        0        0        0      F
## 4        0        0        0      M
## 5        1        0        1      M
## 6        0        3        1      F
## 7        1        1        1      M
## 8        0        0        0      M
## 9        0        1        1      M

dffcomplete <- dffb %>% 
  group_by(gender) %>% 
  summarize("complete" = sum(complete)) # group by gender for complete

dffcomplete

## # A tibble: 2 x 2
##   gender complete
##   <fct>     <int>
## 1 F             1
## 2 M             4

dffcomplete <- data.frame(dffcomplete) # make df of prior group

dfc <- join(gcount, dffcomplete) # join the complete and gender count dataframes

## Joining by: gender

dfc$pct <- dfc$complete / dfc$count # create vector for % of each gender to have washed

dfc

##   gender count complete       pct
## 1      F     3        1 0.3333333
## 2      M     6        4 0.6666667

Conclusion

Employees clearly need more direction in how to fill out these forms. The most successful method would probably to provide some sort of incentive for correctly filling it out.

Thursdays have the most dirty mugs–should make sure to keep on top of it these days.

Men and woman are equally likely (about 2/3) to clean either the sides or the sink.

Men are more likely to completely fill the form.