- **2e.** Check that in both `sprint.m.df` and `sprint.w.df`, the values in `City.Date` column
only once (i.e., there are no duplicated values).
Do this in a way that you find suitable, but when you Knit this Lab,
the results that demonstrate this claim should be visible in the HTML file.

```{r}

duplicate.bool = FALSE

for (i in (length(sprint.m.df["Date"]) - 1)) {

  if (sprint.m.df["Date"][i] == sprint.m.df["Date"][i + 1]) {

    duplicate.bool = TRUE

  }
}

duplicate.bool

```

Merging data
===

- **3a.** In preparation of merging `sprint.m.df` and `sprint.w.df`, we first
want to find all the sprints that occur in the same race in both data frames.
Specifically, remove all the rows in `sprint.m.df` that have a `City.Date`
that does not occur in `sprint.w.df`. Likewise, remove all the rows in
`sprint.w.df` that have a `City.Date`
that does not occur in `sprint.m.df`.
Then, remove the `City` and `Date` columns in both data frames.
(Hint: You might be interested in the `%in%` function in R. Try looking this up
to see what it does.)
In the end, both `sprint.m.df` and `sprint.w.df` should have 385 rows and 7 columns.
Print out the first 3 lines of `sprint.m.df` and `sprint.w.df`
afterwards.

```{r}

in.both = vector(mode="numeric", length=0)

for (i in (length(sprint.m.df["City.Date"]))) {
  temp = which(sprint.w.df$City.Date == sprint.m.df["City.Date"][i])

  in.both = in.both + temp

}

sprint.w.df = sprint.w.df[-in.both]

head(sprint.m.df, n = 3)
head(sprint.w.df, n = 3)

```

- **3b.** We now will complete the manual merge of `sprint.m.df` and `sprint.w.df`.
Here are the sequence steps to do: First, check the order of values in `City.Date` in
`sprint.m.df` match exactly with those in `sprint.w.df`. Second, use the `cbind()`
function appropriately to create a new data frame `sprint.df` that has 13 columns.
The first column should be `City.Date`, the next 6 columns should contain all the
remaining columns from `sprint.m.df`, and the last 6 columns should contain all the
remaining columns form `sprint.w.df`. Of course, each row should correspond to
sprints from the same `City.Date`. Print out the first 3 lines of `sprint.df`
afterwards.

```{r}

idential(sprint.m.df["City.Date"], sprint.w.df["City.Date"])

sprint.df = cbind(sprint.m.df, sprint.w.df, by = "City.Date")

head(sprint.df, n = 3)

```