# Keys and fast binary search based subset

#### 2016-12-02

This vignette is aimed at those who are already familiar with data.table syntax, its general form, how to subset rows in `i`, select and compute on columns, add/modify/delete columns by reference in `j` and group by using `by`. If you’re not familiar with these concepts, please read the “Introduction to data.table” and “Reference semantics” vignettes first.

## Data

We will use the same `flights` data as in the “Introduction to data.table” vignette.

    flights <- fread("flights14.csv")
    head(flights)
    #    year month day dep_delay arr_delay carrier origin dest air_time distance hour
    # 1: 2014     1   1        14        13      AA    JFK  LAX      359     2475    9
    # 2: 2014     1   1        -3        13      AA    JFK  LAX      363     2475   11
    # 3: 2014     1   1         2         9      AA    JFK  LAX      351     2475   19
    # 4: 2014     1   1        -8       -26      AA    LGA  PBI      157     1035    7
    # 5: 2014     1   1         2         1      AA    JFK  LAX      350     2475   13
    # 6: 2014     1   1         4         0      AA    EWR  LAX      339     2454   18
    dim(flights)
    # [1] 253316     11

## Introduction

In this vignette, we will

- first introduce the concept of `key` in data.table, and set and use keys to perform fast binary search based subsets in `i`,

- see that we can combine key based subsets along with `j` and `by` in the exact same way as before,

- look at other additional useful arguments - `mult` and `nomatch`,

- and finally conclude by looking at the advantage of setting keys - perform fast binary search based subsets and compare with the traditional vector scan approach.

## 1. Keys

### a) What is a key?

In the “Introduction to data.table” vignette, we saw how to subset rows in `i` using logical expressions, row numbers and using `order()`. In this section, we will look at another way of subsetting incredibly fast - using keys.

But first, let’s start by looking at data.frames. All data.frames have a row names attribute. Consider the data.frame `DF` below.

    set.seed(1L)
    DF = data.frame(ID1 = sample(letters[1:2], 10, TRUE),
                    ID2 = sample(1:3, 10, TRUE),
                    val = sample(10),
                    stringsAsFactors = FALSE,
                    row.names = sample(LETTERS[1:10]))
    DF
    #   ID1 ID2 val
    # C   a   3   5
    # D   a   1   6
    # E   b   2   4
    # G   a   1   2
    # B   b   1  10
    # H   a   2   8
    # I   b   1   9
    # F   b   2   1
    # J   a   3   7
    # A   b   2   3

    rownames(DF)
    #  [1] "C" "D" "E" "G" "B" "H" "I" "F" "J" "A"

We can subset a particular row using its row name as shown below:

    DF["C", ]
    #   ID1 ID2 val
    # C   a   3   5

i.e., row names are more or less an index to rows of a data.frame. However,

1. Each row is limited to exactly one row name.

   But, a person (for example) has at least two names - a first and a second name. It is useful to organise a telephone directory by surname then first name.

2. And row names should be unique.

       rownames(DF) = sample(LETTERS[1:5], 10, TRUE)
       # Warning: non-unique values when setting 'row.names': 'C', 'D'
       # Error in `row.names<-.data.frame`(`*tmp*`, value = value): duplicate 'row.names' are not allowed

Now let’s convert it to a data.table.

    DT = as.data.table(DF)
    DT
    #     ID1 ID2 val
    #  1:   a   3   5
    #  2:   a   1   6
    #  3:   b   2   4
    #  4:   a   1   2
    #  5:   b   1  10
    #  6:   a   2   8
    #  7:   b   1   9
    #  8:   b   2   1
    #  9:   a   3   7
    # 10:   b   2   3

    rownames(DT)
    #  [1] "1"  "2"  "3"  "4"  "5"  "6"  "7"  "8"  "9"  "10"

- Note that row names have been reset.

- data.tables never uses row names. Since data.tables inherit from data.frames, it still has the row names attribute. But it never uses them. We’ll see in a moment as to why.

  If you would like to preserve the row names, use `keep.rownames = TRUE` in `as.data.table()` - this will create a new column called `rn` and assign row names to this column.

Instead, in data.tables we set and use `keys`. Think of a `key` as supercharged rownames.

#### Keys and their properties

1. We can set keys on multiple columns and the column can be of different types – integer, numeric, character, factor, integer64 etc. list and complex types are not supported yet.

2. Uniqueness is not enforced, i.e., duplicate key values are allowed. Since rows are sorted by key, any duplicates in the key columns will appear consecutively.

3. Setting a `key` does two things:

   1. physically reorders the rows of the data.table by the column(s) provided by reference, always in increasing order.

   2. marks those columns as key columns by setting an attribute called `sorted` to the data.table.

   Since the rows are reordered, a data.table can have at most one key because it can not be sorted in more than one way.

#

For the rest of the vignette, we will work with `flights` data set.

### b) Set, get and use keys on a data.table

#### – How can we set the column `origin` as key in the data.table `flights`?

    setkey(flights, origin)
    head(flights)
    #    year month day dep_delay arr_delay carrier origin dest air_time distance hour
    # 1: 2014     1   1         4         0      AA    EWR  LAX      339     2454   18
    # 2: 2014     1   1        -5       -17      AA    EWR  MIA      161     1085   16
    # 3: 2014     1   1       191       185      AA    EWR  DFW      214     1372   16
    # 4: 2014     1   1        -1        -2      AA    EWR  DFW      214     1372   14
    # 5: 2014     1   1        -3       -10      AA    EWR  MIA      154     1085    6
    # 6: 2014     1   1         4       -17      AA    EWR  DFW      215     1372    9

    ## alternatively we can provide character vectors to the function 'setkeyv()'
    # setkeyv(flights, "origin") # useful to program with

####

- You can use the function `setkey()` and provide the column names (without quoting them). This is helpful during interactive use.

- Alternatively you can pass a character vector of column names to the function `setkeyv()`. This is particularly useful while designing functions to pass columns to set key on as function arguments.

- Note that we did not have to assign the result back to a variable. This is because like the `:=` function we saw in the “Introduction to data.table” vignette, `setkey()` and `setkeyv()` modify the input data.table by reference. They return the result invisibly.

- The data.table is now reordered (or sorted) by the column we provided - `origin`. Since we reorder by reference, we only require additional memory of one column of length equal to the number of rows in the data.table, and is therefore very memory efficient.

- You can also set keys directly when creating data.tables using the `data.table()` function using `key` argument. It takes a character vector of column names.

#### set* and `:=`:

In data.table, the `:=` operator and all the `set*` (e.g., `setkey`, `setorder`, `setnames` etc..) functions are the only ones which modify the input object by reference.

#

Once you key a data.table by certain columns, you can subset by querying those key columns using the `.()` notation in `i`. Recall that `.()` is an alias to `list()`.

#### – Use the key column `origin` to subset all rows where the origin airport matches “JFK”

    flights[.("JFK")]
    #        year month day dep_delay arr_delay carrier origin dest air_time distance hour
    #     1: 2014     1   1        14        13      AA    JFK  LAX      359     2475    9
    #     2: 2014     1   1        -3        13      AA    JFK  LAX      363     2475   11
    #     3: 2014     1   1         2         9      AA    JFK  LAX      351     2475   19
    #     4: 2014     1   1         2         1      AA    JFK  LAX      350     2475   13
    #     5: 2014     1   1        -2       -18      AA    JFK  LAX      338     2475   21
    #    ---
    # 81479: 2014    10  31        -4       -21      UA    JFK  SFO      337     2586   17
    # 81480: 2014    10  31        -2       -37      UA    JFK  SFO      344     2586   18
    # 81481: 2014    10  31         0       -33      UA    JFK  LAX      320     2475   17
    # 81482: 2014    10  31        -6       -38      UA    JFK  SFO      343     2586    9
    # 81483: 2014    10  31        -6       -38      UA    JFK  LAX      323     2475   11

    ## alternatively
    # flights[J("JFK")] (or)
    # flights[list("JFK")]

####

- The key column has already been set to `origin`. So it is sufficient to provide the value, here “JFK”, directly. The `.()` syntax helps identify that the task requires looking up the value “JFK” in the key column of data.table (here column `origin` of `flights` data.table).

- The row indices corresponding to the value “JFK” in `origin` is obtained first. And since there is no expression in `j`, all columns corresponding to those row indices are returned.

- On single column key of character type, you can drop the `.()` notation and use the values directly when subsetting, like subset using row names on data.frames.

      flights["JFK"]              ## same as flights[.("JFK")]

- We can subset any amount of values as required

      flights[c("JFK", "LGA")]    ## same as flights[.(c("JFK", "LGA"))]

  This returns all columns corresponding to those rows where `origin` column matches either “JFK” or “LGA”.

#### – How can we get the column(s) a data.table is keyed by?

Using the function `key()`.

    key(flights)
    # [1] "origin"

####

- It returns a character vector of all the key columns.

- If no key is set, it returns `NULL`.

### c) Keys and multiple columns

To refresh, keys are like supercharged row names. We can set key on multiple columns and they can be of multiple types.

#### – How can I set keys on both `origin` and `dest` columns?

    setkey(flights, origin, dest)
    head(flights)
    #    year month day dep_delay arr_delay carrier origin dest air_time distance hour
    # 1: 2014     1   2        -2       -25      EV    EWR  ALB       30      143    7
    # 2: 2014     1   3        88        79      EV    EWR  ALB       29      143   23
    # 3: 2014     1   4       220       211      EV    EWR  ALB       32      143   15
    # 4: 2014     1   4        35        19      EV    EWR  ALB       32      143    7
    # 5: 2014     1   5        47        42      EV    EWR  ALB       26      143    8
    # 6: 2014     1   5        66        62      EV    EWR  ALB       31      143   23

    ## or alternatively
    # setkeyv(flights, c("origin", "dest")) # provide a character vector of column names

    key(flights)
    # [1] "origin" "dest"

####

- It sorts the data.table first by the column `origin` and then by `dest` by reference.

#### – Subset all rows using key columns where first key column `origin` matches “JFK” and second key column `dest` matches “MIA”

    flights[.("JFK", "MIA")]
    #       year month day dep_delay arr_delay carrier origin dest air_time distance hour
    #    1: 2014     1   1        -1       -17      AA    JFK  MIA      161     1089   15
    #    2: 2014     1   1         7        -8      AA    JFK  MIA      166     1089    9
    #    3: 2014     1   1         2        -1      AA    JFK  MIA      164     1089   12
    #    4: 2014     1   1         6         3      AA    JFK  MIA      157     1089    5
    #    5: 2014     1   1         6       -12      AA    JFK  MIA      154     1089   17
    #   ---
    # 2746: 2014    10  31        -1       -22      AA    JFK  MIA      148     1089   16
    # 2747: 2014    10  31        -3       -20      AA    JFK  MIA      146     1089    8
    # 2748: 2014    10  31         2       -17      AA    JFK  MIA      150     1089    6
    # 2749: 2014    10  31        -3       -12      AA    JFK  MIA      150     1089    5
    # 2750: 2014    10  31        29         4      AA    JFK  MIA      146     1089   19

#### How does the subset work here?

- It is important to undertand how this works internally. “JFK” is first matched against the first key column `origin`. And within those matching rows, “MIA” is matched against the second key column `dest` to obtain row indices where both `origin` and `dest` match the given values.

- Since no `j` is provided, we simply return all columns corresponding to those row indices.

#

#### – Subset all rows where just the first key column `origin` matches “JFK”

    key(flights)
    # [1] "origin" "dest"

    flights[.("JFK")] ## or in this case simply flights["JFK"], for convenience
    #        year month day dep_delay arr_delay carrier origin dest air_time distance hour
    #     1: 2014     1   1        10         4      B6    JFK  ABQ      280     1826   20
    #     2: 2014     1   2       134       161      B6    JFK  ABQ      252     1826   22
    #     3: 2014     1   7         6         6      B6    JFK  ABQ      269     1826   20
    #     4: 2014     1   8        15       -15      B6    JFK  ABQ      259     1826   20
    #     5: 2014     1   9        45        32      B6    JFK  ABQ      267     1826   20
    #    ---
    # 81479: 2014    10  31         0       -18      DL    JFK  TPA      142     1005    8
    # 81480: 2014    10  31         1        -8      B6    JFK  TPA      149     1005   19
    # 81481: 2014    10  31        -2       -22      B6    JFK  TPA      145     1005   14
    # 81482: 2014    10  31        -8        -5      B6    JFK  TPA      149     1005    9
    # 81483: 2014    10  31        -4       -18      B6    JFK  TPA      145     1005    8

####

- Since we did not provide any values for the second key column `dest`, it just matches “JFK” against the first key column `origin` and returns all the matched rows.

#### – Subset all rows where just the second key column `dest` matches “MIA”

    flights[.(unique(origin), "MIA")]
    #       year month day dep_delay arr_delay carrier origin dest air_time distance hour
    #    1: 2014     1   1        -5       -17      AA    EWR  MIA      161     1085   16
    #    2: 2014     1   1        -3       -10      AA    EWR  MIA      154     1085    6
    #    3: 2014     1   1        -5        -8      AA    EWR  MIA      157     1085   11
    #    4: 2014     1   1        43        42      UA    EWR  MIA      155     1085   15
    #    5: 2014     1   1        60        49      UA    EWR  MIA      162     1085   21
    #   ---
    # 9924: 2014    10  31       -11        -8      AA    LGA  MIA      157     1096   13
    # 9925: 2014    10  31        -5       -11      AA    LGA  MIA      150     1096    9
    # 9926: 2014    10  31        -2        10      AA    LGA  MIA      156     1096    6
    # 9927: 2014    10  31        -2       -16      AA    LGA  MIA      156     1096   19
    # 9928: 2014    10  31         1       -11      US    LGA  MIA      164     1096   15

#### What’s happening here?

- Read this again. The value provided for the second key column “MIA” has to find the matching values in `dest` key column on the matching rows provided by the first key column `origin`. We can not skip the values of key columns before. Therefore we provide all unique values from key column `origin`.

- “MIA” is automatically recycled to fit the length of `unique(origin)` which is 3.

## 2) Combining keys with `j` and `by`

All we have seen so far is the same concept – obtaining row indices in `i`, but just using a different method – using `keys`. It shouldn’t be surprising that we can do exactly the same things in `j` and `by` as seen from the previous vignettes. We will highlight this with a few examples.

### a) Select in `j`

#### – Return `arr_delay` column as a data.table corresponding to `origin = "LGA"` and `dest = "TPA"`.

    key(flights)
    # [1] "origin" "dest"
    flights[.("LGA", "TPA"), .(arr_delay)]
    #       arr_delay
    #    1:         1
    #    2:        14
    #    3:       -17
    #    4:        -4
    #    5:       -12
    #   ---
    # 1848:        39
    # 1849:       -24
    # 1850:       -12
    # 1851:        21
    # 1852:       -11

####

- The row indices corresponding to `origin == "LGA"` and `dest == "TPA"` are obtained using key based subset.

- Once we have the row indices, we look at `j` which requires only the `arr_delay` column. So we simply select the column `arr_delay` for those row indices in the exact same way as we have seen in Introduction to data.table vignette.

- We could have returned the result by using `with = FALSE` as well.

      flights[.("LGA", "TPA"), "arr_delay", with = FALSE]

### b) Chaining

#### – On the result obtained above, use chaining to order the column in decreasing order.

    flights[.("LGA", "TPA"), .(arr_delay)][order(-arr_delay)]
    #       arr_delay
    #    1:       486
    #    2:       380
    #    3:       351
    #    4:       318
    #    5:       300
    #   ---
    # 1848:       -40
    # 1849:       -43
    # 1850:       -46
    # 1851:       -48
    # 1852:       -49

### c) Compute or do in `j`

#### – Find the maximum arrival delay correspondong to `origin = "LGA"` and `dest = "TPA"`.

    flights[.("LGA", "TPA"), max(arr_delay)]
    # [1] 486

####

- We can verify that the result is identical to first value (486) from the previous example.

### d) sub-assign by reference using `:=` in `j`

We have seen this example already in the Reference semantics vignette. Let’s take a look at all the `hours` available in the `flights` data.table:

    # get all 'hours' in flights
    flights[, sort(unique(hour))]
    #  [1]  0  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

We see that there are totally `25` unique values in the data. Both 0 and 24 hours seem to be present. Let’s go ahead and replace 24 with 0, but this time using key.

    setkey(flights, hour)
    key(flights)
    # [1] "hour"
    flights[.(24), hour := 0L]
    #         year month day dep_delay arr_delay carrier origin dest air_time distance hour
    #      1: 2014     4  15       598       602      DL    EWR  ATL      104      746    0
    #      2: 2014     5  22       289       267      DL    EWR  ATL      102      746    0
    #      3: 2014     7  14       277       253      DL    EWR  ATL      101      746    0
    #      4: 2014     2  14       128       117      EV    EWR  BDL       27      116    0
    #      5: 2014     6  17       127       119      EV    EWR  BDL       24      116    0
    #     ---
    # 253312: 2014     8   3         1       -13      DL    JFK  SJU      196     1598    0
    # 253313: 2014    10   8         1         1      B6    JFK  SJU      199     1598    0
    # 253314: 2014     7  14       211       219      B6    JFK  SLC      282     1990    0
    # 253315: 2014     7   3       440       418      FL    LGA  ATL      107      762    0
    # 253316: 2014     6  13       300       280      DL    LGA  PBI      140     1035    0
    key(flights)
    # NULL

####

- We first set `key` to `hour`. This reorders `flights` by the column `hour` and marks that column as the `key` column.

- Now we can subset on `hour` by using the `.()` notation. We subset for the value 24 and obtain the corresponding row indices.

- And on those row indices, we replace the `key` column with the value `0`.

- Since we have replaced values on the key column, the data.table `flights` isn’t sorted by `hour` anymore. Therefore, the key has been automatically removed by setting to NULL.

#

Now, there shouldn’t be any 24 in the `hour` column.

    flights[, sort(unique(hour))]
    #  [1]  0  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

### e) Aggregation using `by`

Let’s set the key back to `origin, dest` first.

    setkey(flights, origin, dest)
    key(flights)
    # [1] "origin" "dest"

#### – Get the maximum departure delay for each `month` corresponding to `origin = "JFK"`. Order the result by `month`

    ans <- flights["JFK", max(dep_delay), keyby = month]
    head(ans)
    #    month   V1
    # 1:     1  881
    # 2:     2 1014
    # 3:     3  920
    # 4:     4 1241
    # 5:     5  853
    # 6:     6  798
    key(ans)
    # [1] "month"

####

- We subset on the `key` column origin to obtain the row indices corresponding to “JFK”.

- Once we obtain the row indices, we only need two columns - `month` to group by and `dep_delay` to obtain `max()` for each group. data.table’s query optimisation therefore subsets just those two columns corresponding to the row indices obtained in `i`, for speed and memory efficiency.

- And on that subset, we group by month and compute `max(dep_delay)`.

- We use `keyby` to automatically key that result by month. Now we understand what that means. In addition to ordering, it also sets month as the `key` column.

## 3) Additional arguments - `mult` and `nomatch`

### a) The mult argument

We can choose, for each query, if “all” the matching rows should be returned, or just the “first” or “last” using the `mult` argument. The default value is “all” - what we’ve seen so far.

#### – Subset only the first matching row from all rows where `origin` matches “JFK” and `dest` matches “MIA”

    flights[.("JFK", "MIA"), mult = "first"]
    #    year month day dep_delay arr_delay carrier origin dest air_time distance hour
    # 1: 2014     1   1         6         3      AA    JFK  MIA      157     1089    5

#### – Subset only the last matching row of all the rows where `origin` matches “LGA”, “JFK”, “EWR” and `dest` matches “XNA”

    flights[.(c("LGA", "JFK", "EWR"), "XNA"), mult = "last"]
    #    year month day dep_delay arr_delay carrier origin dest air_time distance hour
    # 1: 2014     5  23       163       148      MQ    LGA  XNA      158     1147   18
    # 2:   NA    NA  NA        NA        NA      NA    JFK  XNA       NA       NA   NA
    # 3: 2014     2   3       231       268      EV    EWR  XNA      184     1131   12

####

- The query “JFK”, “XNA” doesn’t match any rows in `flights` and therefore returns `NA`.

- Once again, the query for second key column `dest`, “XNA”, is recycled to fit the length of the query for first key column `origin`, which is of length 3.

### b) The nomatch argument

We can choose if queries that do not match should return `NA` or be skipped altogether using the `nomatch` argument.

#### – From the previous example, Subset all rows only if there’s a match

    flights[.(c("LGA", "JFK", "EWR"), "XNA"), mult = "last", nomatch = 0L]
    #    year month day dep_delay arr_delay carrier origin dest air_time distance hour
    # 1: 2014     5  23       163       148      MQ    LGA  XNA      158     1147   18
    # 2: 2014     2   3       231       268      EV    EWR  XNA      184     1131   12

####

- Default value for `nomatch` is `NA`. Setting `nomatch = 0L` skips queries with no matches.

- The query “JFK”, “XNA” doesn’t match any rows in flights and therefore is skipped.

## 4) binary search vs vector scans

We have seen so far how we can set and use keys to subset. But what’s the advantage? For example, instead of doing:

    # key by origin,dest columns
    flights[.("JFK", "MIA")]

we could have done:

    flights[origin == "JFK" & dest == "MIA"]

One advantage very likely is shorter syntax. But even more than that, binary search based subsets are incredibly fast.

### a) Performance of binary search approach

To illustrate, let’s create a sample data.table with 20 million rows and three columns and key it by columns `x` and `y`.

    set.seed(2L)
    N = 2e7L
    DT = data.table(x = sample(letters, N, TRUE),
                    y = sample(1000L, N, TRUE),
                  val = runif(N), key = c("x", "y"))
    print(object.size(DT), units = "Mb")
    # 381.5 Mb

    key(DT)
    # [1] "x" "y"

`DT` is ~380MB. It is not really huge, but this will do to illustrate the point.

From what we have seen in the Introduction to data.table section, we can subset those rows where columns `x = "g"` and `y = 877` as follows:

    ## (1) Usual way of subsetting - vector scan approach
    t1 <- system.time(ans1 <- DT[x == "g" & y == 877L])
    t1
    #    user  system elapsed
    #   0.172   0.016   0.188
    head(ans1)
    #    x   y       val
    # 1: g 877 0.3946652
    # 2: g 877 0.9424275
    # 3: g 877 0.7068512
    # 4: g 877 0.6959935
    # 5: g 877 0.9673482
    # 6: g 877 0.4842585
    dim(ans1)
    # [1] 761   3

Now let’s try to subset by using keys.

    ## (2) Subsetting using keys
    t2 <- system.time(ans2 <- DT[.("g", 877L)])
    t2
    #    user  system elapsed
    #   0.000   0.004   0.000
    head(ans2)
    #    x   y       val
    # 1: g 877 0.3946652
    # 2: g 877 0.9424275
    # 3: g 877 0.7068512
    # 4: g 877 0.6959935
    # 5: g 877 0.9673482
    # 6: g 877 0.4842585
    dim(ans2)
    # [1] 761   3

    identical(ans1$val, ans2$val)
    # [1] TRUE

- The speedup is ~188x!

### b) Why does keying a data.table result in blazing fast susbets?

To understand that, let’s first look at what vector scan approach (method 1) does.

#### Vector scan approach:

- The column `x` is searched for the value “g” row by row, on all 20 million of them. This results in a logical vector of size 20 million, with values `TRUE, FALSE or NA` corresponding to `x`’s value.

- Similarly, the column `y` is searched for `877` on all 20 million rows one by one, and stored in another logical vector.

- Element wise `&` operations are performed on the intermediate logical vectors and all the rows where the expression evaluates to `TRUE` are returned.

This is what we call a vector scan approach. And this is quite inefficient, especially on larger tables and when one needs repeated subsetting, because it has to scan through all the rows each time.

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Now let us look at binary search approach (method 2). Recall from Properties of key - setting keys reorders the data.table by key columns. Since the data is sorted, we don’t have to scan through the entire length of the column! We can instead use binary search to search a value in `O(log n)` as opposed to `O(n)` in case of vector scan approach, where `n` is the number of rows in the data.table.

#### Binary search approach:

Here’s a very simple illustration. Let’s consider the (sorted) numbers shown below:

    1, 5, 10, 19, 22, 23, 30

Suppose we’d like to find the matching position of the value 1, using binary search, this is how we would proceed - because we know that the data is sorted.

- Start with the middle value = 19. Is 1 == 19? No. 1 < 19.

- Since the value we’re looking for is smaller than 19, it should be somewhere before 19. So we can discard the rest of the half that are >= 19.

- Our set is now reduced to 1, 5, 10. Grab the middle value once again = 5. Is 1 == 5? No. 1 < 5.

- Our set is reduced to 1. Is 1 == 1? Yes. The corresponding index is also 1. And that’s the only match.

A vector scan approach on the other hand would have to scan through all the values (here, 7).

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It can be seen that with every search we reduce the number of searches by half. This is why binary search based subsets are incredibly fast. Since rows of each column of data.tables have contiguous locations in memory, the operations are performed in a very cache efficient manner (also contributes to speed).

In addition, since we obtain the matching row indices directly without having to create those huge logical vectors (equal to the number of rows in a data.table), it is quite memory efficient as well.

## Summary

In this vignette, we have learnt another method to subset rows in `i` by keying a data.table. Setting keys allows us to perform blazing fast subsets by using binary search. In particular, we have seen how to

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- set key and subset using the key on a data.table.

- subset using keys which fetches row indices in `i`, but much faster.

- combine key based subsets with `j` and `by`. Note that the `j` and `by` operations are exactly the same as before.

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Key based subsets are incredibly fast and are particularly useful when the task involves repeated subsetting. But it may not be always desirable to set key and physically reorder the data.table. In the next vignette, we will address this using a new feature – secondary indexes.