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.
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
# <int> <int> <int> <int> <int> <char> <char> <char> <int> <int> <int>
# 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
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.
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
# I a 1 10
# D a 3 9
# G a 1 4
# A a 1 7
# B a 1 1
# E b 1 8
# C b 2 3
# J b 1 2
# F b 1 5
# H a 2 6
rownames(DF)
# [1] "I" "D" "G" "A" "B" "E" "C" "J" "F" "H"
We can subset a particular row using its row name as shown below:
DF["C", ]
# ID1 ID2 val
# C b 2 3
i.e., row names are more or less an index to rows of a data.frame. However,
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.
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 `.rowNamesDF<-`(x, 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
# <char> <int> <int>
# 1: a 1 10
# 2: a 3 9
# 3: a 1 4
# 4: a 1 7
# 5: a 1 1
# 6: b 1 8
# 7: b 2 3
# 8: b 1 2
# 9: b 1 5
# 10: a 2 6
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.
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.
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.
Setting a key does two things:
a. physically reorders the rows of the data.table by the column(s) provided by reference, always in increasing order.
b. 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.
origin as key in the data.table flights?setkey(flights, origin)
head(flights)
# Key: <origin>
# year month day dep_delay arr_delay carrier origin dest air_time distance hour
# <int> <int> <int> <int> <int> <char> <char> <char> <int> <int> <int>
# 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 “Reference semantics” 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.
:=: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().
origin to subset all rows where the origin airport matches “JFK”flights[.("JFK")]
# Key: <origin>
# year month day dep_delay arr_delay carrier origin dest air_time distance hour
# <int> <int> <int> <int> <int> <char> <char> <char> <int> <int> <int>
# 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”.
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.
To refresh, keys are like supercharged row names. We can set key on multiple columns and they can be of multiple types.
origin and dest columns?setkey(flights, origin, dest)
head(flights)
# Key: <origin, dest>
# year month day dep_delay arr_delay carrier origin dest air_time distance hour
# <int> <int> <int> <int> <int> <char> <char> <char> <int> <int> <int>
# 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"
origin and then by dest by reference.origin matches “JFK” and second key column dest matches “MIA”flights[.("JFK", "MIA")]
# Key: <origin, dest>
# year month day dep_delay arr_delay carrier origin dest air_time distance hour
# <int> <int> <int> <int> <int> <char> <char> <char> <int> <int> <int>
# 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
It is important to understand 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.
origin matches “JFK”key(flights)
# [1] "origin" "dest"
flights[.("JFK")] ## or in this case simply flights["JFK"], for convenience
# Key: <origin, dest>
# year month day dep_delay arr_delay carrier origin dest air_time distance hour
# <int> <int> <int> <int> <int> <char> <char> <char> <int> <int> <int>
# 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
dest, it just matches “JFK” against the first key column origin and returns all the matched rows.dest matches “MIA”flights[.(unique(origin), "MIA")]
# Key: <origin, dest>
# year month day dep_delay arr_delay carrier origin dest air_time distance hour
# <int> <int> <int> <int> <int> <char> <char> <char> <int> <int> <int>
# 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
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.
j and byAll 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.
jarr_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
# <int>
# 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]
flights[.("LGA", "TPA"), .(arr_delay)][order(-arr_delay)]
# arr_delay
# <int>
# 1: 486
# 2: 380
# 3: 351
# 4: 318
# 5: 300
# ---
# 1848: -40
# 1849: -43
# 1850: -46
# 1851: -48
# 1852: -49
jorigin = "LGA" and dest = "TPA".flights[.("LGA", "TPA"), max(arr_delay)]
# [1] 486
:= in jWe 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]
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 any more. 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
byLet’s set the key back to origin, dest first.
setkey(flights, origin, dest)
key(flights)
# [1] "origin" "dest"
month corresponding to origin = "JFK". Order the result by monthans <- flights["JFK", max(dep_delay), keyby = month]
head(ans)
# Key: <month>
# month V1
# <int> <int>
# 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.
mult and nomatchWe 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.
origin matches “JFK” and dest matches “MIA”flights[.("JFK", "MIA"), mult = "first"]
# Key: <origin, dest>
# year month day dep_delay arr_delay carrier origin dest air_time distance hour
# <int> <int> <int> <int> <int> <char> <char> <char> <int> <int> <int>
# 1: 2014 1 1 6 3 AA JFK MIA 157 1089 5
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
# <int> <int> <int> <int> <int> <char> <char> <char> <int> <int> <int>
# 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.
We can choose if queries that do not match should return NA or be skipped altogether using the nomatch argument.
flights[.(c("LGA", "JFK", "EWR"), "XNA"), mult = "last", nomatch = NULL]
# year month day dep_delay arr_delay carrier origin dest air_time distance hour
# <int> <int> <int> <int> <int> <char> <char> <char> <int> <int> <int>
# 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 = NULL skips queries with no matches.
The query “JFK”, “XNA” doesn’t match any rows in flights and therefore is skipped.
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.
As the time goes data.table gets new optimization and currently the latter call is automatically optimized to use binary search.
To use slow vector scan key needs to be removed.
setkey(flights, NULL)
flights[origin == "JFK" & dest == "MIA"]
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))
print(object.size(DT), units = "Mb")
# 381.5 Mb
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:
key(DT)
# NULL
## (1) Usual way of subsetting - vector scan approach
t1 <- system.time(ans1 <- DT[x == "g" & y == 877L])
t1
# user system elapsed
# 0.189 0.015 0.204
head(ans1)
# x y val
# <char> <int> <num>
# 1: g 877 0.57059767
# 2: g 877 0.74859806
# 3: g 877 0.03616756
# 4: g 877 0.28087868
# 5: g 877 0.83727299
# 6: g 877 0.43867189
dim(ans1)
# [1] 762 3
Now let’s try to subset by using keys.
setkeyv(DT, c("x", "y"))
key(DT)
# [1] "x" "y"
## (2) Subsetting using keys
t2 <- system.time(ans2 <- DT[.("g", 877L)])
t2
# user system elapsed
# 0.000 0.000 0.001
head(ans2)
# Key: <x, y>
# x y val
# <char> <int> <num>
# 1: g 877 0.57059767
# 2: g 877 0.74859806
# 3: g 877 0.03616756
# 4: g 877 0.28087868
# 5: g 877 0.83727299
# 6: g 877 0.43867189
dim(ans2)
# [1] 762 3
identical(ans1$val, ans2$val)
# [1] TRUE
To understand that, let’s first look at what vector scan approach (method 1) does.
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.
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.
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).
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.
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
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.
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.