bar
Syntax
bar(X, interval, [closed=’left’])
Related function: dailyAlignedBar
Arguments
X is an integral/temporal scalar or vector.
interval is an integral/DURATION type scalar greater than 0 or a vector of the same length as X.
When interval is of type DURATION, the following time units are supported (case-sensitive): w, d, H, m, s, ms, us, ns.
Note: As time units y and M are not supported in interval, to group X by year or month, convert the data format of X with function month or year. Specify the interval as an integer for calculation. You can refer to Example 2.
closed is a string. This optional parameter can specified as ‘left’ or ‘right’, indicating whether an element of X that is divisible by interval is the left boundary (the first element of the group) or the right boundary(the last element of the group) of a group. The default value is ‘left’.
closed = ‘left’: X-(X % interval), indicating that the value with a remainder of 0 is specified as the left boundary of a group.
closed = ‘right’: iif((X % interval) == 0, X, X + (interval-(X % interval))), indicating that the value with a remainder of 0 is specified as the right boundary of a group.
Details
bar can group X based on the length specified by interval. Return a vector with the same length as X.
Examples
$ bar(100,3); // 100-(100%3)=100-1=99
99
$ bar(0..15, 3)
[0,0,0,3,3,3,6,6,6,9,9,9,12,12,12,15]
$ x=[7,4,5,8,9,3,3,5,2,6,12,1,0,-5,32]
$ bar(x, 5)
[5,0,5,5,5,0,0,5,0,5,10,0,0,-5,30]
$ t=table(2021.01.01T01:00:00..2021.01.01T01:00:29 as time, rand(1.0, 30) as x)
$ select max(x) from t group by bar(time,5s)
bar_time |
max_x |
|---|---|
2021.01.01T01:00:00 |
0.539024 |
2021.01.01T01:00:05 |
0.793327 |
2021.01.01T01:00:10 |
0.958522 |
2021.01.01T01:00:15 |
0.96987 |
2021.01.01T01:00:20 |
0.827086 |
2021.01.01T01:00:25 |
0.617353 |
In the following example, to group data by every 3 months, convert X with the month function, and specify duration as an integer in bar.
$ t=table(take(2018.01.01T01:00:00+1..10,10) join take(2018.02.01T02:00:00+1..10,10) join take(2018.03.01T08:00:00+1..10,10) join take(2018.04.01T08:00:00+1..10,10) join take(2018.05.01T08:00:00+1..10, 10) as time, rand(1.0, 50) as x)
$ select max(x) from t group by bar(month(time), 3);
bar |
max_x |
|---|---|
2018.01M |
0.9868 |
2018.04M |
0.9243 |
The following example groups data by week and calculates the maximum values for each week. Depending on parameter closed, the results are different.
$ t=table(2022.01.01 + 1..20 as time, rand(100, 20) as x)
time |
x |
|---|---|
2022.01.02 |
6 |
2022.01.03 |
29 |
2022.01.04 |
71 |
2022.01.05 |
56 |
2022.01.06 |
93 |
2022.01.07 |
34 |
2022.01.08 |
77 |
2022.01.09 |
18 |
2022.01.10 |
62 |
2022.01.11 |
33 |
2022.01.12 |
34 |
2022.01.13 |
64 |
2022.01.14 |
80 |
2022.01.15 |
63 |
2022.01.16 |
17 |
2022.01.17 |
66 |
2022.01.18 |
85 |
2022.01.19 |
27 |
2022.01.20 |
77 |
2022.01.21 |
27 |
$ select max(x) from t group by bar(time, 7d);
bar_time |
max_x |
|---|---|
2021.12.30 |
71 |
2022.01.06 |
93 |
2022.01.13 |
85 |
2022.01.20 |
77 |
$ print select max(x) from t group by bar(time, 7d, closed='right');
bar_time |
max_x |
|---|---|
2021.01.06 |
93 |
2022.01.13 |
77 |
2022.01.20 |
85 |
2022.01.27 |
27 |
When calculating 1-minute OHLC bars, the data type needs to be converted to LONG if n needs to be converted to NANOTIMESTAMP, otherwise an integer overflow will occur.
1$ n = 1000000
2$ nano = (09:30:00.000000000 + rand(long(6.5*60*60*1000000000), n)).sort!()
3$ price = 100+cumsum(rand(0.02, n)-0.01)
4$ volume = rand(1000, n)
5$ symbol = rand(`600519`000001`600000`601766, n)
6$ tradeNano = table(symbol, nano, price, volume).sortBy!(`symbol`nano)
7$ undef(`nano`price`volume`symbol)
8$ barMinutes = 7
9$ itv = barMinutes*60*long(1000000000)
10
11$ OHLC_nano=select first(price) as open, max(price) as high, min(price) as low, last(price) as close, sum(volume) as volume from tradeNano group by symbol, bar(nano, itv) as barStart