I have two tables in BigQuery that I am trying to merge. For the purpose of explanation, let us name the two tables as A and B. So, we merge B into A. Also, I have a primary key called id based on which I am performing the merge. Now, both of them have a column (let us name it as X for explanation purposes) which is of the type ARRAY. My main intention is to replace the array data in A with that of B if the arrays are not equal in both the table. How can I do that. I did find posts on SO and other sites but none of them are working in my usecase.
A B
---------- ----------
id | x id | x
---------- ----------
1 | [1,2] 1 | [1,2]
---------- ----------
2 | [3] 2 | [4, 5]
The result of the merge should be
A
----------
id | x
----------
1 | [1,2]
----------
2 | [4,5]
How can I achieve the above result. Any leads will be very helpful. Also, if there are some other posts that address the above scenario directly, please point me to them
Edits:
I tried the following:
merge A as main_table
using B as updated_table
on main_table.id = updated_taable.id
when matched and main_table.x != updated_table.x then update set main_table.x = updated_table.x
when not matched then
insert(id, x) values (updated_table.id, updated_table.x)
;
Hope, this helps.
I cannot direclty use a compare operator over array right. My use case is that only update values when they are not equal. So, i cannot use something like != directly. This is the main problem
You can use to_json_string function to compare two arrays "directly"
to_json_string(main_table.x) != to_json_string(updated_table.x)
I have a result set that looks like
{add=[44961373 (1645499799657512961), 44961374 (1645499799658561538), 44962094 (1645499799659610114), 44962095 (1645499799659610117), 44962096 (1645499799660658689), 44962097 (1645499799660658691), 44962098 (1645499799661707264), 44962099 (1645499799661707267), 44962100 (1645499799662755840), 44962101 (1645499799662755843), ... (592 adds)]}
If the add=[ array has more than 10 elements in it. Then it will put (x adds) at the end of the statement to show how many actual adds there were. IF it has less than 10, then it wont put the (x adds) statement. I am wanting timechart and also single value these outputs to a dashboard(separate modules).
I can get one or the other but I would like to use from logic to figure out which one to report.
index="index" host="host*" path=/update | eval count=mvcount(add) | stats count
will get the count of the array
index="index" host="host*" path=/update | stats sum(Adds)
will get the value of the (x adds). Adds is a 'extracted field'.
How do I get either or? If add array >10, use sum(Adds), in the same breath.
index="index" host="host*" path=/update | eval count=mvcount(add)
| eval first_ten="{add=[".mvjoin(mvindex(add,0,9), ",")." (" (count-10)." adds)}"
| eval msg=if(count<10,_raw,first_ten)
You can do something like this. Get the count of adds, create a new string with the first 10 elements only, with the count-10 adds message at the end. Then, depending on the actual count, either use the original (_raw), or the new message.
I need to remove multiple substrings from a column of strings. I have ~200k rows, as well as a large bag of custom "stopwords."
Specifically, I have a table (NameTable), with a Names column full of multiple-word strings. Only a small subset of the words in each string is relevant for what I'm doing next. names has a trigram/GIN index. Is there any way to speed up a query like the following, but with a much larger string of target words? I'm using PG10.
...
Create Index Names_trgm on NameTable using gin(Name gin_trgm_ops);
Update NameTable
set CleanedName = regexp_replace(Name, '( fuzzy| wuzzy| was| a | bear | the |
first| time | yossarian| saw | the | chaplain | he | fell| madly| in | love|
with| him)',' ','g');
I have a question about elisp. For example:
(setq trees '(maple oak pine birch))
-> (maple oak pine birch)
(setcdr (nthcdr 2 trees) nil)
-> nil
trees
-> (maple oak pine)
I thought (nthcdr 2 trees) returns a new list - (pine birch) and put the list into the setcdr expression, which should not change the value of trees. Could anyone explain it to me?
If you read the documentation string for nthcdr, you'll see that it just returns a pointer to the "nth" "cdr" - which is a pointer into the original list. So you're modifying the original list.
Doc string:
Take cdr N times on LIST, return the result.
Edit Wow, "pointer" seems to stir up confusion. Yes, Lisp has pointers.
Just look at the box diagrams used to explain list structure in lisp (here's Emacs's documentation on that very thing):
--- --- --- --- --- ---
| | |--> | | |--> | | |--> nil
--- --- --- --- --- ---
| | |
| | |
--> rose --> violet --> buttercup
Look at all those arrows, they almost look like they're .... pointing to things. When you take the cdr of a list, you get what the 2nd box refers to (aka "points" to), be it an atom, a string, or another cons cell. Heck, check it out on Wikipedia's entry for CAR and CDR.
If it feels better to call it a reference, use that terminology.
cdr certainly does NOT return a copy of what it refers to, which is what was confusing RNAer.
R: "Look, in the sky! The Lambda Signal! A citizen is in trouble Lambda Man!"
LM: "I see it! And I've got just the box art they need."
In Lisps, lists are singly-linked data structures, comprised of elements called cons cells. Each of these cells is a structure that consists of
a pointer to a value
a pointer to the next cell
These are called the car and cdr respectively for historical reasons. Here's the traditional box art representing a 3-element list:
Structure: (car . cdr -)--->(car . cdr -)--->(car . cdr)
| | | |
v v v v
Values: 1 2 3 nil
The car and cdr functions allow you to work with lists from this low level of abstraction, and return the values of the respective cells. Thus, car returns the 'value' of the cell, and cdr dereferences to the remainder of the list. nthcdr is a generalisation on top of cdr for convenience.
The value returned by cdr is a reference to a raw data structure, which is mutable at this level. If you change the value of a cons-cell's cdr, you are changing the underlying structure of the list.
Given:
let A = '(1 2) ~= (1 . -)-->(2 . nil)
let B = '(3 4) ~= (3 . -)-->(4 . nil)
Setting the cdr of (cdr A) to B will destructively concatenate A and B such that A is now the structure below:
A B
(1 . -)-->(2 . -)-->(3 . -)-->(4 . nil)
As we've shown, a nil value in a cell's cdr represents the end of the list - there's nothing more that can be traversed. If we set the cdr of A to nil, we lobotomise the list, such that A is now
A
(1 . nil) <- [Not pointing to anything - the rest of the list shall go wanting]
This is pretty much what you've done - you've mutated the underlying data structures using low-level functions. :) By setting one of the cell's cdrs to nil, you've trimmed the end off your list.
This is called 'mutation'. And is present everywhere apart from Haskell.
There are functions that mutate the data structures and functions that duplicate it.
Taking an example of Fibonacci Series from the Clojure Wiki, the Clojure code is :
(def fib-seq
(lazy-cat [0 1] (map + (rest fib-seq) fib-seq)))
If you were to think about this starting from the [0 1], how does it work ? Would be great if there are suggestions on the thought process that goes into thinking in these terms.
As you noted, the [0 1] establishes the base cases: The first two values in the sequence are zero, then one. After that, each value is to be the sum of the previous value and the value before that one. Hence, we can't even compute the third value in the sequence without having at least two that come before it. That's why we need two values with which to start off.
Now look at the map form. It says to take the head items from two different sequences, combine them with the + function (adding multiple values to produce one sum), and expose the result as the next value in a sequence. The map form is zipping together two sequences — presumably of equal length — into one sequence of the same length.
The two sequences fed to map are different views of the same basic sequence, shifted by one element. The first sequence is "all but the first value of the base sequence". The second sequence is the base sequence itself, which, of course, includes the first value. But what should the base sequence be?
The definition above said that each new element is the sum of the previous (Z - 1) and the predecessor to the previous element (Z - 2). That means that extending the sequence of values requires access to the previously computed values in the same sequence. We definitely need a two-element shift register, but we can also request access to our previous results instead. That's what the recursive reference to the sequence called fib-seq does here. The symbol fib-seq refers to a sequence that's a concatenation of zero, one, and then the sum of its own Z - 2 and Z - 1 values.
Taking the sequence called fib-seq, drawing the first item yields the first element of the [0 1] vector — zero. Drawing the second item yields the second element of the vector — one. Upon drawing the third item, we consult the map to generate a sequence and use that as the remaining values. The sequence generated by map here starts out with the sum of the first item of "the rest of" [0 1], which is one, and the first item of [0 1], which is zero. That sum is one.
Drawing the fourth item consults map again, which now must compute the sum of the second item of "the rest of" the base sequence, which is the one generated by map, and the second item of the base sequence, which is the one from the vector [0 1]. That sum is two.
Drawing the fifth item consults map, summing the third item of "the rest of" the base sequence — again, the one resulting from summing zero and one — and the third item of the base sequence — which we just found to be two.
You can see how this is building up to match the intended definition for the series. What's harder to see is whether drawing each item is recomputing all the preceding values twice — once for each sequence examined by map. It turns out there's no such repetition here.
To confirm this, augment the definition of fib-seq like this to instrument the use of function +:
(def fib-seq
(lazy-cat [0 1]
(map
(fn [a b]
(println (format "Adding %d and %d." a b))
(+ a b))
(rest fib-seq) fib-seq)))
Now ask for the first ten items:
> (doall (take 10 fib-seq))
Adding 1 and 0.
Adding 1 and 1.
Adding 2 and 1.
Adding 3 and 2.
Adding 5 and 3.
Adding 8 and 5.
Adding 13 and 8.
Adding 21 and 13.
(0 1 1 2 3 5 8 13 21 34)
Notice that there are eight calls to + to generate the first ten values.
Since writing the preceding discussion, I've spent some time studying the implementation of lazy sequences in Clojure — in particular, the file LazySeq.java — and thought this would be a good place to share a few observations.
First, note that many of the lazy sequence processing functions in Clojure eventually use lazy-seq over some other collection. lazy-seq creates an instance of the Java type LazySeq, which models a small state machine. It has several constructors that allow it to start in different states, but the most interesting case is the one that starts with just a reference to a nullary function. Constructed that way, the LazySeq has neither evaluated the function nor found a delegate sequence (type ISeq in Java). The first time one asks the LazySeq for its first element — via first — or any successors — via next or rest — it evaluates the function, digs down through the resulting object to peel away any wrapping layers of other LazySeq instances, and finally feeds the innermost object through the java function RT#seq(), which results in an ISeq instance.
At this point, the LazySeq has an ISeq to which to delegate calls on behalf of first, next, and rest. Usually the "head" ISeq will be of type Cons, which stores a constant value in its "first" (or "car") slot and another ISeq in its "rest" (or "cdr") slot. That ISeq in its "rest" slot can in turn be a LazySeq, in which case accessing it will again require this same evaluation of a function, peeling away any lazy wrappers on the return value, and passing that value through RT#seq() to yield another ISeq to which to delegate.
The LazySeq instances remain linked together, but having forced one (through first, next, or rest) causes it to delegate straight through to some non-lazy ISeq thereafter. Usually that forcing evaluates a function that yields a Cons bound to first value and its tail bound to another LazySeq; it's a chain of generator functions that each yield one value (the Cons's "first" slot) linked to another opportunity to yield more values (a LazySeq in the Cons's "rest" slot).
Tying this back, in the Fibonacci Sequence example above, map will take each of the nested references to to fib-seq and walk them separately via repeated calls to rest. Each such call will transform at most one LazySeq holding an unevaluated function into a LazySeq pointing to something like a Cons. Once transformed, any subsequent accesses will quickly resolve to the Conses — where the actual values are stored. When one branch of the map zipping walks fib-seq one element behind the other, the values have already been resolved and are available for constant-time access, with no further evaluation of the generator function required.
Here are some diagrams to help visualize this interpretation of the code:
+---------+
| LazySeq |
fib-seq | fn -------> (fn ...)
| sv |
| s |
+---------+
+---------+
| LazySeq |
fib-seq | fn -------> (fn ...) -+
| sv <------------------+
| s |
+---------+
+---------+
| LazySeq |
fib-seq | fn |
| sv -------> RT#seq() -+
| s <------------------+
+---------+
+---------+ +------+
| LazySeq | | ISeq |
fib-seq | fn | | |
| sv | | |
| s ------->| |
+---------+ +------+
+---------+ +--------+ +------+
| LazySeq | | Cons | | ISeq |
fib-seq | fn | | first ---> 1 | |
| sv | | more -------->| |
| s ------->| | | |
+---------+ +--------+ +------+
+---------+ +--------+ +---------+
| LazySeq | | Cons | | LazySeq |
fib-seq | fn | | first ---> 1 | fn -------> (fn ...)
| sv | | more -------->| sv |
| s ------->| | | s |
+---------+ +--------+ +---------+
As map progresses, it hops from LazySeq to LazySeq (and hence Cons to Cons), and the rightmost edge only expands the first time one calls first, next, or rest on a given LazySeq.
My Clojure is a bit rusty, but this seems to be a literal translation of the famous Haskell one-liner:
fibs = 0 : 1 : zipWith (+) fibs (tail fibs)
[I'm going to be using pseudo-Haskell, because it's a little bit more succinct.]
The first thing you need to do, is simply let laziness sink in. When you look at a definition like this:
zeroes = 0 : zeroes
Your immediate gut reaction as a strict programmer would be "ZOMG infinite loop! Must fix bug!" But it isn't an infinite loop. This is a lazy infinite loop. If you do something stupid like
print zeroes
Then, yes, there will be an infinite loop. But as long as you simply use a finite number of elements, you will never notice that the recursion doesn't actually terminate. This is really hard to get. I still don't.
Laziness is like the monetary system: it's based on the assumption that the vast majority of people never use the vast majority of their money. So, when you put $1000 in the bank, they don't keep it in their safe. They lend it to someone else. Actually, they leverage the money, which means that they lend $5000 to someone else. They only ever need enough money so that they can quickly reshuffle it so that it's there when you are looking at it, giving you the appearance that they actually keep your money.
As long as they can manage to always give out money when you walk up to an ATM, it doesn't actually matter that the vast majority of your money isn't there: they only need the small amount you are withdrawing at the point in time when you are making your withdrawal.
Laziness works the same: whenever you look at it, the value is there. If you look at the first, tenth, hundreth, quadrillionth element of zeroes, it will be there. But it will only be there, if and when you look at it, not before.
That's why this inifintely recursive definition of zeroes works: as long as you don't try to look at the last element (or every element) of an infinite list, you are safe.
Next step is zipWith. (Clojure's map is just a generalization of what in other programming languages are usually three different functions: map, zip and zipWith. In this example, it is used as zipWith.)
The reason why the zip family of functions is named that way, is because it actually works like a zipper, and that is also how to best visualize it. Say we have some sporting event, where every contestant gets two tries, and the score from both tries is added up to give the end result. If we have two sequences, run1 and run2 with the scores from each run, we can calculate the end result like so:
res = zipWith (+) run1 run2
Assuming our two lists are (3 1 6 8 6) and (4 6 7 1 3), we line both of those lists up side by side, like the two halves of a zipper, and then we zip them together using our given function (+ in this case) to yield a new sequence:
3 1 6 8 6
+ + + + +
4 6 7 1 3
= = = = =
7 7 13 9 9
Contestant #3 wins.
So, what does our fib look like?
Well, it starts out with a 0, then we append a 1, then we append the sum of the infinite list with the infinite list shifted by one element. It's easiest to just draw that out:
the first element is zero:
0
the second element is one:
0 1
the third element is the first element plus the first element of the rest (i.e. the second element). We visualize this again like a zipper, by putting the two lists on top of each other.
0 1
+
1
=
1
Now, the element that we just computed is not just the output of the zipWith function, it is at the same time also the input, as it gets appended to both lists (which are actually the same list, just shifted by one):
0 1 1
+ +
1 1
= =
1 2
and so forth:
0 1 1 2
+ + +
1 1 2
= = =
1 2 3
0 1 1 2 3 ^
+ + + + |
1 1 2 3 ^ |
= = = = | |
1 2 3 5---+---+
Or if you draw it a little bit differently and merge the result list and the second input list (which really are the same, anyway) into one:
0 1 1 2 3 ^
+ + + + + |
1 = 1 = 2 = 3 = 5---+
That's how I visualize it, anyway.
As for how this works:
Each term of the fibonacci series is obviously the result of adding the previous two terms.
That's what map is doing here, map applies + to each element in each sequence until one of the sequences runs out (which they won't in this case, of course). So the result is a sequence of numbers that result from adding one term in the sequence to the next term in the sequence. Then you need lazy-cat to give it a starting point and make sure the function only returns what it's asked for.
The problem with this implementation is that fib-seq is holding onto the whole sequence for as long as the fib-seq is defined, so it will eventually run you out of memory.
Stuart Halloway's book spends some time on dissecting different implementations of this function, I think the most interesting one is below (it's Christophe Grande's):
(defn fibo []
(map first (iterate (fn [[a b]] [b (+ a b)]) [0 1])))
Unlike the posted implementation previously read elements of the sequence have nothing holding onto them so this one can keep running without generating an OutOfMemoryError.
How to get thinking in these terms is a harder question. So far for me it's a matter of getting acquainted with a lot of different ways of doing things and trying them out, while in general looking for ways to apply the existing function library in preference to using recursion and lazy-cat. But in some cases the recursive solution is really great, so it depends on the problem. I'm looking forward to getting the Joy of Clojure book, because I think it will help me a lot with this issue.