I have already seen many threads with examples of how to do this, the problem is, I still can't do it.
All the examples have tables with extra data. For example somethings like this
lines = {
luaH_set = 10,
luaH_get = 24,
luaH_present = 48,
}
or this,
obj = {
{ N = 'Green1' },
{ N = 'Green' },
{ N = 'Sky blue99' }
}
I can code in a few languages but I'm very new to Lua, and tables are really confusing to me. I can't seem to work out how to adapt the code in the examples to be able to sort a simple table.
This is my table:
local players = {"barry", "susan", "john", "wendy", "kevin"}
I want to sort these names alphabetically. I understand that Lua tables don't preserve order, and that's what's confusing me. All I essentially care about doing is just printing these names in alphabetical order, but I feel I need to learn this properly and know how to index them in the right order to a new table.
The examples I see are like this:
local function cmp(a, b)
a = tostring(a.N)
b = tostring(b.N)
local patt = '^(.-)%s*(%d+)$'
local _,_, col1, num1 = a:find(patt)
local _,_, col2, num2 = b:find(patt)
if (col1 and col2) and col1 == col2 then
return tonumber(num1) < tonumber(num2)
end
return a < b
end
table.sort(obj, cmp)
for i,v in ipairs(obj) do
print(i, v.N)
end
or this:
function pairsByKeys (t, f)
local a = {}
for n in pairs(t) do table.insert(a, n) end
table.sort(a, f)
local i = 0 -- iterator variable
local iter = function () -- iterator function
i = i + 1
if a[i] == nil then return nil
else return a[i], t[a[i]]
end
end
return iter
end
for name, line in pairsByKeys(lines) do
print(name, line)
end
and I'm just absolutely thrown by this as to how to do the same thing for a simple 1D table.
Can anyone please help me to understand this? I know if I can understand the most basic example, I'll be able to teach myself these harder examples.
local players = {"barry", "susan", "john", "wendy", "kevin"}
-- sort ascending, which is the default
table.sort(players)
print(table.concat(players, ", "))
-- sort descending
table.sort(players, function(a,b) return a > b end)
print(table.concat(players, ", "))
Here's why:
Your table players is a sequence.
local players = {"barry", "susan", "john", "wendy", "kevin"}
Is equivalent to
local players = {
[1] = "barry",
[2] = "susan",
[3] = "john",
[4] = "wendy",
[5] = "kevin",
}
If you do not provide keys in the table constructor, Lua will use integer keys automatically.
A table like that can be sorted by its values. Lua will simply rearrange the index value pairs in respect to the return value of the compare function. By default this is
function (a,b) return a < b end
If you want any other order you need to provide a function that returs true if element a comes befor b
Read this https://www.lua.org/manual/5.4/manual.html#pdf-table.sort
table.sort
Sorts the list elements in a given order, in-place, from list[1] to
list[#list]
This example is not a "list" or sequence:
lines = {
luaH_set = 10,
luaH_get = 24,
luaH_present = 48,
}
Which is equivalent to
lines = {
["luaH_set"] = 10,
["luaH_get"] = 24,
["luaH_present"] = 48,
}
it only has strings as keys. It has no order. You need a helper sequence to map some order to that table's element.
The second example
obj = {
{ N = 'Green1' },
{ N = 'Green' },
{ N = 'Sky blue99' }
}
which is equivalent to
obj = {
[1] = { N = 'Green1' },
[2] = { N = 'Green' },
[3] = { N = 'Sky blue99' },
}
Is a list. So you could sort it. But sorting it by table values wouldn't make too much sense. So you need to provide a function that gives you a reasonable way to order it.
Read this so you understand what a "sequence" or "list" is in this regard. Those names are used for other things as well. Don't let it confuse you.
https://www.lua.org/manual/5.4/manual.html#3.4.7
It is basically a table that has consecutive integer keys starting at 1.
Understanding this difference is one of the most important concepts while learning Lua. The length operator, ipairs and many functions of the table library only work with sequences.
This is my table:
local players = {"barry", "susan", "john", "wendy", "kevin"}
I want to sort these names alphabetically.
All you need is table.sort(players)
I understand that LUA tables don't preserve order.
Order of fields in a Lua table (a dictionary with arbitrary keys) is not preserved.
But your Lua table is an array, it is self-ordered by its integer keys 1, 2, 3,....
To clear up the confusing in regards to "not preserving order": What's not preserving order are the keys of the values in the table, in particular for string keys, i.e. when you use the table as dictionary and not as array. If you write myTable = {orange="hello", apple="world"} then the fact that you defined key orange to the left of key apple isn't stored. If you enumerate keys/values using for k, v in pairs(myTable) do print(k, v) end then you'd actually get apple world before orange hello because "apple" < "orange".
You don't have this problem with numeric keys though (which is what the keys by default will be if you don't specify them - myTable = {"hello", "world", foo="bar"} is the same as myTable = {[1]="hello", [2]="world", foo="bar"}, i.e. it will assign myTable[1] = "hello", myTable[2] = "world" and myTable.foo = "bar" (same as myTable["foo"]). (Here, even if you would get the numeric keys in a random order - which you don't, it wouldn't matter since you could still loop through them by incrementing.)
You can use table.sort which, if no order function is given, will sort the values using < so in case of numbers the result is ascending numbers and in case of strings it will sort by ASCII code:
local players = {"barry", "susan", "john", "wendy", "kevin"}
table.sort(players)
-- players is now {"barry", "john", "kevin", "susan", "wendy"}
This will however fall apart if you have mixed lowercase and uppercase entries because uppercase will go before lowercase due to having lower ASCII codes, and of course it also won't work properly with non-ASCII characters like umlauts (they will go last) - it's not a lexicographic sort.
You can however supply your own ordering function which receives arguments (a, b) and needs to return true if a should come before b. Here an example that fixes the lower-/uppercase issues for example, by converting to uppercase before comparing:
table.sort(players, function (a, b)
return string.upper(a) < string.upper(b)
end)
I want arrange this list by list[i][1].
List lst =
[
['a',2],
['b',5],
['c',3],
['d',1],
];
So it will be changed like this.
[
['d',1],
['a',2],
['c',3],
['b',5],
];
How can i do this?
Is it possible to make it by combining with for loop and sort()?
Use the sort function:
lst.sort((a, b) => a[1].compareTo(b[1]));
Now lst is sorted.
I have the following data in a guava Multiset. Each item is the combined string of 3 items separated by a ':'. I know all the values for each of the slots. I'm using the values to generate a data file for an interactive graph (by stuffing the split values into an object and then using Gson to print the object).
What's the best way to grab the cumulative count for all items that match just one, one:two, or one:two:three of the substrings? I keep going round and round with streams, forEach, maps and filters, but can't seem to write an elegant set of loops. Any suggestions or examples would be helpful.
Executive:Healthcare:United States x 5
Executive:Healthcare:Malaysia x 2
Executive:Financials:United States x 1
FinancialHealth:Technology:Malaysia x 3
FinancialHealth:Technology:United States x 2
FinancialHealth:Energy:United States x 1
Executive = 8
FinancialHealth = 6
Executive:Heathcare = 7
Executive:Financials = 1
FinancialHealth:Technology = 5
FinancialHealth:Energy = 1
Executive:Healthcare:United States = 5
etc.
Streams can help a great deal here, and it is not even difficult.
We need to take three steps in a stream:
allTheStrings.stream()
// First, we will multiply each string "A:B:C" using `flatMap`
// so that the stream contains "A", "A:B", and "A:B:C":
.flatMap(s -> Stream.of(s.substring(0, s.indexOf(":")),
s.substring(0, s.lastIndexOf(":")),
s))
// next, we are going to summarize multiple occurrences
// of the strings using a groupingBy collector:
.collect(Collectors.groupingBy(Function.identity(),
// This would return a Map<String, List<String>> containing each unique
// string mapped to its occurrences. But because you don't need the
// single occurrences, but instead just their number, we add a step
// to the collect which will make it return a Map<String, Long>
Collectors.counting()))
So, as a full example:
Stream.of("Executive:Healthcare:United States", "Executive:Healthcare:United States",
"Executive:Healthcare:United States", "Executive:Healthcare:United States",
"Executive:Healthcare:United States", "Executive:Healthcare:Malaysia",
"Executive:Healthcare:Malaysia", "Executive:Financials:United States",
"FinancialHealth:Technology:Malaysia", "FinancialHealth:Technology:Malaysia",
"FinancialHealth:Technology:Malaysia", "FinancialHealth:Technology:United States",
"FinancialHealth:Technology:United States", "FinancialHealth:Energy:United States")
.flatMap(s -> Stream.of(s.substring(0, s.indexOf(":")), s.substring(0, s.lastIndexOf(":")), s))
.collect(Collectors.groupingBy(Function.identity(), Collectors.counting()))
.entrySet()
.forEach(System.out::println);
will output
Executive=8
Executive:Healthcare=7
FinancialHealth:Technology=5
FinancialHealth=6
FinancialHealth:Energy=1
FinancialHealth:Technology:Malaysia=3
FinancialHealth:Energy:United States=1
Executive:Healthcare:United States=5
Executive:Financials:United States=1
FinancialHealth:Technology:United States=2
Executive:Healthcare:Malaysia=2
Executive:Financials=1
In Neo4j, I create a small graph with 4 nodes, some of which are linked to some others:
CREATE
(a:Room {name:"A"})
-[:DOOR]->
(b:Room {name:"B"})
-[:DOOR]->
(c:Room {name:"C"})
-[:DOOR]->
(d:Room {name:"D"}),
a-[:DOOR]->c,
a-[:DOOR]->d,
b-[:DOOR]->a
RETURN a,b,c,d
I want to find which rooms do not have a door between them. I'm hoping for an output something like this:
{"B": ["D"], "C": ["A", "B"], "D": ["A", "B", "C"]}
I can do this for one given starting point...
MATCH (b), (r)
WHERE b.name = "B"
AND NOT (b)-[:DOOR]->(r)
AND b <> r
RETURN r
// Returns Room D
Here's my cargo-cult pseudo code for iterating through each possible pair of nodes:
MATCH rooms = (r)
SET output = {}
FOREACH (
room IN nodes(rooms),
exit IN nodes(rooms),
missing = [],
output[room.name] = missing
|
IF room <> exit AND NOT room-[:DOOR]->(exit)
THEN missing = missing + exit
)
RETURN output
Please help me to understand how to formulate this correctly in Cypher.
The WHERE clause takes relationship patterns and you can use the NOT function to filter on the absence of a relationship.
MATCH (a:Room), (b:Room)
WHERE NOT a-[:DOOR]-b AND a <> b
RETURN a, b
Here's the section in the docs.
I have a code where I have the following frame and filter rows as follows:
let dfff=
[ "year" => series [ 1 => 1990.0; 2 => 1991.00; 3 => 1992.0; 4 => 1993.0]
"gold" => series [ 1 => 10.0; 2 => 10.00; 3 => 15.0; 4 => 20.0]
"silver" => series [ 1 => 20.0; 2 => 30.00; 3 => 45.0; 4 => 55.0] ]
|> frame
let dfff2 = dfff |> Frame.filterRows (fun key row -> row?year <= 1992.0 )
Why do I have to write key in
Frame.filterRows (fun key row -> row?year <= 1992.0)
if my function only depends on row? What role does key play here? I will appreciate if anybody could explain me the logic. Thanks!
In Deedle, frames have row keys and column keys. In your case, you have Frame<int, string> meaning that the row keys are integers (just numbers) and column keys are strings (column names) - but you might also have dates or other more interesting things as row keys.
The filterRows function gives you the row key together with row data. The key parameter is just the row key - in your case, this is (uninteresting) int index, but it might be e.g. useful date in other scenarios.
F# lets you write _ to explicitly ignore the value:
let dfff2 = dfff |> Frame.filterRows (fun _ row -> row?year <= 1992.0 )
In the Series module, we have Series.filter and Series.filterValues where the first one gives you key & value and the second one gives you just the value. So, we could follow the same pattern and add Frame.filterRowValues.
This would actually be quite easy, so if you want to contribute, please send a pull request with a change somewhere around here :-).