Imagine you have the following sentences:
I am happy.
I am very cool.
I am very happy.
You're looking for a sentence that satisfies the following condition:
NOT(very OR happy)
It's clear that nor sentence 1 nor sentence 2 satisfies this condition. However, does sentence 3 satisfy this condition? Since it doesn't contain "very OR happy", but "very AND happy".
It is satisfying the condition 3 as well ie, it contains both. "very OR happy" is also true and "very AND happy" is also true.
In the logical statement if out of the two statements in OR condition the first is true and the next statement is not evaluated. However in case of AND if the first statement evaluates to false then the next statement is not evaluated.
The OR condition for a and b means at least one of them is true.
I am very happy.
Very = true, Happy = true
Put that into NOT(very OR happy) gives you:
NOT(true OR true)
Which is equal to:
NOT(true)
Which is equal to:
false
Related
I am new learner of RobotFramework, I tried to run this code, but this error shows: Keyword 'BuiltIn.Exit For Loop If' expected 1 argument, got 2. Thank you in advance!
Select the Card
[arguments] ${cardName}
${elements} = Get WebElements css:.card-title
${index}= Set Variable 1
FOR ${element} IN #{elements}
Exit For Loop If '${cardName}' == '${element.text}'
${index}= Evaluate ${index} + 1
END
You have two spaces after ==. Robot uses two or more spaces as argument separators so it sees '${cardName}' == as one argument and '${element.text}' as a second argument.
The solution is to make sure the entire expression doesn't have any sequences of two or more spaces.
Exit For Loop If '${cardName}' == '${element.text}'
I have a list of indexes and trying to populate a column 'Type' for these rows only.
What I tried to do:
index_list={1,5,9,10,13}
df.loc[index_list,'Type']=="gain+loss"
Output:
1 False
5 False
9 False
10 False
13 False
But the output just gives the list with all False instead of populating these rows.
Thanks for any advice.
You need to put a single equal instead of double equal. In python, and in most progamming languages, == is the comparison operator. In your case you need the assignment operator =.
So the following code will do what you want :
index_list={1,5,9,10,13}
df.loc[index_list,'Type']="gain+loss"
I have a table which has a boolean column, this column is used to filter some responses. I'm in the need to return a response as a tuple as {claimed, unclaimed} (Imagine the table is called winnings)
While working on it I've done two separate queries to return claimed then unclaimed rows and manually constructing the response, then I went with returning all rows without checking the boolean column and splitting it outside of the query. Now I'm wondering if there's a way I can run a single query on the same table and return both claimed and unclaimed as separate results mainly for performance hoping it runs better. I've tried doing it with joins but its returning a list of two items tuples like:
[{claimed, unclaimed}, {claimed, unclaimed}]...
While I want:
{claimed, unclaimed}
# OR
[{claimed, unclaimed}]
At most, no more tuples. Note that I'm not running the raw queries but using a library so excuse if the terminology is not right.
This is the last query I ran:
SELECT w0."claimed", w1."claimed"
FROM "winnings" AS w0
INNER JOIN "winnings" AS w1 ON TRUE
WHERE (w0."claimed" AND NOT (w1."claimed"))
LIMIT 10;
EDIT: More details.
When I run the query from above this is the result I get:
=> SELECT w0."claimed", w1."claimed" FROM "winnings" AS w0 INNER JOIN "winnings" AS w1 ON TRUE WHERE (w0."claimed" AND NOT (w1."claimed")) LIMIT 10;
claimed | claimed
---------+---------
t | f
t | f
t | f
t | f
t | f
t | f
t | f
t | f
t | f
t | f
(10 rows)
This is converted to the following on Elixir which is the language I'm using:
[
true: false,
true: false,
true: false,
true: false,
true: false,
true: false,
true: false,
true: false,
true: false,
true: false
]
This is a keyword list which internally is a list of tuples as [{true, false}, {true, false}] - I want: [{[true, true], [false, false]}]
Means that I want 2 lists, each list with their respective rows, only claimed on one and only unclaimed on the other one.
I don't really mind the type it outputs as long as it includes two lists with their rows how I said.
To get the first column from a list of rows, you can use Enum.map/2 to get the first element of each tuple:
Enum.map(rows, &elem(&1, 0))
If you're a newcomer to elixir, the & syntax may be a bit confusing. That code is shorthand for
Enum.map(rows, fn field -> elem(field, 0) end)
You could make that into a function that does that for all your columns like this:
def columnize(rows = [first_row | _]) when is_tuple(first_row) do
for column <- 1..tuple_size(first_row), do: Enum.map(rows, &elem(&1, column - 1))
end
def columnize([]) do
[]
end
hd/1 is a function used to get the first tuple in the list. The = [first_row | _] part guarantees the argument is a list with at least one element; the when is_tuple(first_row) assures at least the first row is a tuple.
If you'd like to know more about the for, it's a comprehension.
Note that this function assumes all rows have the same number of columns and are tuples (which should be true in the results of a query; but may not be in other cases), and will throw an ArgumentError if the first row is wider than any other row. It will also cause errors if other elements of the list are not tuples.
This feels very unidiomatic though. Is this maybe an XY problem?
I would like to add a column based on another column and fill it with all the values that do NOT contain "jpg"
so the negation of this:
filter(value.split(","), v, v.contains("jpg")).join("|")
How can I write "does not contain"?
contains gives a boolean output i.e. true or false. So we have:
v = "picture.jpg" -> v.contains("jpg") = TRUE
v = "picture.gif" -> v.contains("jpg") = FALSE
filter finds all values in an array which return TRUE for whatever condition you use in the filter. There are a couple of ways you could filter an array to find the values that don't contain a string, but using contains the simplest is probably to use not to reverse the result of your condition:
filter(value.split(","), v, not(v.contains("jpg"))).join("|")
Recently I found an error which was due to the way I constructed an IF OR statement.
The original statement:
If a = 1 OR 2 OR 3 Then
`Execute code 1`
`Else if a = 4 OR 5 OR 6 Then`
`Execute code 2`
`Else`
`Do nothing`
The corrected code:
If a = 1 OR a = 2 OR a = 3 Then
`Execute code 1`
`Else if a = 4 OR a = 5 OR a = 6 Then`
`Execute code 2`
`Else`
`Do nothing`
So, the issue was the first part of the IF statement would always evaluate true regardless of the value of a, because it also evaluates the boolean value of the number 2, which is true. So simply spelling out that each number should be compared to the value of a fixes that issue.
My questions are:
1. When using an if statement and you have a comparative operation ( if a =1) followed by "OR 2", in what situation would you actually want to look at the boolean value of the number versus comparing the number against the value of the previously referenced variable?
Is there a way to identify if statements present in the code which will logically ALWAYS be true (and probably have a error)?
Edit: the_lotus pointed out that the boolean value of each number was not being evaluated but bitwise operations were performed between all the values and then a boolean evaluation was performed on the result. In this particular case it may be possible for it to evaluate either true or false depending on the value of a, though I am still interested in identifying if statements which always evaluate true.
As per your Edit, since you want to check if the value of A falls within a range, what you could do is store the ranges as an array and perform IndexOf to see if the value of A is in the collection.
Here is a quick example:
Dim a As Integer = 4
Dim collection1() As Integer = {1, 2, 3}
Dim collection2() As Integer = {4, 5, 6}
If Array.IndexOf(collection1, a) > -1 Then
Console.WriteLine("a = 1, 2, or 3")
ElseIf Array.IndexOf(collection2, a) > -1 Then
Console.WriteLine("a = 4, 5, or 6")
Else
Console.WriteLine("a is not a value between 1-6")
End If
Fiddle: Live Demo