I am trying to use the formula below to set conditions in LibreOffice but I keep getting an error. What am I doing wrong with the statement below:
=IF(G2<=2,'negative',IF(OR(G2>2 & G2<=3,'neutral',IF(OR(G2>=4,'positive))))))
Thanks
It seems, that in your formula is missing the last ':
'positive))))))
should be 'positive'))))))
Also the
&
is string-concatenation in LibreOffice, so you need here the equivalent to OR() and that is AND().
But you can simplify your formula to
=IF(G2<=2,'negative',IF(AND(G2>2,G2<=3),'neutral','positive'))
The first test is if the number is lower than 2 (negative),
the second test is if the number is between 2 and 3 (neutral)
and then there is no further test needed as it is the only remainig possiblity.
For a different locale, a slightly shorter, and I'd say simpler, version that also avoids the need for OR/AND:
=IF(G2<=2,"negative",IF(G2<=3,"neutral","positive"))
Once <=2 first test is handled (either but outputting negative or by proceeding to the 'result if FALSE') there is no longer the possibility of 2 or less, so the AND is not necessary.
The above though does fill a gap left by OP between 3 and 4.
Related
In the Running Total Fields, how do you set up a Distinct Count that includes blank values as one of your conditions?
Just found a solution. It's:
isnull({table.column})
Before asking, that's what I tried, but it wasn't working. So for people like me who tried that and it didn't work, that's because you have multiple conditions in your Running Total, and for whatever reason it only works when you edit your syntax and place that near the top of your conditions instead of the bottom. Don't know the reason, but it's working now.
I'd recommend replacing the field you're currently totaling with a new Formula. Something that doesn't ever come up blank, like:
If ISNULL {yourfieldhere} THEN "Blank" ELSE {yourfieldhere}
or if it's an empty string:
If {yourfieldhere}="" THEN "Blank" ELSE {yourfieldhere}
You can replace "Blank" with whatever suits you, even just an empty " " space or 0. But then it's at least something distinct to be counted.
I am doing binary classification using vowpal-wabbit. A particular record(set of features) has 10 zeroes and 5 ones. So, I am creating two lines in vowpal-format
-1 10 `50 |f f1
1 5 `50 |f f1
Since the prediction(probability) for both these records would be same, I want to keep the same tag, so that I can dedupe the predictions({tag,prediction}) later and join with my original raw-data.
Is it possible to keep the same tag for more than one record in vowpal-wabbit?
First, the syntax above isn't correct
To be identified as such, tags should either:
Touch the | separator (no space between them) OR
The leading quote, needs to be a simple quote, not a backquote, by convention.
(or both).
Otherwise you get:
warning: `50 is not a good float, replacing with 0
warning: `50 is not a good float, replacing with 0
Which hints that vw interprets these "tags" as prediction-base.
For details, see Input format in the official documentation
Once the example is fixed to the correct syntax:
-1 10 '50|f f1
1 5 '50|f f1
Which runs fine, we can answer the question:
Is it possible to keep the same tag for more than one record in vowpal-wabbit?
Yes, you can. The tag is merely a simple way to connect input and output (when predictions are involved), there's no check for uniqueness anywhere. If you duplicate tags on input, you'll simply get the same duplicate tags on prediction output as well.
More notes:
Even if two examples are identical, you may get different predictions, if the model has changed somewhat in between them. Remember vw is an online learner, so the model can continuously change with each example unless you add the -t (test-only, don't learn) option.
Features whose value is zero are ignored, so you can drop them. The standard way in vw to say this is 'positive' and this is 'negative' is to use the values {+1, -1}. This is true for both labels and input features.
I know that there are hash functions that from a variable length input can give a fixed output. To take the simplest one, using the module of ten no matter how big is the input number I will always get an output between 0 and 9.
I need to do have from an unknown password, a variable length output. My first thought was to use the module, increasing the prim number as much as many digits I need to have as output.
My problems are:
I must handle short passwords as well as I would with long passwords;
I don't know how long should the output be before writing the program, and even though I would know after the user has set the password I may need to change it if he modifies the file.
My first thought was using a simple function and modify it based on my needs.
If I have to hash 123 but I need to have 5 characters as output, that's what I would do:
I add 2 zeros on the right, changing the input to 12300;
I take the lowest 5 digits prime number (10007);
And I then I have my hash doing 12300 % 10007 = 02293.
But since I would probably need output in the order of hundreds if not thousands I'm pretty sure module is not the solution to my problem.
I could also try to create my own hash function, but I have no idea how to verify if it works or if it's trash.
Are there some common solutions in literature for this kind of problem?
I am posting for a sanity check so please forgive me if this sounds a little basic. I am trying to learn more about encryption so I figured a good starter project would be implement the Sha-1 hashing algorithm. I found a walk-through and have hit a point where I do not know if the walk-through is wrong or my understanding of bitness/rotation/binary operations is wrong.
From the document:
Step 11.2: Put them together
After completing one of the four functions above, each variable will move on to this step before restarting the loop with the next
word. For this step we are going to create a new variable called
'temp' and set it equal to: (A left rotate 5) + F + E + K + (the
current word).
Notice that other than the left rotate the only operation we're doing is basic addition. Addition in binary is about as simple as it
can be.
We'll use the results from the last word(79) as an example for this step.
A lrot 5:
00110001000100010000101101110100
F:
10001011110000011101111100100001
A lrot 5 + F
Out:
110111100110100101110101010010101
Notice that the result of this operation is one bit longer than the two inputs. This is just like adding 5 and 6, you will need a new
place value to represent the answer. For everything to work out
properly we will need to truncate that extra bit eventually. However,
we do not want to do that until the end!
This does not quite work out. What I think would happen is:
A = 00110001000100010000101101110100
F = 10001011110000011101111100100001
A Left rotate 5 = 00100010001000010110111010000110
(A Left Rotate 5) + F = 10101101111000110100110110100111 (which is still 32 bits)
What I need is just another set of eyes on this to say "Yes krtzer, you are correct and this document is wrong" Or "Your understanding of bitness, endianness, and/or bit rotation is wrong, this is how it works".
Right now I am not sure if my integer representation is wrong (the spec says use U32s, but this section says that I need to keep track of the extra bits), the endianness of my program is messing up the rotation (I use little endian) or there is something else.
Any experience or insight will be appreciated!
You are correct in your understanding of how everything works. The problem was with the article (which I wrote). An extra digit must always be added in the beginning of step 11.2, whether it's necessary or not, and if it's not necessary, it should be set to 1.
The article now reads:
Notice that the result of this operation is one bit longer than the two inputs. After each iteration the new word should be one bit longer than the last. Sometimes this will be a necessary carrier bit (like the extra place value you need to represent the result of adding the two single digit numbers 5 and 6 in base 10), and when that's not needed you must simply prepend a 1. For everything to work out properly we will need to truncate that extra bit eventually; however, we do not want to do that until the end!
The article was also unclear about the fact that an already rotated A was being shown in the example.
My teacher says our homework program must handle "an arbitrary number of input lines". It seems pretty arbitrary to only accept one line, but is it arbitrary enough? My roommate said seven is more a arbitrary number than one, and maybe he's right. But I just have no idea how to measure the arbitrariness of a number and Google doesn't seem to help.
UPDATE:
It sounds like maybe the best thing to do is accepty any given number of input lines, and hope the prof can see that that makes a lot more sense than insisting that the user just give you one specific arbitrary number of input lines. Especially since we weren't instructed to notify the user about what the arbitrary number is. You can't just make the user guess, that's crazy.
"Arbitrary" doesn't mean you get to pick a random number to accept. It means that it should handle an input with any number of lines.
So if someone decides to give your program an input with 0 lines, 1 line, 2 lines... n lines, then it should still do the right thing (and not crash).
Arbitrary means it could be ANY number. 0, 1, 7, 100124453225.
I would probably test for 0 and display some sort of error in that case since it's supposed to have SOME text. Other than that so long as there are more lines your program should keep doing whatever it's designed to do.
Typically when teachers indicate that a program should accept arbitrary amounts of input they are indicating to you that you should consider corner cases which you may not have thought about, one of the most common being no input at all which can often cause errors in programs if the programmer hasn't considered this case.
The point of the word is to emphasize that your program should be able to handle different inputs instead of simply crashing unless input comes in a certain quantity or is formatted in a specific way.