I initially learned that if I want to see if a cell has any contents to use if(A1<>"",.... But as I receive more and more assistance on SO, it seems most people use if(LEN(A1),... Is there a difference? Did I learn the wrong information? Should I ever opt for one over the other or just always use LEN from now on?
pretty much the same result. difference is:
LEN(A1) - checks if A1 has a length
A1<>"" - checks if A1 is not equal to "empty"
then there is a length of the formula itself (some prefer to save 1 extra character space):
A1<>"" has 6 characters compared to LEN(A1) 7 characters
the superiority of LEN comes when you need to check for character count like:
=IF(LEN(A1)=4, TRUE, FALSE)
eg. output TRUE only if A1 value has exactly 4 characters
Related
I am trying to right a script for a current project but I need it to note display the number in the tenth position.
For example,
A1 = 9
A2 = 3
=SUM(A1+A2)
Would be
12
How ever how can I get it to display only 2?
The numbers will change with each entry. Sometimes it could be 9+45= 54 however I would want it to display only 4. Not sure how I would go about making it take away the number only in the tenth position.
You probably want something like:
=MOD(A1+A2,10)
Though I believe OpenCalc will use the semicolon as a parameter delimiter.
You can trasnform number to string then take last number. Something like this:
C1 = SUM(A1 + A2)
RIGHT(TEXT(C1,0),1)
The result is
2
You are looking for the Units position, not the Tens. The Units is always the furthest right character so just extract it right from there. No need to convert to string first
=RIGHT(SUM(A1:A2),1)
My question is a continuation of this one. Basically, I have a table of words like so:
HAT18178_890909.098070313.1
HAT18178_890909.098070313.2
HAT18178_890909.143412462.1
HAT18178_890909.143412462.2
For my purposes, I do not need the terminal .1 or .2 for this set of names. I can manually write the following regex (using Python syntax):
r = re.compile('(.*\.\d+)\.\d+')
However, I cannot guarantee that my next set of names will have a similar structure where the final 2 characters will be discardable - it could be 3 characters (i.e. .12) and the separator could change as well (i.e. . to _).
What is the appropriate way to either explicitly learn a regex or to determine which characters are unnecessary?
It's an interesting problem.
X y
HAT18178_890909.098070313.1 HAT18178_890909.098070313
HAT18178_890909.098070313.2 HAT18178_890909.098070313
HAT18178_890909.143412462.1 HAT18178_890909.143412462
HAT18178_890909.143412462.2 HAT18178_890909.143412462
The problem is that there is not a single solution but many.
Even for a human it is not clear what the regex should be that you want.
Based on this data, I would think the possibilities to learn are:
Just match a fixed width of 25: .{25}
Fixed first part: HAT18178_890909.
Then:
There's only 2 varying numbers on each single spot (as you show 2 cases).
So e.g. [01] (either 0 or 1), [94] the next spot and so on would be a good solution.
The obvious one would be \d+
But it could also be \d{9}
You see, there are multiple correct answers.
These regexes would still work if the second point would be an underscore instead.
My conclusion:
The problem is that it is much more work to prepare the data for machine learning than it is to create a regex. If you want to be sure you cover everything, you need to have complete data, so then a regex is probably less effort.
You could split on non-alphanumeric characters;
[^a-zA-Z0-9']+
That would get you, in this case, few strings like this:
HAT18178
890909
098070313
1
From there on you can simply discard the last one if that's never necessary, and continue on processing the first sequences
I want to write a program that can give me all 4 letter words (from the dictionary or outside the dictionary). I code in C++. And by far, I've reached nowhere.
I'm simply a beginner in C++, I can apply the logic but I'm not introduced to advanced features in C++. It doesn't matter if it takes a long time for this program to end execution, I just want the solution.
For example:
abcd
king
ngik
cbda
play
lpay
payl
and so on (just a few of the millions of outputs I hope this program to output).
NOTE: The words generated need not make sense and I do not want to discard any combinations, I want it all.
I suggest looping say i from 0 to 26^4 - 1, each time outputting 'A' + i / (26*26*26), 'A' + i / (26*26) % 26, 'A' + i / 26 % 26, and 'A' + i % 26, then a newline.
Make a array that has all the possible letters in it(add numbers and symbols if you want).
Then use four nested for loops that loop a number from 0 to the length of the array.
Lets say that the loop number variables are a,b,c,d.
In the inner loop(the last one) you can output it as array[a] + array[b] + array[c] + array[d]
This gives all the possible combinations where you can add numbers and symbols aswell.
use recursion
explore DFS a 26-ary tree and output the letter every time you go down on the corresponding branch. depth = 4.
P.S. recursive algorithms eat stack memory like hell... so be sure your machine let programs have enough stack mem. You don't wanna run this on a pic micro with only 3-level call stack :-)
As of right now, I decided to take a dictionary and iterate through the entire thing. Every time I see a newline, I make a string containing from that newline to the next newline, then I do string.find() to see if that English word is somewhere in there. This takes a VERY long time, each word taking about 1/2-1/4 a second to verify.
It is working perfectly, but I need to check thousands of words a second. I can run several windows, which doesn't affect the speed (Multithreading), but it still only checks like 10 a second. (I need thousands)
I'm currently writing code to pre-compile a large array containing every word in the English language, which should speed it up a lot, but still not get the speed I want. There has to be a better way to do this.
The strings I'm checking will look like this:
"hithisisastringthatmustbechecked"
but most of them contained complete garbage, just random letters.
I can't check for impossible compinations of letters, because that string would be thrown out because of the 'tm', in between 'thatmust'.
You can speed up the search by employing the Knuth–Morris–Pratt (KMP) algorithm.
Go through every dictionary word, and build a search table for it. You need to do it only once. Now your search for individual words will proceed at faster pace, because the "false starts" will be eliminated.
There are a lot of strategies for doing this quickly.
Idea 1
Take the string you are searching and make a copy of each possible substring beginning at some column and continuing through the whole string. Then store each one in an array indexed by the letter it begins with. (If a letter is used twice store the longer substring.
So the array looks like this:
a - substr[0] = "astringthatmustbechecked"
b - substr[1] = "bechecked"
c - substr[2] = "checked"
d - substr[3] = "d"
e - substr[4] = "echecked"
f - substr[5] = null // since there is no 'f' in it
... and so forth
Then, for each word in the dictionary, search in the array element indicated by its first letter. This limits the amount of stuff that has to be searched. Plus you can't ever find a word beginning with, say 'r', anywhere before the first 'r' in the string. And some words won't even do a search if the letter isn't in there at all.
Idea 2
Expand upon that idea by noting the longest word in the dictionary and get rid of letters from those strings in the arrays that are longer than that distance away.
So you have this in the array:
a - substr[0] = "astringthatmustbechecked"
But if the longest word in the list is 5 letters, there is no need to keep any more than:
a - substr[0] = "astri"
If the letter is present several times you have to keep more letters. So this one has to keep the whole string because the "e" keeps showing up less than 5 letters apart.
e - substr[4] = "echecked"
You can expand upon this by using the longest words starting with any particular letter when condensing the strings.
Idea 3
This has nothing to do with 1 and 2. Its an idea that you could use instead.
You can turn the dictionary into a sort of regular expression stored in a linked data structure. It is possible to write the regular expression too and then apply it.
Assume these are the words in the dictionary:
arun
bob
bill
billy
body
jose
Build this sort of linked structure. (Its a binary tree, really, represented in such a way that I can explain how to use it.)
a -> r -> u -> n -> *
|
b -> i -> l -> l -> *
| | |
| o -> b -> * y -> *
| |
| d -> y -> *
|
j -> o -> s -> e -> *
The arrows denote a letter that has to follow another letter. So "r" has to be after an "a" or it can't match.
The lines going down denote an option. You have the "a or b or j" possible letters and then the "i or o" possible letters after the "b".
The regular expression looks sort of like: /(arun)|(b(ill(y+))|(o(b|dy)))|(jose)/ (though I might have slipped a paren). This gives the gist of creating it as a regex.
Once you build this structure, you apply it to your string starting at the first column. Try to run the match by checking for the alternatives and if one matches, more forward tentatively and try the letter after the arrow and its alternatives. If you reach the star/asterisk, it matches. If you run out of alternatives, including backtracking, you move to the next column.
This is a lot of work but can, sometimes, be handy.
Side note I built one of these some time back by writing a program that wrote the code that ran the algorithm directly instead of having code looking at the binary tree data structure.
Think of each set of vertical bar options being a switch statement against a particular character column and each arrow turning into a nesting. If there is only one option, you don't need a full switch statement, just an if.
That was some fast character matching and really handy for some reason that eludes me today.
How about a Bloom Filter?
A Bloom filter, conceived by Burton Howard Bloom in 1970 is a
space-efficient probabilistic data structure that is used to test
whether an element is a member of a set. False positive matches are
possible, but false negatives are not; i.e. a query returns either
"inside set (may be wrong)" or "definitely not in set". Elements can
be added to the set, but not removed (though this can be addressed
with a "counting" filter). The more elements that are added to the
set, the larger the probability of false positives.
The approach could work as follows: you create the set of words that you want to check against (this is done only once), and then you can quickly run the "in/not-in" check for every sub-string. If the outcome is "not-in", you are safe to continue (Bloom filters do not give false negatives). If the outcome is "in", you then run your more sophisticated check to confirm (Bloom filters can give false positives).
It is my understanding that some spell-checkers rely on bloom filters to quickly test whether your latest word belongs to the dictionary of known words.
This code was modified from How to split text without spaces into list of words?:
from math import log
words = open("english125k.txt").read().split()
wordcost = dict((k, log((i+1)*log(len(words)))) for i,k in enumerate(words))
maxword = max(len(x) for x in words)
def infer_spaces(s):
"""Uses dynamic programming to infer the location of spaces in a string
without spaces."""
# Find the best match for the i first characters, assuming cost has
# been built for the i-1 first characters.
# Returns a pair (match_cost, match_length).
def best_match(i):
candidates = enumerate(reversed(cost[max(0, i-maxword):i]))
return min((c + wordcost.get(s[i-k-1:i], 9e999), k+1) for k,c in candidates)
# Build the cost array.
cost = [0]
for i in range(1,len(s)+1):
c,k = best_match(i)
cost.append(c)
# Backtrack to recover the minimal-cost string.
costsum = 0
i = len(s)
while i>0:
c,k = best_match(i)
assert c == cost[i]
costsum += c
i -= k
return costsum
Using the same dictionary of that answer and testing your string outputs
>>> infer_spaces("hithisisastringthatmustbechecked")
294.99768817854056
The trick here is finding out what threshold you can use, keeping in mind that using smaller words makes the cost higher (if the algorithm can't find any usable word, it returns inf, since it would split everything to single-letter words).
In theory, I think you should be able to train a Markov model and use that to decide if a string is probably a sentence or probably garbage. There's another question about doing this to recognize words, not sentences: How do I determine if a random string sounds like English?
The only difference for training on sentences is that your probability tables will be a bit larger. In my experience, though, a modern desktop computer has more than enough RAM to handle Markov matrices unless you are training on the entire Library of Congress (which is unnecessary- even 5 or so books by different authors should be enough for very accurate classification).
Since your sentences are mashed together without clear word boundaries, it's a bit tricky, but the good news is that the Markov model doesn't care about words, just about what follows what. So, you can make it ignore spaces, by first stripping all spaces from your training data. If you were going to use Alice in Wonderland as your training text, the first paragraph would, perhaps, look like so:
alicewasbeginningtogetverytiredofsittingbyhersisteronthebankandofhavingnothingtodoonceortwiceshehadpeepedintothebookhersisterwasreadingbutithadnopicturesorconversationsinitandwhatistheuseofabookthoughtalicewithoutpicturesorconversation
It looks weird, but as far as a Markov model is concerned, it's a trivial difference from the classical implementation.
I see that you are concerned about time: Training may take a few minutes (assuming you have already compiled gold standard "sentences" and "random scrambled strings" texts). You only need to train once, you can easily save the "trained" model to disk and reuse it for subsequent runs by loading from disk, which may take a few seconds. Making a call on a string would take a trivially small number of floating point multiplications to get a probability, so after you finish training it, it should be very fast.
I have homework due that states that I need to write a program that generates the first 15 letters of the english alphabet. I can't delcare and set 15 different variables or constants. The letters must be displayed in a number of columns initially set by the user. the numbers have to be aligned in columns. Can anyone help? Maximum number of columns is 7 and the minimum is 1.
Here's some pseudo-code to get you started. Read it, understand it, then try to implement it.
get numcols from user
if numcols < 1 or numcols > 7:
print error and exit
ch = 'a'
for count = 1 to 15:
output ch followed by space
add 1 to ch
if count is an integral multiplier of numcols:
output newline
endif
endfor
if numcols is not equal to 3 or 5:
output newline
endif
It's pitched at about the level of your homework (no fancy stuff and the smallest hint of awkwardness) and should map reasonably well into C code.
As part of this implementation, you should research:
the fact that character constants like 'a' are really integers in disguise.
remainder or modulus (%) operators and how/why they are useful here.
getting user input with scanf.
putchar for outputting characters.
why you have that final if statement :-)
Here is a hint:
ASCII code of A is 65, B is 66, C is 67 and so on. You can do it in a loop starting from 65 and going on for 15 iterations.
This can be done with two nested loops, one for the vertical and one for the horizontal. since the numbers are in sequence in value you can increment the variable for the character each time.
I don't want to give away more than that unless another user says I should. I've already given a lot of help and I'm sure you can figure out the rest.
If you feel you need more help I'll try to not give too much but explain more.