I have a column of numbers in excel, with positives and negatives. It is an accounting book. I need to eliminate cells that sums to zero. It means I want to remove the subset, so the rest of element can not form any subset to sum to zero. I think this problem is to find the largest subset sum. By remove/eliminate, I mean to mark them in excel.
For example:
a set {1,-1,2,-2,3,-3,4,-4,5,-5,6,7,8,9},
I need a function that find subset {1,-1,2,-2,3,-3,4,-4,5,-5} and mark each element.
This suggestion may be a little heavy-handed, but it should be able to handle a broad class of problems -- like when one credit may be zeroed out by more than one debit (or vice versa) -- if that's what you want. Like you asked for, it will literally find the largest subset that sums to zero:
Enter your numbers in column A, say in the range A1:A14.
In column B, beside your numbers, enter 0 in each of the cells B1:B14. Eventually, these cells will be set to 1 if the corresponding number in column A is selected, or 0 if it isn't.
In cell C1, enter the formula =A1*B1. Copy the formula down to cells C2:C14.
At the bottom of column B, in cell B15, enter the formula =SUM(B1:B14). This formula calculates the count of your numbers that are selected.
At the bottom of column C, in cell C15, enter the formula =SUM(C1:C14). This formula calculates the sum of your numbers that are selected.
Activate the Solver Add-In and use it for the steps that follow.
Set the objective to maximize the value of cell $B$15 -- in other words, to maximize the count of your numbers that are selected (that is, to find the largest subset).
Set the following three constraints to require the values in cells B1:B14 (that indicate whether or not each of your numbers is selected) to be 0 or 1: a) $B$1:$B$14 >= 0, b) $B$1:$B$14 <= 1, and, c) $B$1:$B$14 = integer.
Set the following constraint to require the selected numbers to add up to 0: $C$15 = 0.
Use the Solver Add-In to solve the problem.
Hope this helps.
I think that you need to better define your problem because as it is currently stated there is no clear answer.
Here's why. Take this set of numbers:
{ -9, -5, -1, 6, 7, 10 }
There are 64 possible subsets - including the empty set - and of these three have zero sums:
{ -9, -1, 10 }, { -5, -1, 6 } & { }
There are two possible "biggest" zero-sum subsets.
If you remove either of these you end up with either of:
{ -5, 6, 7 } or { -9, 7, 10 }
Neither of these sum to zero, but there's no rule to determine which subset to pick.
You could decide to remove the "merged" set of zero sum subsets. This would leave you with:
{ 7 }
But does that make sense in your accounting package?
Equally you could just decide to eliminate only pairs of matching positive & negative numbers, but many transactions would involve triples (i.e. sale = cost + tax).
I'm not sure your question can be answered unless you describe your requirements more clearly.
Related
I have a data frame, lets say xyz. I have written code to find out the % of null values each column possess in the dataframe. my code below:
round(100*(xyz.isnull().sum()/len(xyz.index)), 2)
let say i got following results:
abc 26.63
def 36.58
ghi 78.46
I want to drop column ghi because it has more than 70% of null values.
I achieved it using the following code:
xyz = xyz.drop(xyz.loc[:,round(100*(xyz.isnull().sum()/len(xyz.index)), 2)>70].columns, 1)
but , i did not understand how does this code works, can anyone please explain it?
the code is doing the following:
xyz.drop( [...], 1)
removes the specified elements for a given axis, either by row or by column. In this particular case, df.drop( ..., 1) means you're dropping by axis 1, i.e, column
xyz.loc[:, ... ].columns
will return a list with the column names resulting from your slicing condition
round(100*(xyz.isnull().sum()/len(xyz.index)), 2)>70
this instruction is counting the number of nulls, adding them up and normalizing by the number of rows, effectively computing the percentage of nan in each column. Then, the amount is rounded to have only 2 decimal positions and finally you return True is the number of nan is more than 70%. Hence, you get a mapping between columns and a True/False array.
Putting everything together: you're first producing a Boolean array that marks which columns have more than 70% nan, then, using .loc you use Boolean indexing to look only at the columns you want to drop ( nan % > 70%), then using .columns you recover the name of such columns, which then are used by the .drop instruction.
Hopefully this clear things up!
If you code is hard to understand , you can just check dropna with thresh, since pandas already cover this case.
df=df.dropna(axis=1,thresh=round(len(df)*0.3))
I can't understand why to use two values inside the boundary when using Boundary Value Analysis.
For instance, the program has the requirement: 1) Values between 1 and 100 are true, otherwise false.
func calc(x):
if (x >= 1 and x <= 100):
return True
else:
return False
A lot of books (Pressman, for instance) say you have to use the inputs 0, 1, 2, 99, 100 and 101 to test such program.
So, my question is: Why does use the inputs '2' and '99'?
I try to make a program with a fault that the test case set (0, 1, 2, 99, 100 and 101) expose a fail and the test case set (0, 1, 100, 101) does not expose it.
I can't make such program.
Could you make such program?
If not, it is a waste of resource create redundant test cases '2' and '99'.
The basic requirement is to have +-1 of the boundary values. So to test values for a range of 1-100
One test case for exact boundary values of input domains each means
1 and 100.
One test case for just below boundary value of input domains each
means 0 and 99.
One test case for just above boundary values of input domains each
means 2 and 101.
To answer your question - Why does use the inputs '2' and '99'? It is because if you are following BVA, you are checking both the limits (upper as well as lower) of the range to ensure that the software is behaving correctly. However, there are no hard and fast rules. If the range is big enough, then you should have more test points. You can also test the middle values as part of BVA.
Also, you can use Switch Case statements to create a program or multiple Ifs.
There is an inconsistency with dataframes that I cant explain. In the following, I'm not looking for a workaround (already found one) but an explanation of what is going on under the hood and how it explains the output.
One of my colleagues which I talked into using python and pandas, has a dataframe "data" with 12,000 rows.
"data" has a column "length" that contains numbers from 0 to 20. she wants to divided the dateframe into groups by length range: 0 to 9 in group 1, 9 to 14 in group 2, 15 and more in group 3. her solution was to add another column, "group", and fill it with the appropriate values. she wrote the following code:
data['group'] = np.nan
mask = data['length'] < 10;
data['group'][mask] = 1;
mask2 = (data['length'] > 9) & (data['phraseLength'] < 15);
data['group'][mask2] = 2;
mask3 = data['length'] > 14;
data['group'][mask3] = 3;
This code is not good, of course. the reason it is not good is because you dont know in run time whether data['group'][mask3], for example, will be a view and thus actually change the dataframe, or it will be a copy and thus the dataframe would remain unchanged. It took me quit sometime to explain it to her, since she argued correctly that she is doing an assignment, not a selection, so the operation should always return a view.
But that was not the strange part. the part the even I couldn't understand is this:
After performing this set of operation, we verified that the assignment took place in two different ways:
By typing data in the console and examining the dataframe summary. It told us we had a few thousand of null values. The number of null values was the same as the size of mask3 so we assumed the last assignment was made on a copy and not on a view.
By typing data.group.value_counts(). That returned 3 values: 1,2 and 3 (surprise) we then typed data.group.value_counts.sum() and it summed up to 12,000!
So by method 2, the group column contained no null values and all the values we wanted it to have. But by method 1 - it didnt!
Can anyone explain this?
see docs here.
You dont' want to set values this way for exactly the reason you pointed; since you don't know if its a view, you don't know that you are actually changing the data. 0.13 will raise/warn that you are attempting to do this, but easiest/best to just access like:
data.loc[mask3,'group'] = 3
which will guarantee you inplace setitem
I basically need the answer to this SO question that provides a power-law distribution, translated to T-SQL for me.
I want to pull a last name, one at a time, from a census provided table of names. I want to get roughly the same distribution as occurs in the population. The table has 88,799 names ranked by frequency. "Smith" is rank 1 with 1.006% frequency, "Alderink" is rank 88,799 with frequency of 1.7 x 10^-6. "Sanders" is rank 75 with a frequency of 0.100%.
The curve doesn't have to fit precisely at all. Just give me about 1% "Smith" and about 1 in a million "Alderink"
Here's what I have so far.
SELECT [LastName]
FROM [LastNames] as LN
WHERE LN.[Rank] = ROUND(88799 * RAND(), 0)
But this of course yields a uniform distribution.
I promise I'll still be trying to figure this out myself by the time a smarter person responds.
Why settle for the power-law distribution when you can draw from the actual distribution ?
I suggest you alter the LastNames table to include a numeric column which would contain a numeric value representing the actual number of indivuduals with a name that is more common. You'll probably want a number on a smaller but proportional scale, say, maybe 10,000 for each percent of representation.
The list would then look something like:
(other than the 3 names mentioned in the question, I'm guessing about White, Johnson et al)
Smith 0
White 10,060
Johnson 19,123
Williams 28,456
...
Sanders 200,987
..
Alderink 999,997
And the name selection would be
SELECT TOP 1 [LastName]
FROM [LastNames] as LN
WHERE LN.[number_described_above] < ROUND(100000 * RAND(), 0)
ORDER BY [number_described_above] DESC
That's picking the first name which number does not exceed the [uniform distribution] random number. Note how the query, uses less than and ordering in desc-ending order; this will guaranty that the very first entry (Smith) gets picked. The alternative would be to start the series with Smith at 10,060 rather than zero and to discard the random draws smaller than this value.
Aside from the matter of boundary management (starting at zero rather than 10,060) mentioned above, this solution, along with the two other responses so far, are the same as the one suggested in dmckee's answer to the question referenced in this question. Essentially the idea is to use the CDF (Cumulative Distribution function).
Edit:
If you insist on using a mathematical function rather than the actual distribution, the following should provide a power law function which would somehow convey the "long tail" shape of the real distribution. You may wan to tweak the #PwrCoef value (which BTW needn't be a integer), essentially the bigger the coeficient, the more skewed to the beginning of the list the function is.
DECLARE #PwrCoef INT
SET #PwrCoef = 2
SELECT 88799 - ROUND(POWER(POWER(88799.0, #PwrCoef) * RAND(), 1.0/#PwrCoef), 0)
Notes:
- the extra ".0" in the function above are important to force SQL to perform float operations rather than integer operations.
- the reason why we subtract the power calculation from 88799 is that the calculation's distribution is such that the closer a number is closer to the end of our scale, the more likely it is to be drawn. The List of family names being sorted in the reverse order (most likely names first), we need this substraction.
Assuming a power of, say, 3 the query would then look something like
SELECT [LastName]
FROM [LastNames] as LN
WHERE LN.[Rank]
= 88799 - ROUND(POWER(POWER(88799.0, 3) * RAND(), 1.0/3), 0)
Which is the query from the question except for the last line.
Re-Edit:
In looking at the actual distribution, as apparent in the Census data, the curve is extremely steep and would require a very big power coefficient, which in turn would cause overflows and/or extreme rounding errors in the naive formula shown above.
A more sensible approach may be to operate in several tiers i.e. to perform an equal number of draws in each of the, say, three thirds (or four quarters or...) of the cumulative distribution; within each of these parts list, we would draw using a power law function, possibly with the same coeficient, but with different ranges.
For example
Assuming thirds, the list divides as follow:
First third = 425 names, from Smith to Alvarado
Second third = 6,277 names, from to Gainer
Last third = 82,097 names, from Frisby to the end
If we were to need, say, 1,000 names, we'd draw 334 from the top third of the list, 333 from the second third and 333 from the last third.
For each of the thirds we'd use a similar formula, maybe with a bigger power coeficient for the first third (were were are really interested in favoring the earlier names in the list, and also where the relative frequencies are more statistically relevant). The three selection queries could look like the following:
-- Random Drawing of a single Name in top third
-- Power Coef = 12
SELECT [LastName]
FROM [LastNames] as LN
WHERE LN.[Rank]
= 425 - ROUND(POWER(POWER(425.0, 12) * RAND(), 1.0/12), 0)
-- Second third; Power Coef = 7
...
WHERE LN.[Rank]
= (425 + 6277) - ROUND(POWER(POWER(6277.0, 7) * RAND(), 1.0/7), 0)
-- Bottom third; Power Coef = 4
...
WHERE LN.[Rank]
= (425 + 6277 + 82097) - ROUND(POWER(POWER(82097.0, 4) * RAND(), 1.0/4), 0)
Instead of storing the pdf as rank, store the CDF (the sum of all frequencies until that name, starting from Aldekirk).
Then modify your select to retrieve the first LN with rank greater than your formula result.
I read the question as "I need to get a stream of names which will mirror the frequency of last names from the 1990 US Census"
I might have read the question a bit differently than the other suggestions and although an answer has been accepted, and a very through answer it is, I will contribute my experience with the Census last names.
I had downloaded the same data from the 1990 census. My goal was to produce a large number of names to be submitted for search testing during performance testing of a medical record app. I inserted the last names and the percentage of frequency into a table. I added a column and filled it with a integer which was the product of the "total names required * frequency". The frequency data from the census did not add up to exactly 100% so my total number of names was also a bit short of the requirement. I was able to correct the number by selecting random names from the list and increasing their count until I had exactly the required number, the randomly added count never ammounted to more than .05% of the total of 10 million.
I generated 10 million random numbers in the range of 1 to 88799. With each random number I would pick that name from the list and decrement the counter for that name. My approach was to simulate dealing a deck of cards except my deck had many more distinct cards and a varing number of each card.
Do you store the actual frequencies with the ranks?
Converting the algebra from that accepted answer to MySQL is no bother, if you know what values to use for n. y would be what you currently have ROUND(88799 * RAND(), 0) and x0,x1 = 1,88799 I think, though I might misunderstand it. The only non-standard maths operator involved from a T-SQL perspective is ^ which is just POWER(x,y) == x^y.
I am trying to figure out a calculation I can perform in C# to determine the rows per column. Let's say I know I am going to have 3 columns and my record count is 46. I know that I can mod the results to get a remainder, but I would like something more efficient than what I have tried. So I know I will have 16 rows per column with a remainder of 14 for the last column, but what is the best way to loop through the resutls and keep counts.
Integer divsion will give you the number of complete rows (46 / 3 = 15). You then check the modulus to see if you have any leftover (46 Mod 3 = 1; yep, you have one column to put in a final extra row.)
To loop through, just check the modulus of the current record index (zero-based) with your column count. That modulus is the (zero-based) column index. If it equals 0, you start a new row.
But from your question, it sounds like you already got this far. So am I misunderstanding the question?