I've been learning TSQL and need some help with a conversion CPU MIPS into PERCENTAGE.
I've built my code to get some data that I'm expecting. In addition to this, I want to add a column to my code which is to get the CPU%. I have a column that gives me TOTALCPU MIPS and want to use this in the code but in the form of percentage. Example, I have these values in my TOTAL CPU Column:
1623453.66897
0
0
2148441.01573933
3048946.946314
I want to convert these values into percentage and use them. I couldn't find much info on the internet.
Appreciate your response.
I assume that you have 5 numeric quantities (2 of them being zero) and you want to find the percentage that corresponds to each of them out of the addition of the five quantities. Is it so?
To find the percentage of a particular number in the addition you multiply the number by 100 and divide by the addition, the result is the percentage that that number is in relation with the addition.
The sum: 6820841.631023
The percentage of the first number (of MIPS):
1623453.668970 * 100 / 6820841.631023 = 23.80136876 =>
23.80136876% is the percentage of CPU used by the first program.
To give the answer some SQL looking, refering to Mips_Table as the view/table that contains the MIPs data:
select mips, mips/totMips*100 Pct_CPU
from Mips_Table,
(select sum(mips) TotMips from Mips_Table) k
Related
Help with homework problem: "Let us define the "data science experience" of a given person as the person's largest score among Regression, Classification, and Clustering. Compute the average data science experience among all MSIS students."
Beginner to coding. I am trying to figure out how to check amongst columns and compare those columns to each other for the largest value. And then take the average of those found values.
I greatly appreciate your help in advance!
Picture of the sample data set: 1: https://i.stack.imgur.com/9OSjz.png
Provided Code:
import pandas as pd
df = pd.read_csv("cleaned_survey.csv", index_col=0)
df.drop(['ProgSkills','Languages','Expert'],axis=1,inplace=True)
Sample Data:
What I have tried so far:
df[data_science_experience]=df[["Regression","Classification","Clustering"]].values.max()
df['z']=df[['Regression','Classification','Clustering']].apply(np.max,axis=1)
df[data_science_experience]=df[["Regression","Classification","Clustering"]].apply(np.max,axis=1)
If you want to get the highest score of column 'hw1' you can get it with:
pd['hw1'].max(). this gives you a series of all the values in that column and max returns the maximum. for average use mean:
pd['hw1'].mean()
if you want to find the maximum of multiple columns, you can use:
maximum_list = list()
for col in pd.columns:
maximum_list.append(pd[col].max)
max = maximum_list.max()
avg = maximum_list.mean()
hope this helps.
First, you want to get only the rows with MSIS in the Program column. That can be done in the following way:
df[df['Program'] == 'MSIS']
Next, you want to get only the Regression, Classification and Clustering columns. The previous query filtered only rows; we can add to that, like this:
df.loc[df['Program'] == 'MSIS', ['Regression', 'Classification', 'Clustering']]
Now, for each row remaining, we want to take the maximum. That can be done by appending .max(axis=1) to the previous line (axis=1 because we want the maximum of each row, not each column).
At this point, we should have a DataFrame where each row represents the highest score of the three categories for each student. Now, all that's left to do is take the mean, which can be done with .mean(). The full code should therefore look like this:
df.loc[df['Program'] == 'MSIS', ['Regression', 'Classification', 'Clustering']].max(axis=1).mean()
For example - Column ABC number(12,4).
My value is 2.7487 and if divided by 2 it's 1.3744 (rounding with 4).
1.3744 + 1.3744 = 2.7488
How do I get a result - 2.7487 as the original.
You cannot. That's normal behavior of finite arithmetic (integer and floating point).
Even if you add large number of decimal places, you'll still lose precision.
This is a typical case of the rounding of an invoice. For example you have an invoice for a total of $100.00 that you need to divide in 3 items. You get three items of $100.00 / 3.0 each one. That is:
$33.33 Item #1.
$33.33 Item #2.
$33.33 Item #3.
For a total of... $99.99! (not the $100.00 that you expected).
The known solution for ages now is to adjust one or more of the values to add or decrease it by one cent. In this case, you could add 1 cent to the last item, to get:
$33.33 Item #1.
$33.33 Item #2.
$33.34 Item #3.
For a total of... $100.00! Perfect.
There are multiple possible combinations (all valid) to adjust those values. That's how it's done in accounting.
I have this TypeError as per below, I have checked my df and it all contains numbers only, can this be caused when I converted to numpy array? After the conversion the array has items like
[Timestamp('1993-02-11 00:00:00') 28.1216 28.3374 ...]
Any suggestion how to solve this, please?
df:
Date Open High Low Close Volume
9 1993-02-11 28.1216 28.3374 28.1216 28.2197 19500
10 1993-02-12 28.1804 28.1804 28.0038 28.0038 42500
11 1993-02-16 27.9253 27.9253 27.2581 27.2974 374800
12 1993-02-17 27.2974 27.3366 27.1796 27.2777 210900
X = np.array(df.drop(['High'], 1))
X = preprocessing.scale(X)
TypeError: float() argument must be a string or a number
While you're saying that your dataframe "all contains numbers only", you also note that the first column consists of datetime objects. The error is telling you that preprocessing.scale only wants to work with float values.
The real question, however, is what you expect to happen to begin with. preprocessing.scale centers values on the mean and normalizes the variance. This is such that measured quantities are all represented on roughly the same footing. Now, your first column tells you what dates your data correspond to, while the rest of the columns are numeric data themselves. Why would you want to normalize the dates? How would you normalize the dates?
Semantically speaking, I believe you should leave your dates alone. Whatever post-processing you're planning to perform on your numerical data, the normalized data should still be parameterized by the original dates. If you want to process your dates too, you need to come up with an explicit way to handle your dates to something numeric (say, elapsed time from a given date in given units).
So I believe you should drop your dates from your processing round altogether, and start with
X = df.drop(['Date','High'], 1).as_matrix()
I have two values for minimum 1D and maximum 1.5D for hardware engagement. I also have table of hardware length and number.
See snapshot:
I need a formula in which it will search for minimum hardware length and respective number which is available in table (in above problem two hardware’s are within range 0.250 and 0.313, I need minimum of two). Finally, display in output cell.
This may well need some adjustment (for example what would be the result desired from .3 and .375) but hopefully gives an indication. Assumes the columns of the main table are named Size and HNum and the inputs are in A1 and B1:
Hardware size:
=INDEX(Size,IF(MATCH(A1,Size,1)+1<MATCH(B1,Size,1),MATCH(A1,Size,1)+1,MATCH(A1,Size,1)))
Hardware number:
=INDEX(HNum,IF(MATCH(A1,Size,1)+1<MATCH(B1,Size,1),MATCH(A1,Size,1)+1,MATCH(A1,Size,1)))
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.