Right now, I see there are quick ways to get things like Sum/Avg/Max/Etc. for two or more rows or columns when building a table in GoodData.
quick total options
I am building a little table that shows last week and the week prior, and I'm trying to show the delta between them.
So if the first column is 100 and the second is 50, I want '-50'
If the first column is 25 and the second is 100, i want '75'
Is there an easy way to do this?
Let’s consider, that the first column contains result of calculating of metric #1 and the second column contains result of calculating of metric #2, you can simply create a metric #3, which would be defined as the (metric #1 - metric #2) or vice versa.
Related
total beginner here. If my question is irrelevant, apologies in advance, I'll remove it. So, I have a question : using pandas, I want to calculate an evolution ratio for a week data compared with the previous rolling 4 weeks mean data.
df['rolling_mean_fourweeks'] = df.rolling(4).mean().round(decimals=1)
from here I wanna create a new column for the evolution ratio based on the week data compared with the row of the rolling mean at the previous week.
what is the best way to go here? (I don't have big data) I have tried unsuccessfully with .shift() but am very foreign to .shift()... I should get NAN for week 3 (fourth week) and ~47% for fifth week.
Any suggestion for retrieving the value at row with step -1?
Thanks and have a good day!
Your idea about using shift can perfectly work. The shift(x) function simply shifts a series (a full column in your case) of x steps.
A simple way to check if the rolling_mean_fourweeks is a good predictor can be to shift Column1 and then check how it differs from rolling_mean_fourweeks:
df['column1_shifted'] = df['Column1'].shift(-1)
df['rolling_accuracy'] = ((df['column1_shifted']-df['rolling_mean_fourweeks'])
/df['rolling_mean_fourweeks'])
resulting in:
Lets say i have Dataframe, which has 200 values, prices for products. I want to run some operation on this dataframe, like calculate average price for last 10 prices.
The way i understand it, right now pandas will go through every single row and calculate average for each row. Ie first 9 rows will be Nan, then from 10-200, it would calculate average for each row.
My issue is that i need to do a lot of these calculations and performance is an issue. For that reason, i would want to run the average only on say on last 10 values (dont need more) from all values, while i want to keep those values in the dataframe. Ie i dont want to get rid of those values or create new Dataframe.
I just essentially want to do calculation on less data, so it is faster.
Is something like that possible? Hopefully the question is clear.
Building off Chicodelarose's answer, you can achieve this in a more "pandas-like" syntax.
Defining your df as follows, we get 200 prices up to within [0, 1000).
df = pd.DataFrame((np.random.rand(200) * 1000.).round(decimals=2), columns=["price"])
The bit you're looking for, though, would the following:
def add10(n: float) -> float:
"""An exceptionally simple function to demonstrate you can set
values, too.
"""
return n + 10
df["price"].iloc[-12:] = df["price"].iloc[-12:].apply(add10)
Of course, you can also use these selections to return something else without setting values, too.
>>> df["price"].iloc[-12:].mean().round(decimals=2)
309.63 # this will, of course, be different as we're using random numbers
The primary justification for this approach lies in the use of pandas tooling. Say you want to operate over a subset of your data with multiple columns, you simply need to adjust your .apply(...) to contain an axis parameter, as follows: .apply(fn, axis=1).
This becomes much more readable the longer you spend in pandas. 🙂
Given a dataframe like the following:
Price
0 197.45
1 59.30
2 131.63
3 127.22
4 35.22
.. ...
195 73.05
196 47.73
197 107.58
198 162.31
199 195.02
[200 rows x 1 columns]
Call the following to obtain the mean over the last n rows of the dataframe:
def mean_over_n_last_rows(df, n, colname):
return df.iloc[-n:][colname].mean().round(decimals=2)
print(mean_over_n_last_rows(df, 2, "Price"))
Output:
178.67
I have a companies dataset with 35 columns. The companies can belong to one of 8 different groups. How do I for each group create a new dataframe which subtract the mean of the column for that group away from the original value?
Here is an example of part of the dataset.
So for example for row 1 I want to subtract the mean of BANK_AND_DEP for Consumer Markets away from the value of 7204.400207. I need to do this for each column.
I assume this is some kind of combination of a transform and a lambda - but cannot hit the syntax.
Although it might seem counter-intuitive for this to involve a loop at all, looping through the columns themselves allows you to do this as a vectorized operation, which will be quicker than .apply(). For what to subtract by, you'll combine .groupby() and .transform() to get the value you need to subtract from a column. Then, just subtract it.
for column in df.columns:
df['new_'+column] = df[column]-df.groupby('Cluster')['column'].transform('mean')
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()
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.