Np.vstack concatenates along the first dimension. Using your two example arrays: print(a.shape, b.shape) It's easier to understand what np.vstack, np.hstack and np.dstack* do by looking at the. However, I was of the impression that I understood these terms in the context of vstack and hstack just fine.įirst of all, a and b don't have a third axis so how would I stack them along ' the third axis' to begin with? Second of all, assuming a and b are representations of 2D-images, why do I end up with three 2D arrays in the result as opposed to two 2D-arrays 'in sequence'? So either I am really stupid and the meaning of this is obvious or I seem to have some misconception about the terms 'stacking', 'in sequence', 'depth wise' or 'along an axis'. This is a simple way to stack 2D arrays (images) into a single
Takes a sequence of arrays and stack them along the third axis Stack arrays in sequence depth wise (along third axis).
The documentation is rather sparse and just says: Print("the value that separates 0.I have some trouble understanding what numpy's dstack function is actually doing. #the value that separates 0.05 (thus, 5%) of the data from 95% #The p value obtained from the one sample t-test is significant Print("t-value=", tval, "p-value=", pval) Tval = (mean_after_med-mean_before_med)/s #Numpy's var function uses n as divisor as a default but when ddof is set to 1, it uses n-1 as divisor Print('Mean after medication:', mean_after_med) Print('Mean before medication:', mean_before_med) #H1: The medication lowers blood pressureīefore_med = np.array()Īfter_med = np.array()
Np stack python code#
This is my code (where I use var_med = np.var(med_diff, ddof=1) and s = np.sqrt(var_med/n): #H0: The medication has no effect on blood pressure Could somebody confirm this please?ĭisclaimer: I know that there are different modules to do a t-test or to calculate the std with numpy but the prof wants it somehow like this, Computer Linguists, sikes. However, I thought just taking the square root of the variance is enough, without dividing by n again. Presuming that var_med = np.var(med_diff, ddof=1) is correct, do I need to divide the variance by n again under the square root? In my case, the prof has written it like this: s = np.sqrt(var_med/n).Calculating var_med with np.var(med_diff, ddof=1), does it imply that med_diff is divided by n-1 (bcs of ddof=1) by numpy itself? Albeit the confusing documentation on this, I do believe it does that afaik (this is also what I assume for the third question).Is it right that I calculate the variance with the difference between both arrays (because my intuition tells me that I should use the mean difference, ergo mean_after_med - mean_before_med)?.Conduct a one-sided t-test to determine if the medication isĮffective at lowering blood pressure. The alternative hypothesis is that the medication lowers blood pressure. Medication has no effect on blood pressure. Want to know if the medication lowers blood pressure. Measure the blood pressure of 20 participants before and after taking the medication, and Suppose a researcher is studying the effect of a new medication on blood pressure. To calculate the variance and the means, then compare them according to the t-test. For both of these exercise, you get two lists of values that you want to compare.
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