Hi. I am getting issues with the ICC function. It works nicely with the example dataset data = pg.read_dataset('icc')but when I use my data I get the error: AssertionError: Data must have at least 5 non-missing values. I cannot see though any difference in the input data type. What am I doing wrong?
Code example:
n_val = 100
val_a = np.random.rand(1, n_val)
val_b = val_a + 0.6*np.random.rand(1, n_val)
data_icc = pd.DataFrame({'test_run': ['test_a']*n_val + ['test_b']*n_val,
'efr_value': np.concatenate((val_a[0], val_b[0]), axis=0),
'rater': ['A']*2*n_val})
icc = pg.intraclass_corr(data=data_icc, targets='test_run', raters='rater', ratings='efr_value').round(3)You can see an example of the data next:
val_a) and another one for day 2 (retest, or val_b). So, in reality, there are no raters, but just two sets of measurements. Is there any workaround in your function for raters k = 1? Or do you have any idea how to run your ICC for just 2 arrays?
See below (for everyone) how I modified my example code:
n_val = 100
val_a = np.random.rand(1, n_val)
val_b = val_a + 0.6*np.random.rand(1, n_val)
data_icc = pd.DataFrame({'targets': np.concatenate((np.arange(1, n_val+1, 1), np.arange(1, n_val+1, 1)), axis=0),
'efr_value': np.concatenate((val_a[0], val_b[0]), axis=0),
'test_session': ['test_a']*n_val + ['test_b']*n_val})
#Pearson
print('Pearson')
print(pg.corr(x=val_a[0], y=val_b[0], method='pearson'))
#ICC
print('\n\nICC')
print(pg.intraclass_corr(data=data_icc, targets='targets', raters='test_session', ratings='efr_value').round(3))
I'm not sure I understand pairwise_ttests(). See notebook:
https://github.com/FlorinAndrei/soda_pop_coke/blob/main/soda_pop_coke.ipynb
Function call is in cell #27. Dataframe (long form) is in cell #26 (and values are squished with sqrt()). The original wide form is in cell #24. In the wide form, each entry in the 'points' column is a unique pair of lat/long coordinates; the other three columns are distances from each point to three different centroids.
I want to show that, e.g. for the "soda" points, the shortest mean distance is to the "East/West Coasts" centroids.
Why do I get Paired=False from pairwise_ttests()? Does that mean it only compares population means as a group?
My understanding is that, with subject='points', it would somehow figure out that each lat/long pair is repeated 3 times in the 'points' column, and would use that information to do a more in-depth comparison, as opposed to plain group means comparison.
In cell #23 I did confidence intervals for means, each group separately, using compute_bootci(), and that's fine, but I want a more refined test than that.
Maybe I don't truly understand the theory behind pairwise t-tests.
If you clone the repo, it has all the files you need to run the notebook.
https://github.com/FlorinAndrei/soda_pop_coke
Thanks!
It's not clear from the docs how pg.ttest() works for a single sample test. It always requires a second variable y.
I just want to imitate the t.test() function from R, which can be used with a single sample very easily:
t.test(var)
How to get the exact same results from pg?
1 reply
It's not clear to me how to do the one-sample Z test for a proportion. In R, that would be:
prop.test(successes, total, p=ratio)statsmodels provides something similar:
proportions_ztest(successes, total, value=ratio)But the output is poor compared to the R function - there's no confidence interval.
> prop.test(1200, 2500, p=0.44, correct = F)
1-sample proportions test without continuity correction
data: 1200 out of 2500, null probability 0.44
X-squared = 16.234, df = 1, p-value = 5.599e-05
alternative hypothesis: true p is not equal to 0.44
95 percent confidence interval:
0.4604617 0.4995996
sample estimates:
p
0.48> exam = matrix(c(94, 113, 31, 62), nrow = 2)
> stats::prop.test(exam, correct = F)
2-sample test for equality of proportions without continuity correction
data: exam
X-squared = 3.8509, df = 1, p-value = 0.04972
alternative hypothesis: two.sided
95 percent confidence interval:
0.002588801 0.209982628
sample estimates:
prop 1 prop 2
0.7520000 0.6457143pg.linear_regression() seems to return predicted values only when as_dataframe=False. What is the reason?
as_dataframe=True (default), Pingouin only returns a Pandas dataframe with the coefficients, CI, p-values, etc. It is easier to return values of different format/size using a dictionary than a Pandas dataframe, and that's precisely why I've included the as_dataframe=False option for users who may need more exhaustivity in the output
pg.mixed_anova(dv='value', within=['variable_a', 'variable_b'], between = 'group', subject='subject', data=df, effsize="ng2") returns an dtype error. is this me being an idiot or is it a limitation? i couldnt find a direct mention of it anywhere so thought id check before deep diving
p-val column)