def test_friedman(self):
"""Test function friedman"""
df = pd.DataFrame({'DV': np.r_[x, y, z],
'Time': np.repeat(['A', 'B', 'C'], 100),
'Subject': np.tile(np.arange(100), 3)})
friedman(data=df, dv='DV', subject='Subject', within='Time')
summary = friedman(data=df, dv='DV', within='Time', subject='Subject')
# Compare with SciPy built-in function
from scipy import stats
Q, p = stats.friedmanchisquare(x, y, z)
assert np.isclose(Q, summary.at['Friedman', 'Q'])
assert np.isclose(p, summary.at['Friedman', 'p-unc'])
# Test with NaN
df.at[10, 'DV'] = np.nan
friedman(data=df, dv='DV', subject='Subject', within='Time')
def test_friedman(self):
"""Test function friedman"""
df = pd.DataFrame({
'white': {
0: 10,
1: 8,
2: 7,
3: 9,
4: 7,
5: 4,
6: 5,
7: 6,
8: 5,
9: 10,
10: 4,
11: 7
},
'red': {
0: 7,
1: 5,
2: 8,
3: 6,
4: 5,
5: 7,
6: 9,
7: 6,
8: 4,
9: 6,
10: 7,
11: 3
},
'rose': {
0: 8,
1: 5,
2: 6,
3: 4,
4: 7,
5: 5,
6: 3,
7: 7,
8: 6,
9: 4,
10: 4,
11: 3
}
})
# Compare R and SciPy
# >>> friedman.test(data)
Q, p = scipy.stats.friedmanchisquare(*df.to_numpy().T)
assert np.isclose(Q, 2)
assert np.isclose(p, 0.3678794)
# Wide-format
stats = friedman(df)
assert np.isclose(stats.at['Friedman', 'Q'], Q)
assert np.isclose(stats.at['Friedman', 'p-unc'], p)
assert np.isclose(stats.at['Friedman', 'ddof1'], 2)
# Long format
df_long = df.melt(ignore_index=False).reset_index()
stats = friedman(data=df_long,
dv="value",
within="variable",
subject="index")
assert np.isclose(stats.at['Friedman', 'Q'], Q)
assert np.isclose(stats.at['Friedman', 'p-unc'], p)
assert np.isclose(stats.at['Friedman', 'ddof1'], 2)
# Compare Kendall's W
# WARNING: I believe that the value in JASP is wrong (as of Oct 2021), because the W is
# calculated on the transposed dataset. Indeed, to get the correct W / Q / p, one must use:
# >>> library(DescTools)
# >>> KendallW(t(data), correct = T, test = T)
# Which gives the following output:
# Kendall chi - squared = 2, df = 2, subjects = 3, raters = 12, p - value = 0.3679
# alternative hypothesis: Wt is greater 0
# sample estimates: 0.08333333
assert np.isclose(stats.at['Friedman', 'W'], 0.08333333)
# Using the F-test method, which is more conservative
stats_f = friedman(df, method="f")
assert stats_f.at['Friedman', 'p-unc'] > stats.at['Friedman', 'p-unc']
def test_friedman(self):
"""Test function friedman"""
df = pd.DataFrame({
'DV': np.r_[x, y, z],
'Time': np.repeat(['A', 'B', 'C'], 100),
'Subject': np.tile(np.arange(100), 3)
})
summary = friedman(data=df,
dv='DV',
within='Time',
subject='Subject',
method='chisq')
friedman(data=df,
dv='DV',
within='Time',
subject='Subject',
method='f')
# Compare with SciPy built-in function
from scipy import stats
Q, p = stats.friedmanchisquare(x, y, z)
assert np.isclose(Q, summary.at['Friedman', 'Q'])
assert np.isclose(p, summary.at['Friedman', 'p-unc'])
# With Categorical
df['Time'] = df['Time'].astype('category')
df['Time'] = df['Time'].cat.add_categories("Unused")
summary = friedman(data=df,
dv='DV',
within='Time',
subject='Subject',
method='chisq')
friedman(data=df,
dv='DV',
within='Time',
subject='Subject',
method='f')
Q, p = stats.friedmanchisquare(x, y, z)
assert np.isclose(Q, summary.at['Friedman', 'Q'])
assert np.isclose(p, summary.at['Friedman', 'p-unc'])
# Test with NaN
df.at[10, 'DV'] = np.nan
friedman(data=df,
dv='DV',
subject='Subject',
within='Time',
method='chisq')
friedman(data=df,
dv='DV',
within='Time',
subject='Subject',
method='f')
# test Kendall's W
a = np.array([
0.13, 0.51, 0.93, 0.97, 0.24, 0.44, 0.91, 0.15, 0.04, 0.5, 0.6,
0.27, 0.37, 0.03, 0.74, 0.34
])
df = pd.DataFrame({
'DV': a,
'Time': np.repeat(['A', 'B', 'C', 'D'], 4),
'Subject': np.tile(np.arange(4), 4)
})
summary = friedman(data=df,
dv='DV',
subject='Subject',
within='Time',
method='chisq')
assert summary.at['Friedman', 'W'] == 0.325 # R synchrony::kendall.w