def test_cochransq():
# example from dataplot docs, Conovover p. 253
# http://www.itl.nist.gov/div898/software/dataplot/refman1/auxillar/cochran.htm
x = np.array(
[
[1, 1, 1],
[1, 1, 1],
[0, 1, 0],
[1, 1, 0],
[0, 0, 0],
[1, 1, 1],
[1, 1, 1],
[1, 1, 0],
[0, 0, 1],
[0, 1, 0],
[1, 1, 1],
[1, 1, 1],
]
)
res_qstat = 2.8
res_pvalue = 0.246597
assert_almost_equal(cochrans_q(x), [res_qstat, res_pvalue])
# equivalence of mcnemar and cochranq for 2 samples
a, b = x[:, :2].T
assert_almost_equal(mcnemar(a, b, exact=False, correction=False), cochrans_q(x[:, :2]))
def test_cochransq():
#example from dataplot docs, Conovover p. 253
#http://www.itl.nist.gov/div898/software/dataplot/refman1/auxillar/cochran.htm
x = np.array([[1, 1, 1], [1, 1, 1], [0, 1, 0], [1, 1, 0], [0, 0, 0],
[1, 1, 1], [1, 1, 1], [1, 1, 0], [0, 0, 1], [0, 1, 0],
[1, 1, 1], [1, 1, 1]])
res_qstat = 2.8
res_pvalue = 0.246597
assert_almost_equal(cochrans_q(x), [res_qstat, res_pvalue])
#equivalence of mcnemar and cochranq for 2 samples
a, b = x[:, :2].T
assert_almost_equal(mcnemar(a, b, exact=False, correction=False),
cochrans_q(x[:, :2]))
def test_cochransq3():
# another example compared to SAS
# in frequency weight format
dt = [('A', 'S1'), ('B', 'S1'), ('C', 'S1'), ('count', int)]
dta = np.array([('F', 'F', 'F', 6),
('U', 'F', 'F', 2),
('F', 'F', 'U', 16),
('U', 'F', 'U', 4),
('F', 'U', 'F', 2),
('U', 'U', 'F', 6),
('F', 'U', 'U', 4),
('U', 'U', 'U', 6)], dt)
cases = np.array([[0, 0, 0],
[1, 0, 0],
[0, 0, 1],
[1, 0, 1],
[0, 1, 0],
[1, 1, 0],
[0, 1, 1],
[1, 1, 1]])
count = np.array([ 6, 2, 16, 4, 2, 6, 4, 6])
data = np.repeat(cases, count, 0)
res = cochrans_q(data)
assert_allclose(res, [8.4706, 0.0145], atol=5e-5)
def cochranQ():
"""Cochran's Q test: 12 subjects are asked to perform 3 tasks. The outcome of each task is "success" or
"failure". The results are coded 0 for failure and 1 for success. In the example, subject 1 was successful
in task 2, but failed tasks 1 and 3.
Is there a difference between the performance on the three tasks?
"""
tasks = np.array(
[
[0, 1, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0],
[1, 1, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1],
[0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0],
]
)
# I prefer a DataFrame here, as it indicates directly what the values mean
df = pd.DataFrame(tasks.T, columns=["Task1", "Task2", "Task3"])
# --- >>> START stats <<< ---
(Q, pVal) = cochrans_q(df)
# --- >>> STOP stats <<< ---
print("\nCOCHRAN'S Q -----------------------------------------------------")
print("Q = {0:5.3f}, p = {1:5.3f}".format(Q, pVal))
if pVal < 0.05:
print("There is a significant difference between the three tasks.")
def core(x):
'''
x: pd.DataFrame()
'''
n = len(x)
freq = x.apply(lambda a: a.value_counts()).T
perc = freq.apply(lambda a: a / n)
f1 = freq.T
f1 = f1.reset_index()
f1.index = ['频数', '频数']
f1 = f1.reset_index().set_index(['level_0', 'index'])
f2 = perc.T
f2 = f2.reset_index()
f2.index = ['百分比', '百分比']
f2 = f2.reset_index().set_index(['level_0', 'index'])
f = f1.append(f2).T
f.columns.names = [None, None]
z, p = cochrans_q(x)
df = pd.Series({
'样本量': n,
'CochransQ 统计量': z,
'p': p,
'df': x.shape[1] - 1
})
df = df.to_frame().T.set_index('样本量')
res = {'频数分析结果': f, 'CochranQ检验结果': df}
return res
def test_cochransq3():
# another example compared to SAS
# in frequency weight format
dt = [('A', 'S1'), ('B', 'S1'), ('C', 'S1'), ('count', int)]
dta = np.array([('F', 'F', 'F', 6),
('U', 'F', 'F', 2),
('F', 'F', 'U', 16),
('U', 'F', 'U', 4),
('F', 'U', 'F', 2),
('U', 'U', 'F', 6),
('F', 'U', 'U', 4),
('U', 'U', 'U', 6)], dt)
cases = np.array([[0, 0, 0],
[1, 0, 0],
[0, 0, 1],
[1, 0, 1],
[0, 1, 0],
[1, 1, 0],
[0, 1, 1],
[1, 1, 1]])
count = np.array([ 6, 2, 16, 4, 2, 6, 4, 6])
data = np.repeat(cases, count, 0)
res = cochrans_q(data)
assert_allclose(res, [8.4706, 0.0145], atol=5e-5)
def test_cochransq2():
# from an example found on web, verifies 13.286
data = np.array('''
0 0 0 1
0 0 0 1
0 0 0 1
1 1 1 1
1 0 0 1
0 1 0 1
1 0 0 1
0 0 0 1
0 1 0 0
0 0 0 0
1 0 0 1
0 0 1 1'''.split(), int).reshape(-1, 4)
res = cochrans_q(data)
assert_allclose(res, [13.2857143, 0.00405776], rtol=1e-6)
def test_cochransq2():
# from an example found on web, verifies 13.286
data = np.array('''
0 0 0 1
0 0 0 1
0 0 0 1
1 1 1 1
1 0 0 1
0 1 0 1
1 0 0 1
0 0 0 1
0 1 0 0
0 0 0 0
1 0 0 1
0 0 1 1'''.split(), int).reshape(-1, 4)
res = cochrans_q(data)
assert_allclose(res, [13.2857143, 0.00405776], rtol=1e-6)
def cochranQ():
'''Cochran's Q test: 12 subjects are asked to perform 3 tasks. The outcome of each task is "success" or
"failure". The results are coded 0 for failure and 1 for success. In the example, subject 1 was successful
in task 2, but failed tasks 1 and 3.
Is there a difference between the performance on the three tasks?
'''
tasks = np.array([[0,1,1,0,1,0,0,1,0,0,0,0],
[1,1,1,0,0,1,0,1,1,1,1,1],
[0,0,1,0,0,1,0,0,0,0,0,0]])
# I prefer a DataFrame here, as it indicates directly what the values mean
df = pd.DataFrame(tasks.T, columns = ['Task1', 'Task2', 'Task3'])
# --- >>> START stats <<< ---
(Q, pVal) = cochrans_q(df)
# --- >>> STOP stats <<< ---
print('\nCOCHRAN\'S Q -----------------------------------------------------')
print('Q = {0:5.3f}, p = {1:5.3f}'.format(Q, pVal))
if pVal < 0.05:
print("There is a significant difference between the three tasks.")
def test_cochransq3():
# another example compared to SAS
# in frequency weight format
dt = [("A", "S1"), ("B", "S1"), ("C", "S1"), ("count", int)]
dta = np.array(
[
("F", "F", "F", 6),
("U", "F", "F", 2),
("F", "F", "U", 16),
("U", "F", "U", 4),
("F", "U", "F", 2),
("U", "U", "F", 6),
("F", "U", "U", 4),
("U", "U", "U", 6),
],
dt,
)
cases = np.array([[0, 0, 0], [1, 0, 0], [0, 0, 1], [1, 0, 1], [0, 1, 0], [1, 1, 0], [0, 1, 1], [1, 1, 1]])
count = np.array([6, 2, 16, 4, 2, 6, 4, 6])
data = np.repeat(cases, count, 0)
res = cochrans_q(data)
assert_allclose(res, [8.4706, 0.0145], atol=5e-5)
def cochran_q_test(pdf, var_names):
q, p = cochrans_q(pdf[var_names])
return _("Result of Cochran's Q test") + ': <i>Q</i>(%d, <i>N</i> = %d) = %0.3g, %s\n' % \
(len(var_names)-1, len(pdf[var_names[0]]), q, cs_util.print_p(p))
