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[nominal two samples] improve comments #142

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23 changes: 14 additions & 9 deletions nominal/paired_two_sample_test_of_nominal_scale.py
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Original file line number Diff line number Diff line change
Expand Up @@ -12,6 +12,7 @@
class PairedTwoSampleTestOfNominalScale:
def test(self, data):
"""
There is a question which we can answer yes (1) or no (0).
data = {"Before": [1,1,1,1,1,...,0], "After": [1,1,1,1,1,...,0]}

Yes No Total
Expand All @@ -26,10 +27,10 @@ def test(self, data):
number of Yes => No: b
number of No => Yes: c
"""
# check data length is 2
if len(data.keys()) != 2 and len(data[data.keys()[0]]) != len(data[data.keys()[1]]):
print "Please check the components of your data."
print "length of data should be four"
# check if the number of samples are appropriate
if len(data.keys()) != 2 or len(data[data.keys()[0]]) != len(data[data.keys()[1]]):
print ("Please check the components of your data.")
print ("the number of each data should be equal")
sys.exit()
else:
b = 0
Expand All @@ -39,13 +40,17 @@ def test(self, data):
b += 1
elif data[(data.keys())[0]][i] == 0 and data[(data.keys())[1]][i] == 1:
c += 1
# z = abs(b-c)-1 / root(b+c)
# chi2 = pow((abs(b-c)-1), 2.0) / (b+c)
# calculating chi-square value with Yate's continuity correction (イェーツの連続修正)
chi2 = pow((abs(b-c)-1), 2.0) / (b+c)

'''
If there is no consideration on Yate's continuity correction:
chi2 = pow(abs(b-c) - 1.0, 2.0) / (b+c)
'''
p = stats.chi2.pdf(chi2, df=1)
# pdf: probability density function
# cdf: Cumulative distribution function
# https://docs.scipy.org/doc/scipy/reference/generated/scipy.stats.chi2.html
print "chi2 value: {}".format(chi2)
print "p value: {}".format(p)
return p

if __name__ == '__main__':
pass
18 changes: 8 additions & 10 deletions nominal/unpaired_two_sample_test_of_nominal_scale.py
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Original file line number Diff line number Diff line change
Expand Up @@ -6,27 +6,25 @@

class UnpairedTwoSampleTestOfNominalScale:
def test(self, data):
# check data length
if len(data.keys()) != 2:
print "len(data.keys()) should be two"
sys.exit()
elif len(data[(data.keys())[0]]) != len(data[(data.keys())[1]]):
print "len(data[(data.keys())[0]]) and len(data[(data.keys())[1]]) should be same"
# check if the number of samples are appropriate
if len(data.keys()) != 2 or len(data[data.keys()[0]]) != len(data[data.keys()[1]]):
print ("Please check the components of your data.")
print ("the number of each data should be equal")
sys.exit()
else:
"""
Is there any difference between the number of people who satisfies Condition1 and Yes (a) and that of people who satisfies Condition2 and Yes (c)?
Is there any difference between the number of people who satisfies Condition1 as Yes (a) and that of people who satisfies Condition2 as Yes (c)?
data = {"Condition1": [a, b], "Condition2": [c, d]}
OrderedDict([('Illness', [52, 8]), ('Healty', [48, 42])])
ex. OrderedDict([('Illness', [52, 8]), ('Healty', [48, 42])])

Yes No Total <= sum_row
Yes No Total <= sum_row: [a+b, c+d]
--------------------------------------
Condition1 a b a+b
Condition2 c d c+d
--------------------------------------
Total a+c b+d n (= a+b+c+d)
^
|_ sum_column
|_ sum_column: [a+c, b+d]

"""
# calculate n
Expand Down