How do you calculate interquartile range (IQR) correctly using Python?
I am trying to understand the way to compute iqr (interquartile range).
according this, this and this, I tried 3 solutions to do this.
solution_1
a = numpy.array([1, 2, 3, 4, 5, 6, 7])
q1_a = numpy.percentile(a, 25)
q3_a = numpy.percentile(a, 75)
q3_a - q1_a
solution_2
from scipy.stats import iqr
iqr(a)
solution_3
q1_am = np.median(numpy.array([1, 2, 3, 4]))
q3_am = np.median(numpy.array([4, 5, 6, 7]))
q3_am - q1_am
3 of them give the same result 3 which is correct.
when I tried another set of numbers, things were going weird.
both solution_1 and 2 output 0.95 which is not correct.
x = numpy.array([4.1, 6.2, 6.7, 7.1, 7.4, 7.4, 7.9, 8.1])
q1_x = numpy.percentile(x, 25)
q3_x = numpy.percentile(x, 75)
q3_x - q1_x
solution_3 gives 1.2 which is correct
q1_xm = np.median(np.array([4.1, 6.2, 6.7,7.25]))
q3_xm = np.median(np.array([7.25,7.4, 7.9, 8.1]))
q3_xm - q1_xm
What am I missing with the solutions?
any clue would be appreciated.
python numpy scipy
edited Nov 16 '18 at 12:50
tel
6,12811430
asked Nov 16 '18 at 11:54
czlsws
233
add a comment |
I am trying to understand the way to compute iqr (interquartile range).
according this, this and this, I tried 3 solutions to do this.
solution_1
a = numpy.array([1, 2, 3, 4, 5, 6, 7])
q1_a = numpy.percentile(a, 25)
q3_a = numpy.percentile(a, 75)
q3_a - q1_a
solution_2
from scipy.stats import iqr
iqr(a)
solution_3
q1_am = np.median(numpy.array([1, 2, 3, 4]))
q3_am = np.median(numpy.array([4, 5, 6, 7]))
q3_am - q1_am
3 of them give the same result 3 which is correct.
when I tried another set of numbers, things were going weird.
both solution_1 and 2 output 0.95 which is not correct.
x = numpy.array([4.1, 6.2, 6.7, 7.1, 7.4, 7.4, 7.9, 8.1])
q1_x = numpy.percentile(x, 25)
q3_x = numpy.percentile(x, 75)
q3_x - q1_x
solution_3 gives 1.2 which is correct
q1_xm = np.median(np.array([4.1, 6.2, 6.7,7.25]))
q3_xm = np.median(np.array([7.25,7.4, 7.9, 8.1]))
q3_xm - q1_xm
What am I missing with the solutions?
any clue would be appreciated.
python numpy scipy
edited Nov 16 '18 at 12:50
tel
6,12811430
asked Nov 16 '18 at 11:54
czlsws
233
add a comment |
I am trying to understand the way to compute iqr (interquartile range).
according this, this and this, I tried 3 solutions to do this.
solution_1
a = numpy.array([1, 2, 3, 4, 5, 6, 7])
q1_a = numpy.percentile(a, 25)
q3_a = numpy.percentile(a, 75)
q3_a - q1_a
solution_2
from scipy.stats import iqr
iqr(a)
solution_3
q1_am = np.median(numpy.array([1, 2, 3, 4]))
q3_am = np.median(numpy.array([4, 5, 6, 7]))
q3_am - q1_am
3 of them give the same result 3 which is correct.
when I tried another set of numbers, things were going weird.
both solution_1 and 2 output 0.95 which is not correct.
x = numpy.array([4.1, 6.2, 6.7, 7.1, 7.4, 7.4, 7.9, 8.1])
q1_x = numpy.percentile(x, 25)
q3_x = numpy.percentile(x, 75)
q3_x - q1_x
solution_3 gives 1.2 which is correct
q1_xm = np.median(np.array([4.1, 6.2, 6.7,7.25]))
q3_xm = np.median(np.array([7.25,7.4, 7.9, 8.1]))
q3_xm - q1_xm
What am I missing with the solutions?
any clue would be appreciated.
python numpy scipy
edited Nov 16 '18 at 12:50
tel
6,12811430
asked Nov 16 '18 at 11:54
czlsws
233
I am trying to understand the way to compute iqr (interquartile range).
according this, this and this, I tried 3 solutions to do this.
solution_1
a = numpy.array([1, 2, 3, 4, 5, 6, 7])
q1_a = numpy.percentile(a, 25)
q3_a = numpy.percentile(a, 75)
q3_a - q1_a
solution_2
from scipy.stats import iqr
iqr(a)
solution_3
q1_am = np.median(numpy.array([1, 2, 3, 4]))
q3_am = np.median(numpy.array([4, 5, 6, 7]))
q3_am - q1_am
3 of them give the same result 3 which is correct.
when I tried another set of numbers, things were going weird.
both solution_1 and 2 output 0.95 which is not correct.
x = numpy.array([4.1, 6.2, 6.7, 7.1, 7.4, 7.4, 7.9, 8.1])
q1_x = numpy.percentile(x, 25)
q3_x = numpy.percentile(x, 75)
q3_x - q1_x
solution_3 gives 1.2 which is correct
q1_xm = np.median(np.array([4.1, 6.2, 6.7,7.25]))
q3_xm = np.median(np.array([7.25,7.4, 7.9, 8.1]))
q3_xm - q1_xm
What am I missing with the solutions?
any clue would be appreciated.
python numpy scipy
python numpy scipy
edited Nov 16 '18 at 12:50
tel
6,12811430
asked Nov 16 '18 at 11:54
czlsws
233
edited Nov 16 '18 at 12:50
tel
6,12811430
asked Nov 16 '18 at 11:54
czlsws
233
edited Nov 16 '18 at 12:50
tel
6,12811430
edited Nov 16 '18 at 12:50
tel
6,12811430
edited Nov 16 '18 at 12:50
tel
6,12811430
6,12811430
asked Nov 16 '18 at 11:54
czlsws
233
asked Nov 16 '18 at 11:54
czlsws
233
asked Nov 16 '18 at 11:54
czlsws
233
233
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
You'll get your expected result with numpy.percentile if you set interpolation=midpoint:
x = numpy.array([4.1, 6.2, 6.7, 7.1, 7.4, 7.4, 7.9, 8.1])
q1_x = numpy.percentile(x, 25, interpolation='midpoint')
q3_x = numpy.percentile(x, 75, interpolation='midpoint')
print(q3_x - q1_x)
This outputs:
1.2000000000000002
Setting interpolation=midpoint also makes scipy.stats.iqr give the result you wanted:
from scipy.stats import iqr
x = numpy.array([4.1, 6.2, 6.7, 7.1, 7.4, 7.4, 7.9, 8.1])
print(iqr(x, rng=(25,75), interpolation='midpoint'))
which outputs:
1.2000000000000002
See the interpolation parameter in the linked docs for more info on what the option actually does.
answered Nov 16 '18 at 12:44
tel
6,12811430
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
You'll get your expected result with numpy.percentile if you set interpolation=midpoint:
x = numpy.array([4.1, 6.2, 6.7, 7.1, 7.4, 7.4, 7.9, 8.1])
q1_x = numpy.percentile(x, 25, interpolation='midpoint')
q3_x = numpy.percentile(x, 75, interpolation='midpoint')
print(q3_x - q1_x)
This outputs:
1.2000000000000002
Setting interpolation=midpoint also makes scipy.stats.iqr give the result you wanted:
from scipy.stats import iqr
x = numpy.array([4.1, 6.2, 6.7, 7.1, 7.4, 7.4, 7.9, 8.1])
print(iqr(x, rng=(25,75), interpolation='midpoint'))
which outputs:
1.2000000000000002
See the interpolation parameter in the linked docs for more info on what the option actually does.
answered Nov 16 '18 at 12:44
tel
6,12811430
add a comment |
You'll get your expected result with numpy.percentile if you set interpolation=midpoint:
x = numpy.array([4.1, 6.2, 6.7, 7.1, 7.4, 7.4, 7.9, 8.1])
q1_x = numpy.percentile(x, 25, interpolation='midpoint')
q3_x = numpy.percentile(x, 75, interpolation='midpoint')
print(q3_x - q1_x)
This outputs:
1.2000000000000002
Setting interpolation=midpoint also makes scipy.stats.iqr give the result you wanted:
from scipy.stats import iqr
x = numpy.array([4.1, 6.2, 6.7, 7.1, 7.4, 7.4, 7.9, 8.1])
print(iqr(x, rng=(25,75), interpolation='midpoint'))
which outputs:
1.2000000000000002
See the interpolation parameter in the linked docs for more info on what the option actually does.
answered Nov 16 '18 at 12:44
tel
6,12811430
add a comment |
You'll get your expected result with numpy.percentile if you set interpolation=midpoint:
x = numpy.array([4.1, 6.2, 6.7, 7.1, 7.4, 7.4, 7.9, 8.1])
q1_x = numpy.percentile(x, 25, interpolation='midpoint')
q3_x = numpy.percentile(x, 75, interpolation='midpoint')
print(q3_x - q1_x)
This outputs:
1.2000000000000002
Setting interpolation=midpoint also makes scipy.stats.iqr give the result you wanted:
from scipy.stats import iqr
x = numpy.array([4.1, 6.2, 6.7, 7.1, 7.4, 7.4, 7.9, 8.1])
print(iqr(x, rng=(25,75), interpolation='midpoint'))
which outputs:
1.2000000000000002
See the interpolation parameter in the linked docs for more info on what the option actually does.
answered Nov 16 '18 at 12:44
tel
6,12811430
You'll get your expected result with numpy.percentile if you set interpolation=midpoint:
x = numpy.array([4.1, 6.2, 6.7, 7.1, 7.4, 7.4, 7.9, 8.1])
q1_x = numpy.percentile(x, 25, interpolation='midpoint')
q3_x = numpy.percentile(x, 75, interpolation='midpoint')
print(q3_x - q1_x)
This outputs:
1.2000000000000002
Setting interpolation=midpoint also makes scipy.stats.iqr give the result you wanted:
from scipy.stats import iqr
x = numpy.array([4.1, 6.2, 6.7, 7.1, 7.4, 7.4, 7.9, 8.1])
print(iqr(x, rng=(25,75), interpolation='midpoint'))
which outputs:
1.2000000000000002
See the interpolation parameter in the linked docs for more info on what the option actually does.
answered Nov 16 '18 at 12:44
tel
6,12811430
answered Nov 16 '18 at 12:44
tel
6,12811430
answered Nov 16 '18 at 12:44
tel
6,12811430
answered Nov 16 '18 at 12:44
tel
6,12811430
6,12811430
add a comment |
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