How to calculate 1st and 3rd quartiles?

小开

You can use np.percentile to calculate quartiles (including the median):

>>> np.percentile(df.time_diff, 25)  # Q1
0.48333300000000001


>>> np.percentile(df.time_diff, 50)  # median
0.5


>>> np.percentile(df.time_diff, 75)  # Q3
0.51666699999999999

Or all at once:

>>> np.percentile(df.time_diff, [25, 50, 75])
array([ 0.483333,  0.5     ,  0.516667])

小开

最佳答案

By using pandas:

df.time_diff.quantile([0.25,0.5,0.75])




Out[793]:
0.25    0.483333
0.50    0.500000
0.75    0.516667
Name: time_diff, dtype: float64

小开

Using np.percentile.

q75, q25 = np.percentile(DataFrame, [75,25])
iqr = q75 - q25

Answer from How do you find the IQR in Numpy?

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Coincidentally, this information is captured with the describe method:

df.time_diff.describe()


count    5.000000
mean     0.496667
std      0.032059
min      0.450000
25%      0.483333
50%      0.500000
75%      0.516667
max      0.533333
Name: time_diff, dtype: float64

小开

np.percentile DOES NOT calculate the values of Q1, median, and Q3. Consider the sorted list below:

samples = [1, 1, 8, 12, 13, 13, 14, 16, 19, 22, 27, 28, 31]

running np.percentile(samples, [25, 50, 75]) returns the actual values from the list:

Out[1]: array([12., 14., 22.])

However, the quartiles are Q1=10.0, Median=14, Q3=24.5 (you can also use this link to find the quartiles and median online). One can use the below code to calculate the quartiles and median of a sorted list (because of sorting this approach requires O(nlogn) computations where n is the number of items). Moreover, finding quartiles and median can be done in O(n) computations using the Median of medians Selection algorithm (order statistics).

samples = sorted([28, 12, 8, 27, 16, 31, 14, 13, 19, 1, 1, 22, 13])


def find_median(sorted_list):
indices = []


list_size = len(sorted_list)
median = 0


if list_size % 2 == 0:
indices.append(int(list_size / 2) - 1)  # -1 because index starts from 0
indices.append(int(list_size / 2))


median = (sorted_list[indices[0]] + sorted_list[indices[1]]) / 2
pass
else:
indices.append(int(list_size / 2))


median = sorted_list[indices[0]]
pass


return median, indices
pass


median, median_indices = find_median(samples)
Q1, Q1_indices = find_median(samples[:median_indices[0]])
Q3, Q3_indices = find_median(samples[median_indices[-1] + 1:])


quartiles = [Q1, median, Q3]


print("(Q1, median, Q3): {}".format(quartiles))

小开

In my efforts to learn object-oriented programming alongside learning statistics, I made this, maybe you'll find it useful:

samplesCourse = [9, 10, 10, 11, 13, 15, 16, 19, 19, 21, 23, 28, 30, 33, 34, 36, 44, 45, 47, 60]


class sampleSet:
def __init__(self, sampleList):
self.sampleList = sampleList
self.interList = list(sampleList) # interList is sampleList alias; alias used to maintain integrity of original sampleList


def find_median(self):
self.median = 0


if len(self.sampleList) % 2 == 0:
# find median for even-numbered sample list length
self.medL = self.interList[int(len(self.interList)/2)-1]
self.medU = self.interList[int(len(self.interList)/2)]
self.median = (self.medL + self.medU)/2


else:
# find median for odd-numbered sample list length
self.median = self.interList[int((len(self.interList)-1)/2)]
return self.median


def find_1stQuartile(self, median):
self.lower50List = []
self.Q1 = 0


# break out lower 50 percentile from sampleList
if len(self.interList) % 2 == 0:
self.lower50List = self.interList[:int(len(self.interList)/2)]
else:
# drop median to make list ready to divide into 50 percentiles
self.interList.pop(interList.index(self.median))
self.lower50List = self.interList[:int(len(self.interList)/2)]


# find 1st quartile (median of lower 50 percentiles)
if len(self.lower50List) % 2 == 0:
self.Q1L = self.lower50List[int(len(self.lower50List)/2)-1]
self.Q1U = self.lower50List[int(len(self.lower50List)/2)]
self.Q1 = (self.Q1L + self.Q1U)/2


else:
self.Q1 = self.lower50List[int((len(self.lower50List)-1)/2)]


return self.Q1


def find_3rdQuartile(self, median):
self.upper50List = []
self.Q3 = 0


# break out upper 50 percentile from sampleList
if len(self.sampleList) % 2 == 0:
self.upper50List = self.interList[int(len(self.interList)/2):]
else:
self.interList.pop(interList.index(self.median))
self.upper50List = self.interList[int(len(self.interList)/2):]


# find 3rd quartile (median of upper 50 percentiles)
if len(self.upper50List) % 2 == 0:
self.Q3L = self.upper50List[int(len(self.upper50List)/2)-1]
self.Q3U = self.upper50List[int(len(self.upper50List)/2)]
self.Q3 = (self.Q3L + self.Q3U)/2


else:
self.Q3 = self.upper50List[int((len(self.upper50List)-1)/2)]


return self.Q3


def find_InterQuartileRange(self, Q1, Q3):
self.IQR = self.Q3 - self.Q1
return self.IQR


def find_UpperFence(self, Q3, IQR):
self.fence = self.Q3 + 1.5 * self.IQR
return self.fence


samples = sampleSet(samplesCourse)
median = samples.find_median()
firstQ = samples.find_1stQuartile(median)
thirdQ = samples.find_3rdQuartile(median)
iqr = samples.find_InterQuartileRange(firstQ, thirdQ)
fence = samples.find_UpperFence(thirdQ, iqr)


print("Median is: ", median)
print("1st quartile is: ", firstQ)
print("3rd quartile is: ", thirdQ)
print("IQR is: ", iqr)
print("Upper fence is: ", fence)

小开

Building upon or rather correcting a bit on what Babak said....

np.percentile DOES VERY MUCH calculate the values of Q1, median, and Q3. Consider the sorted list below:

s1=[18,45,66,70,76,83,88,90,90,95,95,98]

running np.percentile(s1, [25, 50, 75]) returns the actual values from the list:

[69.  85.5  91.25]

However, the quartiles are Q1=68.0, Median=85.5, Q3=92.5, which is the correct thing to say

What we are missing here is the interpolation parameter of the np.percentile and related functions. By default the value of this argument is linear. This optional parameter specifies the interpolation method to use when the desired quantile lies between two data points i < j:
linear: i + (j - i) * fraction, where fraction is the fractional part of the index surrounded by i and j.
lower: i.
higher: j.
nearest: i or j, whichever is nearest.
midpoint: (i + j) / 2.

Thus running np.percentile(s1, [25, 50, 75], interpolation='midpoint') returns the actual results for the list:

[68. 85.5 92.5]

小开

you can use

df.describe()

which would show the information

小开

If you want to use raw python rather than numpy or panda, you can use the python stats module to find the median of the upper and lower half of the list:

    >>> import statistics as stat
>>> def quartile(data):
data.sort()
half_list = int(len(data)//2)
upper_quartile = stat.median(data[-half_list]
lower_quartile = stat.median(data[:half_list])
print("Lower Quartile: "+str(lower_quartile))
print("Upper Quartile: "+str(upper_quartile))
print("Interquartile Range: "+str(upper_quartile-lower_quartile)


>>> quartile(df.time_diff)

Line 1: import the statistics module under the alias "stat"

Line 2: define the quartile function

Line 3: sort the data into ascending order

Line 4: get the length of half of the list

Line 5: get the median of the lower half of the list

Line 6: get the median of the upper half of the list

Line 7: print the lower quartile

Line 8: print the upper quartile

Line 9: print the interquartile range

Line 10: run the quartile function for the time_diff column of the DataFrame

小开

I also faced a similar problem when trying to find a package that finds quartiles. That's not to say the others are wrong but to say this is how I personally would have defined quartiles. It is similar to Shikar's results with using mid-point but also works on lists that have an odd length. If the quartile position is between lengths, it will use the average of the neighbouring values. (i.e. position always treated as either the exact position or 0.5 of the position)

import math


def find_quartile_postions(size):
if size == 1:
# All quartiles are the first (only) element
return 0, 0, 0
elif size == 2:
# Lower quartile is first element, Upper quartile is second element, Median is average
# Set to 0.5, 0.5, 0.5 if you prefer all quartiles to be the mean value
return 0, 0.5, 1
else:
# Lower quartile is element at 1/4th position, median at 1/2th, upper at 3/4
# Quartiles can be between positions if size + 1 is not divisible by 4
return (size + 1) / 4 - 1, (size + 1) / 2 - 1, 3 * (size + 1) / 4 - 1


def find_quartiles(num_array):
size = len(num_array)
    

if size == 0:
quartiles = [0,0,0]
else:
sorted_array = sorted(num_array)
lower_pos, median_pos, upper_pos = find_quartile_postions(size)


# Floor so can work in arrays
floored_lower_pos = math.floor(lower_pos)
floored_median_pos = math.floor(median_pos)
floored_upper_pos = math.floor(upper_pos)


# If position is an integer, the quartile is the elem at position
# else the quartile is the mean of the elem & the elem one position above
lower_quartile = (sorted_array[floored_lower_pos]
if (lower_pos % 1 == 0)
else (sorted_array[floored_lower_pos] + sorted_array[floored_lower_pos + 1]) / 2
)


median = (sorted_array[floored_median_pos]
if (median_pos % 1 == 0)
else (sorted_array[floored_median_pos] + sorted_array[floored_median_pos + 1]) / 2
)


upper_quartile = (sorted_array[floored_upper_pos]
if (upper_pos % 1 == 0)
else (sorted_array[floored_upper_pos] + sorted_array[floored_upper_pos + 1]) / 2
)


quartiles = [lower_quartile, median, upper_quartile]


return quartiles

小开

try that way:

dfo = sorted(df.time_diff)


n=len(dfo)


Q1=int((n+3)/4)
Q3=int((3*n+1)/4)




print("Q1 position: ", Q1, "Q1 position: " ,Q3)


print("Q1 value: ", dfo[Q1], "Q1 value: ", dfo[Q3])

小开

If you're interested in using JS, I have developed a solution:

var
withThis = (obj, cb) => cb(obj),
sort = array => array.sort((a, b) => a - b),


fractile = (array, parts, nth) => withThis(
(nth * (array.length + 1) / parts),
decimal => withThis(Math.floor(decimal),
even => withThis(sort(array),
sorted => sorted[even - 1] + (
(decimal - even) * (
sorted[even] - sorted[even - 1]
)
)
)
)
),


data = [
78, 72, 74, 79, 74, 71, 75, 74, 72, 68,
72, 73, 72, 74, 75, 74, 73, 74, 65, 72,
66, 75, 80, 69, 82, 73, 74, 72, 79, 71,
70, 75, 71, 70, 70, 70, 75, 76, 77, 67
]


fractile(data, 4, 1) // 1st Quartile is 71
fractile(data, 10, 3) // 3rd Decile is 71.3
fractile(data, 100, 82) // 82nd Percentile is 75.62

You can just copy paste the codes onto your browser and get the exact result. And more about 'Statistics with JS' can be found in https://gist.github.com/rikyperdana/a7349c790cf5b034a1b77db64415e73c/edit

小开

This can be easily done using the python statistics module. https://docs.python.org/3/library/statistics.html

import statistics


time_diff = [0.45,0.483333,0.5,0.516667,0.5333333]
statistics.quantiles(time_diff, method='inclusive')

[0.483333, 0.5, 0.516667]

The above defaults to 4 groups of data (n=4) with 3 split points (1st quartile, median, 3rd quartile), and setting the method to inclusive uses all the data in the list. The output is a list of 1st quartile, median and 3rd quartile.

小开

The main difference of the signatures between numpy.percentile and pandas.quantile: with pandas the q paramter should be given in a scala between [0-1] instead with numpy between [0-100].

Both of them, by default, use a linear interpolation technique to find such quantities. Instead, DataFrame.describe has a less flexible signature and allow to use only the linear one.

In numpy >= 1.22 the parameter interpolation is deprecated and replaced with method.

Here an example of usage with linear interpolation: (default behavior)

import pandas as pd
import numpy as np




s =[18,45,66,70,76,83,88,90,90,95,95,98, 100]
print(pd.DataFrame(s).quantile(q=[.25, .50, .75]))
print(np.percentile(s, q=[25, 50, 75]))
print(pd.DataFrame(s).describe(percentiles=[.25, .5, .75])) # the parameter is redundant, it's the default behavior

Here using the midpoint interpolation:

s_even = [18,45,66,70,76,83,88,90,90,95,95,98]
print(pd.DataFrame(s_even).quantile(q=[.25, .5, .75], interpolation='midpoint'))
print(np.percentile(s_even, q=[25, 50, 75], interpolation='midpoint')) # verion < 1.22
print(np.percentile(s_even, q=[25, 50, 75], method='midpoint')) # version >= 1.22


s_odd = s_even + [100] # made it odd
print(pd.DataFrame(s_odd).quantile(q=[.25, .50, .75], interpolation='midpoint'))
print(np.percentile(s_odd, q=[25, 50, 75], interpolation='midpoint')) # verion < 1.22
print(np.percentile(s_odd, q=[25, 50, 75], method='midpoint')) # version >= 1.22

小开

Full working example:

import numpy as np
sizes_height = np.random.randn(100)
df = pd.DataFrame(sizes_height)
# df = pd.Series(sizes_height)
# x = df.time_diff.quantile(sizes_height)
x = df.describe()
print()
x
0
count  100.000000
mean     0.059808
std      1.012960
min     -2.552990
25%     -0.643857
50%      0.094096
75%      0.737077
max      2.269755