How to calculate the 1st and 3rd quartiles?

Question

I have a DataFrame:

    time_diff   avg_trips
0   0.450000    1.0
1   0.483333    1.0
2   0.500000    1.0
3   0.516667    1.0
4   0.533333    2.0

I want to get 1st quartile, 3rd quartile and median for the column time_diff. To get the med, I use np.median(df["time_diff"].values).

How can I calculate quartiles?

Answer 1

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

Answer 2

You can use np.percentileto calculate quartiles (including 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])

Answer 3

, describe:

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

Answer 4

np.percentile.

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

IQR Numpy?

Answer 5

np.percentile Q1, Q3. :

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

np.percentile(samples, [25, 50, 75]) :

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

, Q1=10.0, Median=14, Q3=24.5 ( , ). (- O(nlogn) , n - ). , O(n) ( ).

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]])
Q2, Q2_indices = find_median(samples[median_indices[-1] + 1:])

quartiles = [Q1, median, Q2]

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

Answer 6

- , , :

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)

Answer 7

, , ...

[np.percentile][1] Q1, Q3. :

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

np.percentile(s1, [25, 50, 75]) :

[69.   85.5  91.25]

Q1 = 68,0, = 85,5, Q3 = 92,5,

np.percentile . linear. , , i & lt; J:
linear: i + (j - i) * , - , i j.
: .
: j.
: i j, , .
: (i + j)/2.

, np.percentile(s1, [25, 50, 75], interpolation='midpoint') :

[68.  85.5 92.5]