LearningAkshat
commented
1 year ago Median
def Median(data): med = [val for val in data]; med.sort(); if even(len(med)): one = med[len(med)/2]; two = med[one + 1]; return med[(one + two)//2]; else: return med[len(med)//2];
Mode:
mode:
mode is the element having highest frequency in the list
list=[1,2,3,4,5,4,3,2,1]
def freqTuple(l,x):
freq=0
for v in l:
if v==x:
freq=freq+1
return x,freq #x is the element of your list
freq is the number of times x occured in the list
(e.g.element x=1 has freq=2 as it occured 2 times in the list)
so a tuple(1,2) is returned
def findMode(l):
t=[freqTuple(l,x) for x in l] #t is a list where its each element is a tuple containing 2 values each
the tuple contains return values of the freqTuple funtion
first value of each tuple contains list elements
second value of tuple contains frequencies of corresponding elements
for e.g. t will be t=[(1,2),(2,2),(3,2),(4,2),(5,1),(4,2),(3,2),(2,2),(1,2)]
a=t[0][1] #algorithm to find the highest number among 2nd elements(frequencies) of tuples
for i in t:
if(i[1]>a): #i[1] here means second element of tuple i
a=i[1] #when for loop ends 'a' points to the highest frequency
(e.g.here a=2 because 2 is the highest number among all 2nd elements of tuples)
c=[]
for i in t:
if(i[1]==a):
c.append(i) #list c contains all tuples of highest frequency numbers(i.e. 2 here)
(e.g. c=[(1,2),(2,2),(3,2),(4,2),(4,2),(3,2),(2,2),(1,2)] )
(x=5 is not present in c as its freq=1 i.e. not highest)
but tuples get repeated in this process
d=[]
for i in c:
if i not in d: #this algorithm removes repeated tuples
d.append(i) #(e.g. d=[(1,2),(2,2),(3,2),(4,2)] )
for i in d:
print(i[0]) #the first element of every tuple is printed as i[0] means first element of tuple i
(e.g. 1,2,3,4 is printed as our list has 4 modes)
print("The mode is")
findMode(list)