Python在数组中找到最常见的值

Question

import numpy as np 
x = ([1,2,3,3])
y = ([1,2,3])
z = ([6,6,1,2,9,9])

(only positive values) In each array i need to return the most common value, or, if values come up the same amount of times - return the minimum. This is home assignment and I can't use anything but numpy.

outputs:

f(x) = 3,
f(y) = 1,
f(z) = 6

Answer 1

It's called a mode function. See https://docs.scipy.org/doc/scipy/reference/generated/scipy.stats.mode.html

Answer 2

for a numpy exclusive solution something like this will work:

occurances = np.bincount(x)
print (np.argmax(occurances))

The above mentioned method won't work if there is a negative number in the list. So in order to account for such an occurrence kindly use:

not_required, counts = np.unique(x, return_counts=True)
x=np.array(x)
if (x >= 0).all():
    print(not_required[np.argmax(counts)])
else:    
    print(not_required[np.argmax(counts)])

Answer 3

Without numpy

n_dict = {}
for k in x:
    try:
        n_dict[k] += 1
    except KeyError:
        n_dict[k] = 1

rev_n_dict = {}
for k in n_dict:
    if n_dict[k] not in rev_n_dict:
        rev_n_dict[n_dict[k]] = [k]
    else:
        rev_n_dict[n_dict[k]].append(k)

local_max = 0
for k in rev_n_dict:
    if k > local_max:
        local_max = k

if len(rev_n_dict[local_max]) > 0:      
    print (min(rev_n_dict[local_max]))
else:
    print (rev_n_dict[local_max])

Answer 4

To add to the previous results, you could use a collections.Counter object:

 my_array =  [3,24,543,3,1,6,7,8,....,223213,13213]
 from collections import Counter
 my_counter = Counter( my_array)
 most_common_value = my_counter.most_common(1)[0][0]

Answer 5

It is quite simple but certainly not pretty. I have used variable names that will be self explanatory along with the comments. Feel free to ask if there is a doubt.

import numpy as np
x=([6,6,1,2,9,9])

def tester(x): 
    not_required, counts = np.unique(x, return_counts=True)
    x=np.array(x)
    if (x >= 0).all():
        highest_occurance=[not_required[np.argmax(counts)]]
        number_of_counts=np.max(counts)

    else:    
        highest_occurance=not_required[np.argmax(counts)]
        number_of_counts=np.max(counts)

    return highest_occurance,number_of_counts

most_abundant,first_test_counts=(tester(x))

new_x=[vals for vals in x if vals not in most_abundant]

second_most_abundant,second_test_counts=(tester(new_x))

if second_test_counts==first_test_counts:
    print("Atleast two elements have the same number of counts",most_abundant," and", second_most_abundant, "have %s"%first_test_counts,"occurances")
else:
    print("%s occurrs for the max of %s times"%(most_abundant,first_test_counts))

we can also loop it to check if there are more than two elements with the same occurrence, instead of using an if else for a specific case of only looking at two elements