Question

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

（仅正值）在每个数组中，我需要返回最常用的值，或者，如果值出现的次数相同，则返回最小值。这是家庭作业，除numpy外我什么也不能使用。

输出：

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

Answer 1

它称为 mode 函数。参见https://docs.scipy.org/doc/scipy/reference/generated/scipy.stats.mode.html

Answer 2

对于一个numpy独占解决方案，类似这样的方法将起作用：

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

如果列表中为负数，则上述方法将无效。因此，为了解决这种情况，请使用：

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

没有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

要添加到以前的结果中，可以使用collections.Counter对象：

 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

这很简单，但肯定不是很漂亮。我使用的变量名将随注释一起自我解释。随时问是否有疑问。

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))

我们还可以循环检查是否有两个以上元素同时出现，而不是在仅查看两个元素的特定情况下使用if else

Python在数组中找到最常见的值

5 个答案: