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Python在数组中找到最常见的值

numpy python

Sergej Fomin · 技术社区 · 7 年前

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,如果值达到相同的次数,则返回最小值。这是家庭作业,除了钱我什么都用不了。

输出:

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

5 回复 | 直到 7 年前

Inder Manali 7 年前

对于Numpy独家解决方案,类似这样的解决方案将起作用:

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

如果列表中有负数,上述方法将不起作用。因此,为了解释这种情况,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)])

blue_note 7 年前

它被称为模式功能。见 https://docs.scipy.org/doc/scipy/reference/generated/scipy.stats.mode.html

madmatrix 7 年前

没有纽比

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])

Learning is a mess 7 年前

要添加到以前的结果,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]

Inder Manali 7 年前

它很简单,但肯定不漂亮我使用了变量名,这些变量名和注释一起都是不言而喻的。如果有疑问,请随时询问。

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))

我们还可以循环它来检查是否有两个以上的元素具有相同的事件,而不是使用if-else来处理只查看两个元素的特定情况