按照lis中值的顺序筛选已排序NumPy数组的行

2024-10-01 22:39:45 发布

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我试图遍历NumPy数组来创建列表列表,但是使用的for循环是按字母顺序而不是按出现的顺序附加到列表列表。你知道吗

以下是我的NumPy数组的一部分,我可以用它作为示例:

tarifas = np.array([['Afganistán', '577.21', '0.9360168799091559', '1.01745744495737'],
                    ['Albania', '5450.0', '1.1439867079655244', '0.9195410037811979'],
                    ['Alemania', '49690', '1.0034542200895549', '0.9873874704432137'],
                    ['Angola', '3670.0', '0.931103978746121', '1.162652536895962'],
                    ['Antigua y Barbuda', '18170', '0.7795684991736309', '0.6399312443495023'],
                    ['Arabia Saudita', '23490', '1.0573676413333202', '0.7477763277701148'],
                    ['Argelia', '4650.0', '0.7969840140783656', '0.5123046862189027'],
                    ['Argentina', '9050.0', '1.3647162509775996', '0.48274125735042017'],
                    ['Armenia', '4450.0', '1.4545784506262867', '1.430465487479917'],
                    ['Australia', '57200', '0.7293018985322222', '1.1744384938116095'],
                    ['Austria', '52470', '1.2396562976033307', '0.8630735107719588'],
                    ['Azerbaiyán', '4780.0', '0.9111186496911305','0.534268284966654']])

我想创建一个列表列表,使用另一个列表进行迭代,其中包含我需要在数组中找到的国家的特定名称,即

list_countries = ["Angola", "Austria", "Argentina", "Albania", "Armenia"]

请注意列表不是按字母顺序排列的,因此列表列表应该遵循这个顺序。迭代后的输出应如下所示:

new_list_of_countries = [['Angola' '3670.0' '0.931103978746121' '1.162652536895962'], 
                         ['Austria' '52470' '1.2396562976033307' '0.8630735107719588'],  
                         ['Argentina' '9050.0' '1.3647162509775996' '0.48274125735042017'], 
                         ['Albania' '5450.0' '1.1439867079655244' '0.9195410037811979'], 
                         ['Armenia' '4450.0' '1.4545784506262867' '1.430465487479917']]

以下是我使用的代码:

tarifas_paises_escogidos = []
for i in tarifas:
    for v in list_countries:
         if str(v) in str(i):
               tarifas_paises_escogidos.append(i)
print(np.array(tarifas_paises_escogidos))

Tags: in列表for顺序数组countrieslistargentina
2条回答

将列表理解与sorted结合使用:

sorted([t for t in tarifas if t[0] in list_countries], 
        key=lambda x: list_countries.index(x[0]))

输出:

[['Angola', '3670.0', '0.931103978746121', '1.162652536895962'],
 ['Austria', '52470', '1.2396562976033307', '0.8630735107719588'],
 ['Argentina', '9050.0', '1.3647162509775996', '0.48274125735042017'],
 ['Albania', '5450.0', '1.1439867079655244', '0.9195410037811979'],
 ['Armenia', '4450.0', '1.4545784506262867', '1.430465487479917']]

一个不使用列表理解:

tarifas_paises_escogidos = []
for t in tarifas:
    # for v in list_countries: You don't need this
    if t[0] in list_countries:
        tarifas_paises_escogidos.append(t)
print(tarifas_paises_escogidos)

产生过滤但未排序的:

[['Albania', '5450.0', '1.1439867079655244', '0.9195410037811979'], 
 ['Angola', '3670.0', '0.931103978746121', '1.162652536895962'], 
 ['Argentina', '9050.0', '1.3647162509775996', '0.48274125735042017'], 
 ['Armenia', '4450.0', '1.4545784506262867', '1.430465487479917'], 
 ['Austria', '52470', '1.2396562976033307', '0.8630735107719588']]

然后进行排序(并重新分配!)地址:

tarifas_paises_escogidos = sorted(tarifas_paises_escogidos, key=lambda x: list_countries.index(x[0]))

这使得上述输出。你知道吗

洞察:

在上面的lambda中,x几乎毫无意义。它只意味着输入lambda得到的内容被定义为x,并用于索引(即x[0])。你知道吗

等同于:

def some_func(x):
    return list_countries.index(x[0])

然后用在sorted

tarifas_paises_escogidos = sorted(tarifas_paises_escogidos, key=some_func)

但是你可能经常发现仅仅为一个用例定义一个函数是非常低效的。这时lambda就开始了:)。你知道吗

由于原始NumPy数组tarifas是按字母顺序排序的,因此可以使用^{}来获取与list_countries对应的索引:

indices = np.searchsorted(tarifas[:, 0], list_countries)
print(indices)
# [ 3 10  7  1  8]

然后使用花式索引(使用数组索引数组)获得所需结果:

result = tarifas[indices]
print(result)
# [['Angola' '3670.0' '0.931103978746121' '1.162652536895962']
#  ['Austria' '52470' '1.2396562976033307' '0.8630735107719588']
#  ['Argentina' '9050.0' '1.3647162509775996' '0.48274125735042017']
#  ['Albania' '5450.0' '1.1439867079655244' '0.9195410037811979']
#  ['Armenia' '4450.0' '1.4545784506262867' '1.430465487479917']]

对于大型数组,这种矢量化方法应该比使用Python的For循环的解决方案快得多。你知道吗

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