Question

我有一个名为eagles的元组列表

eagles= [("NCMS000","NCMS000"),("NCFP000","NCFP000"),("NCMS00D","NCMS00D"),("NCCS000","NCCS000"),("NCCP000","NCCP000"),("NCMN000","NCMN000"),("NCFN000","NCFN000"),("NP000G0","NP000G0"),("NP000G0","NP000G0"),...

和名为Result的列表如下：

['"', '"', 'Fe', '1']
['Hola', 'hola', 'I', '1']
['como', 'como', 'CS', '0.999289']
['estas', 'este', 'DD0FP0', '0.97043']
['Bien', 'bien', 'NP00000', '1']
['gracias', 'gracia', 'NCFP000', '1']
['y', 'y', 'CC', '0.999962']
['tu', 'tu', 'DP2CSS', '1']
['yo', 'yo', 'PP1CSN00', '1']
['estoy', 'estar', 'VAIP1S0', '1']
['bien', 'bien', 'RG', '0.902728']
['huevo', 'huevo', 'NCMS000', '0.916667']
['calcio', 'calcio', 'NCMS000', '1']
['leche', 'leche', 'NCFS000', '1']
['proteina', 'proteina', 'NCFS000', '1']
['Francisco', 'francisco', 'NP00000', '1']
['1999', '1999', 'Z', '1']
['"', '"', 'Fe', '1']

我需要创建一个函数来比较Result列表的第3项和一个连续循环中的eagles 1st项。如果它们匹配，我需要返回一个包含4个元素的列表列表，例如：

 r = [['leche', 'leche', 'NCFS000', '1'],['proteina', 'proteina', 'NCFS000', '1'],['Francisco', 'francisco', 'NP00000', '1']]

到目前为止我做了什么：

def check(lst):
    return [x[2] for x in lst if (x[2] in y[0] for y in eagles)]

IndexError: list index out of range.

我甚至无法从列表中提取第3个元素并将其放在空的

上

e = [x[0] for x in eagles]
r = [item for item in e if item in Result]
rg =[]
for i in Result:
    rg = i[2]

同样的错误

我该怎么办？任何建议都表示赞赏。

Answer 1

首先，最好将eagles列表转换为字典...

>>> eagles = [("NCMS000","NCMS000"), ("NCFP000","NCFP000"), ...]
>>> eagles_dict = dict(eagles)
>>> print eagles_dict
{'NCFP000': 'NCFP000', 'NCMS000': 'NCMS000', ...}

...使查找更简单，更高效。然后你可以使用简单的列表理解，比如......

>>> result = [['"', '"', 'Fe', '1'], ['Hola', 'hola', 'I', '1'], ...]
>>> print [item for item in result if item[2] in eagles_dict]
[['leche', 'leche', 'NCFS000', '1'], ...]

Answer 2

可能有更高效的算法涉及排序，但如果您只是这样做一次或两次：

更新时考虑到您的商品并不总是包含4个元素。

eagles_first_parts = [eagle[0] for eagle in eagles]
r = [item for item in Result if len(item) > 2 and item[2] in eagles_first_parts]

Answer 3

注意：不是写出最有效的代码，而是从你的尝试中得到的东西。我假设结果是列表列表，如：

Result=[['"', '"', 'Fe', '1'],['Hola', 'hola', 'I', '1'],
['como', 'como', 'CS', '0.999289'],
['estas', 'este', 'DD0FP0', '0.97043'],
['Bien', 'bien', 'NP00000', '1'],
['gracias', 'gracia', 'NCFP000', '1'],
['y', 'y', 'CC', '0.999962'],
['tu', 'tu', 'DP2CSS', '1'],
['yo', 'yo', 'PP1CSN00', '1'],
['estoy', 'estar', 'VAIP1S0', '1'],
['bien', 'bien', 'RG', '0.902728'],
['huevo', 'huevo', 'NCMS000', '0.916667'],
['calcio', 'calcio', 'NCMS000', '1'],
['leche', 'leche', 'NCFS000', '1'],
['proteina', 'proteina', 'NCFS000', '1'],
['Francisco', 'francisco', 'NP00000', '1'],
['1999', '1999', 'Z', '1'],
['"', '"', 'Fe', '1']]

现在从你离开的地方开始。

e=[x[0] for x in eagles]

现在，初始化一个空列表r

r=[]

for item in Result:
    for eagle in e:
       if item[2]==eagle:
            r.append(item)
print r

给出输出：

[['gracias', 'gracia', 'NCFP000', '1'],
['huevo', 'huevo', 'NCMS000', '0.916667'],
['calcio', 'calcio', 'NCMS000', '1']]

将元组元素与列表元素相匹配

3 个答案: