python - 如何根据 python 中的浮点值对列表中的元组字符串部分进行排序？

Question

我有一个元组列表，带有 Str 和 float

list = 
[('Trading_State_NZ <= 0.00', 0.0234),
 ('Product_Type_A <= 0.00', 0.045643),
 ('Permanent_Resident_Y <= 1.00', 0.0214),
 ('Applicant_Type_B <= 1.00', 0.0236),
 ('Valuation_Acceptable_Y <= 1.00', 0.0866),
 ('Product_Type_6 <= 0.00', 0.0172),
 ('Trading_State_NT <= 0.00', 0.0553),
 ('Trading_State_TAS <= 0.00', 0.0251),
 ('Property_Acceptable_Y <= 1.00', 0.01588),
 ('Applicant_Type1_at1_miss <= 1.00', 0.0158758),
 ('Product_Type_I <= 0.00', 0.01571),
 ('Employer_Name <= 1.00', 0.0552),
 ('Business_Name <= 0.00', 0.02557),
 ('Product_Type_E <= 0.00', 0.02457),
 ('Product_Type_CTP <= 0.00', 0.02457),
 ('Trading_Postcode <= 1.00', 0.024572),
 ('Trading_State_ts_miss <= 1.00', 0.0785),
 ('Variation_2014 <= 0.00', 0.0694),
 ('Product_Type_C <= 0.00', 0.017145),
 ('Phone_Number <= 0.00', 0.0789678)]

字符串部分前后两段 <= symbol, "Name <= value", 我希望将 0.00 和 1.00 值分开，然后根据浮点值降序排序

list.sort()

list.sort() 基于字符串排序，我需要根据值拆分 str 并排序？

Answer 1

一种方法:

type0 = []
type1 = []
for tup in mytups:
    if '<= 0' in tup[0]:
        type0.append(tup)
    else:
        type1.append(tup)

type0.sort(key=lambda x: x[1], reverse=True)
type1.sort(key=lambda x: x[1], reverse=True)

final_list = type0 + type1

或者使用多级排序(无耻地建立在@Selcuk 的回答之上):

mytups.sort(key=lambda x: (float(x[0].split("<=")[1]), -x[1]))