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使用sorted()按每个元组的第一个值对元组进行排序

sorting python-3.x python

vasili111 · 技术社区 · 6 年前

我需要按每个元组的第一个值对元组列表进行排序。我有密码:

import pandas as pd
import numpy as np

a= np.array([7.33442355e-02, 0.00000000e+00, 6.29090705e-02, 4.11437985e-02, 1.46726854e-02,
       1.41064835e-02, 1.48742809e-02, 1.42575993e-02, 3.66158981e-01,
       5.64191792e-02, 1.08792957e-01, 5.33510773e-02, 1.03114883e-02,
       1.64780994e-01, 6.02803867e-05, 1.96041560e-03, 2.85647464e-03])


b= ['longitude',
    'some_var',
    'latitude',
    'housing_median_age',
    'total_rooms',
    'total_bedrooms',
    'population',
    'households',
    'median_income',
    'rooms_per_hhold',
    'pop_per_hhold',
    'bedrooms_per_room',
    '<1H OCEAN',
    'INLAND',
    'ISLAND',
    'NEAR BAY',
    'NEAR OCEAN']

x=sorted(zip(a, b), reverse=True)

内容 x :看起来像:

 [(0.366158981, 'median_income'),
 (0.164780994, 'INLAND'),
 (0.108792957, 'pop_per_hhold'),
 (0.0733442355, 'longitude'),
 (0.0629090705, 'latitude'),
 (0.0564191792, 'rooms_per_hhold'),
 (0.0533510773, 'bedrooms_per_room'),
 (0.0411437985, 'housing_median_age'),
 (0.0148742809, 'population'),
 (0.0146726854, 'total_rooms'),
 (0.0142575993, 'households'),
 (0.0141064835, 'total_bedrooms'),
 (0.0103114883, '<1H OCEAN'),
 (0.00285647464, 'NEAR OCEAN'),
 (0.0019604156, 'NEAR BAY'),
 (6.02803867e-05, 'ISLAND'),
 (0.0, 'some_var')]

我需要输出如下:

 [(0.0, 'some_var'),
 (6.02803867e-05, 'ISLAND'),
 (0.0019604156, 'NEAR BAY'),
 (0.00285647464, 'NEAR OCEAN'),
 (0.0103114883, '<1H OCEAN'),
 (0.0141064835, 'total_bedrooms'),
 (0.0142575993, 'households'),
 (0.0146726854, 'total_rooms'),
 (0.0148742809, 'population'),
 (0.0411437985, 'housing_median_age'),
 (0.0533510773, 'bedrooms_per_room'),
 (0.0564191792, 'rooms_per_hhold'),
 (0.0629090705, 'latitude'),
 (0.0733442355, 'longitude'),
 (0.108792957, 'pop_per_hhold'),
 (0.164780994, 'INLAND'),
 (0.366158981, 'median_income')]

1 回复 | 直到 6 年前

Jan 6 年前

使用lambda表达式:

lst =  [(0.366158981, 'median_income'),
 (0.164780994, 'INLAND'),
 (0.108792957, 'pop_per_hhold'),
 (0.0733442355, 'longitude'),
 (0.0629090705, 'latitude'),
 (0.0564191792, 'rooms_per_hhold'),
 (0.0533510773, 'bedrooms_per_room'),
 (0.0411437985, 'housing_median_age'),
 (0.0148742809, 'population'),
 (0.0146726854, 'total_rooms'),
 (0.0142575993, 'households'),
 (0.0141064835, 'total_bedrooms'),
 (0.0103114883, '<1H OCEAN'),
 (0.00285647464, 'NEAR OCEAN'),
 (0.0019604156, 'NEAR BAY'),
 (6.02803867e-05, 'ISLAND'),
 (0.0, 'some_var')]

lst = sorted(lst, key=lambda x: x[0])
print(lst)

[(0.0, 'some_var'), (6.02803867e-05, 'ISLAND'), (0.0019604156, 'NEAR BAY'), (0.00285647464, 'NEAR OCEAN'), (0.0103114883, '<1H OCEAN'), (0.0141064835, 'total_bedrooms'), (0.0142575993, 'households'), (0.0146726854, 'total_rooms'), (0.0148742809, 'population'), (0.0411437985, 'housing_median_age'), (0.0533510773, 'bedrooms_per_room'), (0.0564191792, 'rooms_per_hhold'), (0.0629090705, 'latitude'), (0.0733442355, 'longitude'), (0.108792957, 'pop_per_hhold'), (0.164780994, 'INLAND'), (0.366158981, 'median_income')]

或-与 reverse = True :

lst = sorted(lst, key=lambda x: x[0], reverse=True)
print(lst)

会屈服的

[(0.366158981, 'median_income'), (0.164780994, 'INLAND'), (0.108792957, 'pop_per_hhold'), (0.0733442355, 'longitude'), (0.0629090705, 'latitude'), (0.0564191792, 'rooms_per_hhold'), (0.0533510773, 'bedrooms_per_room'), (0.0411437985, 'housing_median_age'), (0.0148742809, 'population'), (0.0146726854, 'total_rooms'), (0.0142575993, 'households'), (0.0141064835, 'total_bedrooms'), (0.0103114883, '<1H OCEAN'), (0.00285647464, 'NEAR OCEAN'), (0.0019604156, 'NEAR BAY'), (6.02803867e-05, 'ISLAND'), (0.0, 'some_var')]

作为 @Barmar 正确地指出,您甚至不需要lambda表达式,因此


   lst = sorted(lst)

完全可以。

lst1 = sorted(lst, key=lambda x: x[0])
lst2 = sorted(lst)
assert lst1 == lst2