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Python实现字典排序、按照list中字典的某个key排序的方法示例

更新时间:2020-06-16 12:48:01 作者:startmvc
本文实例讲述了Python实现字典排序、按照list中字典的某个key排序的方法。分享给大家供大

本文实例讲述了Python实现字典排序、按照list中字典的某个key排序的方法。分享给大家供大家参考,具体如下:

1.给字典按照value按照从大到小排序

排序


dict = {'a':21, 'b':5, 'c':3, 'd':54, 'e':74, 'f':0}
new_dict = sorted(dict.iteritems(), key=lambda d:d[1], reverse = True)
print new_dict

输出:

[('e', 74), ('d', 54), ('a', 21), ('b', 5), ('c', 3), ('f', 0)]

2. python按照list中的字典的某key排序:

例子:


s=[
{"no":28,"score":90},
{"no":25,"score":90},
{"no":1,"score":100},
{"no":2,"score":20},
]
print "original s: ",s
# 单级排序,仅按照score排序
new_s = sorted(s,key = lambda e:e.__getitem__('score'))
print "new s: ", new_s
# 多级排序,先按照score,再按照no排序
new_s_2 = sorted(new_s,key = lambda e:(e.__getitem__('score'),e.__getitem__('no')))
print "new_s_2: ", new_s_2

输出:

original s:  [{'score': 90, 'no': 28}, {'score': 90, 'no': 25}, {'score': 100, 'no': 1}, {'score': 20, 'no': 2}] new s:  [{'score': 20, 'no': 2}, {'score': 90, 'no': 28}, {'score': 90, 'no': 25}, {'score': 100, 'no': 1}] new_s_2:  [{'score': 20, 'no': 2}, {'score': 90, 'no': 25}, {'score': 90, 'no': 28}, {'score': 100, 'no': 1}]

说明

1.new_s和new_s2的区别在于当score均为90的时候,重新按照no排序

2.顺序为从小到大,若在sorted函数的参数加上reverse = True则为从大到小

PS:这里再为大家推荐一款关于排序的演示工具供大家参考:

在线动画演示插入/选择/冒泡/归并/希尔/快速排序算法过程工具: http://tools.jb51.net/aideddesign/paixu_ys