• python实用库记录


    1:Counter对列表进行计数,找重复

    from collections import Counter
    
    a = ["61e7f868eeb03046b96906dd","61e7f861fe380309df61ddf2","61e7f861fe380309df61ddf1",
         "61e7f861fe380309df61ddf0","61e7f85b7258f7696c55f72e","61e7f85b7258f7696c55f72d","61e7f861fe380309df61ddf0"
    ]
    
    b = dict(Counter(a))  #{'61e7f868eeb03046b96906dd': 1, '61e7f861fe380309df61ddf2': 1, '61e7f861fe380309df61ddf1': 1, '61e7f861fe380309df61ddf0': 2, '61e7f85b7258f7696c55f72e': 1, '61e7f85b7258f7696c55f72d': 1}
    print(b)
    # print ([key for key,value in b.items() if value > 1]) #只展示重复元素
    # print ({key:value for key,value in b.items() if value > 1}) #展现重复元素和重复次数
    
    # a = [{'pay_order_no': '164264954668065756', 'is_del': 0}, {'pay_order_no': '164264972082245062', 'is_del': 1}, {'pay_order_no': '164264996092254304', 'is_del': 1}, {'pay_order_no': '164265021689661895', 'is_del': 0}, {'pay_order_no': '164265063783351017', 'is_del': 0}, {'pay_order_no': '164265075567892474', 'is_del': 0}, {'pay_order_no': '164265076330341903', 'is_del': 0}, {'pay_order_no': '164265087134341611', 'is_del': 1}, {'pay_order_no': '164265089388312149', 'is_del': 1}, {'pay_order_no': '164265101568633361', 'is_del': 0}]
    

    2:asq。query

    from asq import query
    
    a = [
        {
            "card_id": 210,
            "donate_count": 6,
            "id": 210,
            "card_name": "这是一张犯人卡",
            "card_num": "AAA6",
            "card_grade_id": 158,
            "card_rarity": 6
        },{
            "card_id": 210,
            "donate_count": 6,
            "id": 210,
            "card_name": "这是一张犯人卡",
            "card_num": "AAA6",
            "card_grade_id": 158,
            "card_rarity": 6
        }, {
            "card_id": 210,
            "donate_count": 6,
            "id": 210,
            "card_name": "这是一张犯人卡",
            "card_num": "AAA6",
            "card_grade_id": 160,
            "card_rarity": 6
        }
    ]
    
    a  = query(a).group_by(lambda x:x["card_grade_id"]).select(lambda x:[x[0]["card_grade_id"],x.sum(lambda x:x["donate_count"])]).to_list()
    print(a) # [158, 12][160, 6]
    

      

     

  • 相关阅读:
    JMeter接口测试系列:Jmeter+jenkins+ant 的自动化构建
    BZOJ1434:[ZJOI2009]染色游戏(博弈论)
    BZOJ4241:历史研究(回滚莫队)
    BZOJ2281:[SDOI2011]黑白棋(博弈论,组合数学,DP)
    BZOJ1188:[HNOI2007]分裂游戏(博弈论)
    BZOJ1022:[SHOI2008]小约翰的游戏John(博弈论)
    BZOJ1406:[AHOI2007]密码箱(数论)
    BZOJ3270:博物馆(高斯消元)
    CF993E:Nikita and Order Statistics(FFT)
    HDU4609:3-idiots(FFT)
  • 原文地址:https://www.cnblogs.com/hero799/p/15842603.html
Copyright © 2020-2023  润新知