如何根据较低的频率进行去重

Question

我有一张像这样的桌子

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我希望能够保持最高频率的品牌ids。例如，在奥迪的情况下，两个in具有相同的频率，因此只保留一个。在梅赛德斯-奔驰的情况下，保留后一个，因为它的频率是7。

这是我的数据框架：

{'Brand': 
 {0: 'audi',
  1: 'audi',
  2: 'bmw',
  3: 'dacia',
  4: 'fiat',
  5: 'ford',
  6: 'ford',
  7: 'honda',
  8: 'honda',
  9: 'hyundai',
  10: 'kia',
  11: 'mercedes-benz',
  12: 'mercedes-benz',
  13: 'nissan',
  14: 'nissan',
  15: 'opel',
  16: 'renault',
  17: 'renault',
  18: 'renault',
  19: 'renault',
  20: 'toyota',
  21: 'toyota',
  22: 'volvo',
  23: 'vw',
  24: 'vw',
  25: 'vw',
  26: 'vw'},
 'id': 
 {0: 'audi_a4_dynamic_2016_otomatik',
  1: 'audi_a6_standart_2015_otomatik',
  2: 'bmw_5 series_executive_2016_otomatik',
  3: 'dacia_duster_laureate_2017_manuel',
  4: 'fiat_egea_easy_2017_manuel',
  5: 'ford_focus_trend x_2015_manuel',
  6: 'ford_focus_trend x_2015_otomatik',
  7: 'honda_civic_eco elegance_2017_otomatik',
  8: 'honda_cr-v_executive_2018_otomatik',
  9: 'hyundai_tucson_elite plus_2017_otomatik',
  10: 'kia_sportage_concept plus_2015_otomatik',
  11: 'mercedes-benz_c-class_amg_2016_otomatik',
  12: 'mercedes-benz_e-class_edition e_2015_otomatik',
  13: 'nissan_qashqai_black edition_2014_manuel',
  14: 'nissan_qashqai_sky pack_2015_otomatik',
  15: 'opel_astra_edition_2016_manuel',
  16: 'renault_clio_joy_2016_manuel',
  17: 'renault_kadjar_icon_2015_otomatik',
  18: 'renault_kadjar_icon_2016_otomatik',
  19: 'renault_mégane_touch_2017_otomatik',
  20: 'toyota_corolla_touch_2015_otomatik',
  21: 'toyota_corolla_touch_2016_otomatik',
  22: 'volvo_s60_advance_2018_otomatik',
  23: 'vw_jetta_comfortline_2013_otomatik',
  24: 'vw_passat_highline_2017_otomatik',
  25: 'vw_tiguan_sport&style_2012_manuel',
  26: 'vw_tiguan_sport&style_2013_manuel'},
 'freq': {0: 4,
  1: 4,
  2: 7,
  3: 4,
  4: 4,
  5: 4,
  6: 4,
  7: 4,
  8: 4,
  9: 4,
  10: 4,
  11: 4,
  12: 7,
  13: 4,
  14: 4,
  15: 4,
  16: 4,
  17: 4,
  18: 4,
  19: 4,
  20: 4,
  21: 4,
  22: 4,
  23: 4,
  24: 7,
  25: 4,
  26: 4}}

编辑:尝试了其中一个答案，获得了额外的标题级别

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Answer 1

您需要对Brand进行pandas.groupby，然后根据最大频率进行聚合。

像这样的东西应该是有效的：

df.groupby('Brand')[['id', 'freq']].agg({'freq': 'max'})

Answer 2

要获得结果，请运行以下命令：

result = df.groupby('Brand', as_index=False).apply(
    lambda grp: grp[grp.freq == grp.freq.max()].iloc[0])