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          【Python】Pandas GroupBy 深度總結(jié)

          共 19034字,需瀏覽 39分鐘

           ·

          2022-06-13 01:05

          今天,我們將探討如何在 Python 的 Pandas 庫中創(chuàng)建 GroupBy 對象以及該對象的工作原理。我們將詳細(xì)了解分組過程的每個步驟,可以將哪些方法應(yīng)用于 GroupBy 對象上,以及我們可以從中提取哪些有用信息

          不要再觀望了,一起學(xué)起來吧

          使用 Groupby 三個步驟

          首先我們要知道,任何 groupby 過程都涉及以下 3 個步驟的某種組合:

          • 根據(jù)定義的標(biāo)準(zhǔn)將原始對象分成組
          • 對每個組應(yīng)用某些函數(shù)
          • 整合結(jié)果

          讓我先來大致瀏覽下今天用到的測試數(shù)據(jù)集

          import pandas as pd
          import numpy as np

          pd.set_option('max_columns'None)

          df = pd.read_csv('complete.csv')
          df = df[['awardYear''category''prizeAmount''prizeAmountAdjusted''name''gender''birth_continent']]
          df.head()

          Output:

          	awardYear	category	prizeAmount	prizeAmountAdjusted	name	gender	birth_continent
          0 2001 Economic Sciences 10000000 12295082 A. Michael Spence male North America
          1 1975 Physics 630000 3404179 Aage N. Bohr male Europe
          2 2004 Chemistry 10000000 11762861 Aaron Ciechanover male Asia
          3 1982 Chemistry 1150000 3102518 Aaron Klug male Europe
          4 1979 Physics 800000 2988048 Abdus Salam male Asia

          將原始對象拆分為組

          在這個階段,我們調(diào)用 pandas DataFrame.groupby() 函數(shù)。我們使用它根據(jù)預(yù)定義的標(biāo)準(zhǔn)將數(shù)據(jù)分組,沿行(默認(rèn)情況下,axis=0)或列(axis=1)。換句話說,此函數(shù)將標(biāo)簽映射到組的名稱。

          例如,在我們的案例中,我們可以按獎項類別對諾貝爾獎的數(shù)據(jù)進(jìn)行分組:

          grouped = df.groupby('category')

          也可以使用多個列來執(zhí)行數(shù)據(jù)分組,傳遞一個列列表即可。讓我們首先按獎項類別對我們的數(shù)據(jù)進(jìn)行分組,然后在每個創(chuàng)建的組中,我們將根據(jù)獲獎年份應(yīng)用額外的分組:

          grouped_category_year = df.groupby(['category''awardYear'])

          現(xiàn)在,如果我們嘗試打印剛剛創(chuàng)建的兩個 GroupBy 對象之一,我們實際上將看不到任何組:

          print(grouped)

          Output:

          <pandas.core.groupby.generic.DataFrameGroupBy object at 0x0000026083789DF0>

          我們要注意的是,創(chuàng)建 GroupBy 對象成功與否,只檢查我們是否通過了正確的映射;在我們顯式地對該對象使用某些方法或提取其某些屬性之前,都不會真正執(zhí)行拆分-應(yīng)用-組合鏈的任何操作

          為了簡要檢查生成的 GroupBy 對象并檢查組的拆分方式,我們可以從中提取組或索引屬性。它們都返回一個字典,其中鍵是創(chuàng)建的組,值是原始 DataFrame 中每個組的實例的軸標(biāo)簽列表(對于組屬性)或索引(對于索引屬性):

          grouped.indices

          Output:

          {'Chemistry': array([  2,   3,   7,   9,  10,  11,  13,  14,  15,  17,  19,  39,  62,
          64, 66, 71, 75, 80, 81, 86, 92, 104, 107, 112, 129, 135,
          153, 169, 175, 178, 181, 188, 197, 199, 203, 210, 215, 223, 227,
          239, 247, 249, 258, 264, 265, 268, 272, 274, 280, 282, 284, 289,
          296, 298, 310, 311, 317, 318, 337, 341, 343, 348, 352, 357, 362,
          365, 366, 372, 374, 384, 394, 395, 396, 415, 416, 419, 434, 440,
          442, 444, 446, 448, 450, 455, 456, 459, 461, 463, 465, 469, 475,
          504, 505, 508, 518, 522, 523, 524, 539, 549, 558, 559, 563, 567,
          571, 572, 585, 591, 596, 599, 627, 630, 632, 641, 643, 644, 648,
          659, 661, 666, 667, 668, 671, 673, 679, 681, 686, 713, 715, 717,
          719, 720, 722, 723, 725, 726, 729, 732, 738, 742, 744, 746, 751,
          756, 759, 763, 766, 773, 776, 798, 810, 813, 814, 817, 827, 828,
          829, 832, 839, 848, 853, 855, 862, 866, 880, 885, 886, 888, 889,
          892, 894, 897, 902, 904, 914, 915, 920, 921, 922, 940, 941, 943,
          946, 947], dtype=int64),
          'Economic Sciences': array([ 0, 5, 45, 46, 58, 90, 96, 139, 140, 145, 152, 156, 157,
          180, 187, 193, 207, 219, 231, 232, 246, 250, 269, 279, 283, 295,
          305, 324, 346, 369, 418, 422, 425, 426, 430, 432, 438, 458, 467,
          476, 485, 510, 525, 527, 537, 538, 546, 580, 594, 595, 605, 611,
          636, 637, 657, 669, 670, 678, 700, 708, 716, 724, 734, 737, 739,
          745, 747, 749, 750, 753, 758, 767, 800, 805, 854, 856, 860, 864,
          871, 882, 896, 912, 916, 924], dtype=int64),
          'Literature': array([ 21, 31, 40, 49, 52, 98, 100, 101, 102, 111, 115, 142, 149,
          159, 170, 177, 201, 202, 220, 221, 233, 235, 237, 253, 257, 259,
          275, 277, 278, 286, 312, 315, 316, 321, 326, 333, 345, 347, 350,
          355, 359, 364, 370, 373, 385, 397, 400, 403, 406, 411, 435, 439,
          441, 454, 468, 479, 480, 482, 483, 492, 501, 506, 511, 516, 556,
          569, 581, 602, 604, 606, 613, 614, 618, 631, 633, 635, 640, 652,
          653, 655, 656, 665, 675, 683, 699, 761, 765, 771, 774, 777, 779,
          780, 784, 786, 788, 796, 799, 803, 836, 840, 842, 850, 861, 867,
          868, 878, 881, 883, 910, 917, 919, 927, 928, 929, 930, 936],
          dtype=int64),
          'Peace': array([ 6, 12, 16, 25, 26, 27, 34, 36, 44, 47, 48, 54, 61,
          65, 72, 78, 79, 82, 95, 99, 116, 119, 120, 126, 137, 146,
          151, 166, 167, 171, 200, 204, 205, 206, 209, 213, 225, 236, 240,
          244, 255, 260, 266, 267, 270, 287, 303, 320, 329, 356, 360, 361,
          377, 386, 387, 388, 389, 390, 391, 392, 393, 433, 447, 449, 471,
          477, 481, 489, 491, 500, 512, 514, 517, 528, 529, 530, 533, 534,
          540, 542, 544, 545, 547, 553, 555, 560, 562, 574, 578, 590, 593,
          603, 607, 608, 609, 612, 615, 616, 617, 619, 620, 628, 634, 639,
          642, 664, 677, 688, 697, 703, 705, 710, 727, 736, 787, 793, 795,
          806, 823, 846, 847, 852, 865, 875, 876, 877, 895, 926, 934, 935,
          937, 944, 948, 949], dtype=int64),
          'Physics': array([ 1, 4, 8, 20, 23, 24, 30, 32, 38, 51, 59, 60, 67,
          68, 69, 70, 74, 84, 89, 97, 103, 105, 108, 109, 114, 117,
          118, 122, 125, 127, 128, 130, 133, 141, 143, 144, 155, 162, 163,
          164, 165, 168, 173, 174, 176, 179, 183, 195, 212, 214, 216, 222,
          224, 228, 230, 234, 238, 241, 243, 251, 256, 263, 271, 276, 291,
          292, 297, 301, 306, 307, 308, 323, 327, 328, 330, 335, 336, 338,
          349, 351, 353, 354, 363, 367, 375, 376, 378, 381, 382, 398, 399,
          402, 404, 405, 408, 410, 412, 413, 420, 421, 424, 428, 429, 436,
          445, 451, 453, 457, 460, 462, 470, 472, 487, 495, 498, 499, 509,
          513, 515, 521, 526, 532, 535, 536, 541, 548, 550, 552, 557, 561,
          564, 565, 566, 573, 576, 577, 579, 583, 586, 588, 592, 601, 610,
          621, 622, 623, 629, 647, 650, 651, 654, 658, 674, 676, 682, 684,
          690, 691, 693, 694, 695, 696, 698, 702, 707, 711, 714, 721, 730,
          731, 735, 743, 752, 755, 770, 772, 775, 781, 785, 790, 792, 797,
          801, 802, 808, 822, 833, 834, 835, 844, 851, 870, 872, 879, 884,
          887, 890, 893, 900, 901, 903, 905, 907, 908, 909, 913, 925, 931,
          932, 933, 938, 942, 945], dtype=int64),
          'Physiology or Medicine': array([ 18, 22, 28, 29, 33, 35, 37, 41, 42, 43, 50, 53, 55,
          56, 57, 63, 73, 76, 77, 83, 85, 87, 88, 91, 93, 94,
          106, 110, 113, 121, 123, 124, 131, 132, 134, 136, 138, 147, 148,
          150, 154, 158, 160, 161, 172, 182, 184, 185, 186, 189, 190, 191,
          192, 194, 196, 198, 208, 211, 217, 218, 226, 229, 242, 245, 248,
          252, 254, 261, 262, 273, 281, 285, 288, 290, 293, 294, 299, 300,
          302, 304, 309, 313, 314, 319, 322, 325, 331, 332, 334, 339, 340,
          342, 344, 358, 368, 371, 379, 380, 383, 401, 407, 409, 414, 417,
          423, 427, 431, 437, 443, 452, 464, 466, 473, 474, 478, 484, 486,
          488, 490, 493, 494, 496, 497, 502, 503, 507, 519, 520, 531, 543,
          551, 554, 568, 570, 575, 582, 584, 587, 589, 597, 598, 600, 624,
          625, 626, 638, 645, 646, 649, 660, 662, 663, 672, 680, 685, 687,
          689, 692, 701, 704, 706, 709, 712, 718, 728, 733, 740, 741, 748,
          754, 757, 760, 762, 764, 768, 769, 778, 782, 783, 789, 791, 794,
          804, 807, 809, 811, 812, 815, 816, 818, 819, 820, 821, 824, 825,
          826, 830, 831, 837, 838, 841, 843, 845, 849, 857, 858, 859, 863,
          869, 873, 874, 891, 898, 899, 906, 911, 918, 923, 939], dtype=int64)}

          要查找 GroupBy 對象中的組數(shù),我們可以從中提取 ngroups 屬性或調(diào)用 Python 標(biāo)準(zhǔn)庫的 len 函數(shù):

          print(grouped.ngroups)
          print(len(grouped))

          Output:

          6
          6

          如果我們需要可視化每個組的所有或部分條目,那么可以遍歷 GroupBy 對象:

          for name, entries in grouped:
              print(f'First 2 entries for the "{name}" category:')
              print(30*'-')
              print(entries.head(2), '\n\n')

          Output:

          First 2 entries for the "Chemistry" category:
          ------------------------------
          awardYear category prizeAmount prizeAmountAdjusted name \
          2 2004 Chemistry 10000000 11762861 Aaron Ciechanover
          3 1982 Chemistry 1150000 3102518 Aaron Klug

          gender birth_continent
          2 male Asia
          3 male Europe

          First 2 entries for the "Economic Sciences" category:
          ------------------------------
          awardYear category prizeAmount prizeAmountAdjusted \
          0 2001 Economic Sciences 10000000 12295082
          5 2019 Economic Sciences 9000000 9000000

          name gender birth_continent
          0 A. Michael Spence male North America
          5 Abhijit Banerjee male Asia

          First 2 entries for the "Literature" category:
          ------------------------------
          awardYear category prizeAmount prizeAmountAdjusted \
          21 1957 Literature 208629 2697789
          31 1970 Literature 400000 3177966

          name gender birth_continent
          21 Albert Camus male Africa
          31 Alexandr Solzhenitsyn male Europe

          First 2 entries for the "Peace" category:
          ------------------------------
          awardYear category prizeAmount prizeAmountAdjusted \
          6 2019 Peace 9000000 9000000
          12 1980 Peace 880000 2889667

          name gender birth_continent
          6 Abiy Ahmed Ali male Africa
          12 Adolfo Pérez Esquivel male South America

          First 2 entries for the "Physics" category:
          ------------------------------
          awardYear category prizeAmount prizeAmountAdjusted name gender \
          1 1975 Physics 630000 3404179 Aage N. Bohr male
          4 1979 Physics 800000 2988048 Abdus Salam male

          birth_continent
          1 Europe
          4 Asia

          First 2 entries for the "Physiology or Medicine" category:
          ------------------------------
          awardYear category prizeAmount prizeAmountAdjusted \
          18 1963 Physiology or Medicine 265000 2839286
          22 1974 Physiology or Medicine 550000 3263449

          name gender birth_continent
          18 Alan Hodgkin male Europe
          22 Albert Claude male Europe

          相反,如果我們想以 DataFrame 的形式選擇單個組,我們應(yīng)該在 GroupBy 對象上使用 get_group() 方法:

          grouped.get_group('Economic Sciences')

          Output:

          	awardYear	category	prizeAmount	prizeAmountAdjusted	name	gender	birth_continent
          0 2001 Economic Sciences 10000000 12295082 A. Michael Spence male North America
          5 2019 Economic Sciences 9000000 9000000 Abhijit Banerjee male Asia
          45 2012 Economic Sciences 8000000 8361204 Alvin E. Roth male North America
          46 1998 Economic Sciences 7600000 9713701 Amartya Sen male Asia
          58 2015 Economic Sciences 8000000 8384572 Angus Deaton male Europe
          … … … … … … … …
          882 2002 Economic Sciences 10000000 12034660 Vernon L. Smith male North America
          896 1973 Economic Sciences 510000 3331882 Wassily Leontief male Europe
          912 2018 Economic Sciences 9000000 9000000 William D. Nordhaus male North America
          916 1990 Economic Sciences 4000000 6329114 William F. Sharpe male North America
          924 1996 Economic Sciences 7400000 9490424 William Vickrey male North America

          按組應(yīng)用函數(shù)

          在拆分原始數(shù)據(jù)并檢查結(jié)果組之后,我們可以對每個組執(zhí)行以下操作之一或其組合:

          • Aggregation(聚合):計算每個組的匯總統(tǒng)計量(例如,組大小、平均值、中位數(shù)或總和)并為許多數(shù)據(jù)點輸出單個數(shù)字
          • Transformation(變換):按組進(jìn)行一些操作,例如計算每個組的z-score
          • Filtration(過濾):根據(jù)預(yù)定義的條件拒絕某些組,例如組大小、平均值、中位數(shù)或總和,還可以包括從每個組中過濾掉特定的行

          Aggregation

          要聚合 GroupBy 對象的數(shù)據(jù)(即按組計算匯總統(tǒng)計量),我們可以在對象上使用 agg() 方法:

          # Showing only 1 decimal for all float numbers
          pd.options.display.float_format = '{:.1f}'.format

          grouped.agg(np.mean)

          Output:

          	awardYear	prizeAmount	prizeAmountAdjusted
          category
          Chemistry 1972.3 3629279.4 6257868.1
          Economic Sciences 1996.1 6105845.2 7837779.2
          Literature 1960.9 2493811.2 5598256.3
          Peace 1964.5 3124879.2 6163906.9
          Physics 1971.1 3407938.6 6086978.2
          Physiology or Medicine 1970.4 3072972.9 5738300.7

          上面的代碼生成一個 DataFrame,其中組名作為其新索引,每個數(shù)字列的平均值作為分組

          我們可以直接在 GroupBy 對象上應(yīng)用其他相應(yīng)的 Pandas 方法,而不僅僅是使用 agg() 方法。最常用的方法是 mean()median()mode()sum()size()count()min()max()std()var()(計算每個的方差 group)、describe()(按組輸出描述性統(tǒng)計信息)和 nunique()(給出每個組中唯一值的數(shù)量)

          grouped.sum()

          Output:

          	awardYear	prizeAmount	prizeAmountAdjusted
          category
          Chemistry 362912 667787418 1151447726
          Economic Sciences 167674 512891000 658373449
          Literature 227468 289282102 649397731
          Peace 263248 418733807 825963521
          Physics 419837 725890928 1296526352
          Physiology or Medicine 431508 672981066 1256687857

          通常情況下我們只對某些特定列或列的統(tǒng)計信息感興趣,因此我們需要指定它們。在上面的例子中,我們絕對不想總結(jié)所有年份,相應(yīng)的我們可能希望按獎品類別對獎品價值求和。為此我們可以選擇 GroupBy 對象的 PrizeAmountAdjusted 列,就像我們選擇 DataFrame 的列,然后對其應(yīng)用 sum() 函數(shù):

          grouped['prizeAmountAdjusted'].sum()

          Output:

          category
          Chemistry 1151447726
          Economic Sciences 658373449
          Literature 649397731
          Peace 825963521
          Physics 1296526352
          Physiology or Medicine 1256687857
          Name: prizeAmountAdjusted, dtype: int64

          對于上面的代碼片段,我們可以在選擇必要的列之前使用對 GroupBy 對象應(yīng)用函數(shù)的等效語法:grouped.sum()['prizeAmountAdjusted']。但是前面的語法更可取,因為它的性能更好,尤其是在大型數(shù)據(jù)集上,效果更為明顯

          如果我們需要聚合兩列或更多列的數(shù)據(jù),我們使用雙方括號:

          grouped[['prizeAmount''prizeAmountAdjusted']].sum()

          Output:

          	prizeAmount	prizeAmountAdjusted
          category
          Chemistry 667787418 1151447726
          Economic Sciences 512891000 658373449
          Literature 289282102 649397731
          Peace 418733807 825963521
          Physics 725890928 1296526352
          Physiology or Medicine 672981066 1256687857

          可以一次將多個函數(shù)應(yīng)用于 GroupBy 對象的一列或多列。為此我們再次需要 agg() 方法和感興趣的函數(shù)列表:

          grouped[['prizeAmount''prizeAmountAdjusted']].agg([np.sum, np.mean, np.std])

          Output:

          	prizeAmount	prizeAmountAdjusted
          sum mean std sum mean std
          category
          Chemistry 667787418 3629279.4 4070588.4 1151447726 6257868.1 3276027.2
          Economic Sciences 512891000 6105845.2 3787630.1 658373449 7837779.2 3313153.2
          Literature 289282102 2493811.2 3653734.0 649397731 5598256.3 3029512.1
          Peace 418733807 3124879.2 3934390.9 825963521 6163906.9 3189886.1
          Physics 725890928 3407938.6 4013073.0 1296526352 6086978.2 3294268.5
          Physiology or Medicine 672981066 3072972.9 3898539.3 1256687857 5738300.7 3241781.0

          此外,我們可以考慮通過傳遞字典將不同的聚合函數(shù)應(yīng)用于 GroupBy 對象的不同列:

          grouped.agg({'prizeAmount': [np.sum, np.size], 'prizeAmountAdjusted': np.mean})

          Output:

          	prizeAmount	prizeAmountAdjusted
          sum size mean
          category
          Chemistry 667787418 184 6257868.1
          Economic Sciences 512891000 84 7837779.2
          Literature 289282102 116 5598256.3
          Peace 418733807 134 6163906.9
          Physics 725890928 213 6086978.2
          Physiology or Medicine 672981066 219 5738300.7

          Transformation

          與聚合方法不同,轉(zhuǎn)換方法返回一個新的 DataFrame,其形狀和索引與原始 DataFrame 相同,但具有轉(zhuǎn)換后的各個值。這里需要注意的是,transformation 一定不能修改原始 DataFrame 中的任何值,也就是這些操作不能原地執(zhí)行

          轉(zhuǎn)換 GroupBy 對象數(shù)據(jù)的最常見的 Pandas 方法是 transform()。例如它可以幫助計算每個組的 z-score:

          grouped[['prizeAmount''prizeAmountAdjusted']].transform(lambda x: (x - x.mean()) / x.std())

          Output:

          	prizeAmount	prizeAmountAdjusted
          0 1.0 1.3
          1 -0.7 -0.8
          2 1.6 1.7
          3 -0.6 -1.0
          4 -0.6 -0.9
          … … …
          945 -0.7 -0.8
          946 -0.8 -1.1
          947 -0.9 0.3
          948 -0.5 -1.0
          949 -0.7 -1.0

          使用轉(zhuǎn)換方法,我們還可以用組均值、中位數(shù)、眾數(shù)或任何其他值替換缺失數(shù)據(jù):

          grouped['gender'].transform(lambda x: x.fillna(x.mode()[0]))

          Output:

          0        male
          1 male
          2 male
          3 male
          4 male
          ...
          945 male
          946 male
          947 female
          948 male
          949 male
          Name: gender, Length: 950, dtype: object

          我們當(dāng)然還可以使用其他一些 Pandas 方法來轉(zhuǎn)換 GroupBy 對象的數(shù)據(jù):bfill()ffill()diff()pct_change()rank()shift()quantile()

          Filtration

          過濾方法根據(jù)預(yù)定義的條件從每個組中丟棄組或特定行,并返回原始數(shù)據(jù)的子集。例如我們可能希望只保留所有組中某個列的值,其中該列的組均值大于預(yù)定義值。在我們的 DataFrame 的情況下,讓我們過濾掉所有組均值小于 7,000,000 的prizeAmountAdjusted 列,并在輸出中僅保留該列:

          grouped['prizeAmountAdjusted'].filter(lambda x: x.mean() > 7000000)

          Output:

          0      12295082
          5 9000000
          45 8361204
          46 9713701
          58 8384572
          ...
          882 12034660
          896 3331882
          912 9000000
          916 6329114
          924 9490424
          Name: prizeAmountAdjusted, Length: 84, dtype: int64

          另一個例子是過濾掉具有超過一定數(shù)量元素的組:

          grouped['prizeAmountAdjusted'].filter(lambda x: len(x) < 100)

          Output:

          0      12295082
          5 9000000
          45 8361204
          46 9713701
          58 8384572
          ...
          882 12034660
          896 3331882
          912 9000000
          916 6329114
          924 9490424
          Name: prizeAmountAdjusted, Length: 84, dtype: int64

          在上述兩個操作中,我們使用了 filter() 方法,將 lambda 函數(shù)作為參數(shù)傳遞。這樣的函數(shù),應(yīng)用于整個組,根據(jù)該組與預(yù)定義統(tǒng)計條件的比較結(jié)果返回 TrueFalse。換句話說,filter()方法中的函數(shù)決定了哪些組保留在新的 DataFrame 中

          除了過濾掉整個組之外,還可以從每個組中丟棄某些行。這里有一些有用的方法是 first()last()nth()。將其中一個應(yīng)用于 GroupBy 對象會相應(yīng)地返回每個組的第一個/最后一個/第 n 個條目:

          grouped.last()

          Output:

          	awardYear	prizeAmount	prizeAmountAdjusted	name	gender	birth_continent
          category
          Chemistry 1911 140695 7327865 Marie Curie female Europe
          Economic Sciences 1996 7400000 9490424 William Vickrey male North America
          Literature 1968 350000 3052326 Yasunari Kawabata male Asia
          Peace 1963 265000 2839286 International Committee of the Red Cross male Asia
          Physics 1972 480000 3345725 John Bardeen male North America
          Physiology or Medicine 2016 8000000 8301051 Yoshinori Ohsumi male Asia

          對于 nth() 方法,我們必須傳遞表示要為每個組返回的條目索引的整數(shù):

          grouped.nth(1)

          Output:

          	awardYear	prizeAmount	prizeAmountAdjusted	name	gender	birth_continent
          category
          Chemistry 1982 1150000 3102518 Aaron Klug male Europe
          Economic Sciences 2019 9000000 9000000 Abhijit Banerjee male Asia
          Literature 1970 400000 3177966 Alexandr Solzhenitsyn male Europe
          Peace 1980 880000 2889667 Adolfo Pérez Esquivel male South America
          Physics 1979 800000 2988048 Abdus Salam male Asia
          Physiology or Medicine 1974 550000 3263449 Albert Claude male Europe

          上面的代碼收集了所有組的第二個條目

          另外兩個過濾每個組中的行的方法是 head()tail(),分別返回每個組的第一/最后 n 行(默認(rèn)為 5):

          grouped.head(3)

          Output:

          	awardYear	category	prizeAmount	prizeAmountAdjusted	name	gender	birth_continent
          0 2001 Economic Sciences 10000000 12295082 A. Michael Spence male North America
          1 1975 Physics 630000 3404179 Aage N. Bohr male Europe
          2 2004 Chemistry 10000000 11762861 Aaron Ciechanover male Asia
          3 1982 Chemistry 1150000 3102518 Aaron Klug male Europe
          4 1979 Physics 800000 2988048 Abdus Salam male Asia
          5 2019 Economic Sciences 9000000 9000000 Abhijit Banerjee male Asia
          6 2019 Peace 9000000 9000000 Abiy Ahmed Ali male Africa
          7 2009 Chemistry 10000000 10958504 Ada E. Yonath female Asia
          8 2011 Physics 10000000 10545557 Adam G. Riess male North America
          12 1980 Peace 880000 2889667 Adolfo Pérez Esquivel male South America
          16 2007 Peace 10000000 11301989 Al Gore male North America
          18 1963 Physiology or Medicine 265000 2839286 Alan Hodgkin male Europe
          21 1957 Literature 208629 2697789 Albert Camus male Africa
          22 1974 Physiology or Medicine 550000 3263449 Albert Claude male Europe
          28 1937 Physiology or Medicine 158463 4716161 Albert Szent-Gy?rgyi male Europe
          31 1970 Literature 400000 3177966 Alexandr Solzhenitsyn male Europe
          40 2013 Literature 8000000 8365867 Alice Munro female North America
          45 2012 Economic Sciences 8000000 8361204 Alvin E. Roth male North America

          整合結(jié)果

          split-apply-combine 鏈的最后一個階段——合并結(jié)果——由Ppandas 在后臺執(zhí)行。它包括獲取在 GroupBy 對象上執(zhí)行的所有操作的輸出并將它們重新組合在一起,生成新的數(shù)據(jù)結(jié)構(gòu),例如 Series 或 DataFrame。將此數(shù)據(jù)結(jié)構(gòu)分配給一個變量,我們可以用它來解決其他任務(wù)

          總結(jié)

          今天我們介紹了使用 pandas groupby 函數(shù)和使用結(jié)果對象的許多知識

          • 分組過程所包括的步驟
          • split-apply-combine 鏈?zhǔn)侨绾我徊揭徊焦ぷ鞯?/section>
          • 如何創(chuàng)建 GroupBy 對象
          • 如何簡要檢查 GroupBy 對象
          • GroupBy 對象的屬性
          • 可應(yīng)用于 GroupBy 對象的操作
          • 如何按組計算匯總統(tǒng)計量以及可用于此目的的方法
          • 如何一次將多個函數(shù)應(yīng)用于 GroupBy 對象的一列或多列
          • 如何將不同的聚合函數(shù)應(yīng)用于 GroupBy 對象的不同列
          • 如何以及為什么要轉(zhuǎn)換原始 DataFrame 中的值
          • 如何過濾 GroupBy 對象的組或每個組的特定行
          • Pandas 如何組合分組過程的結(jié)果
          • 分組過程產(chǎn)生的數(shù)據(jù)結(jié)構(gòu)

          好了,這就是今天分享的全部內(nèi)容,喜歡就點個吧~


          往期精彩回顧




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