我想选择其BETAdn列总和为0的4个资产。大矩阵称为ALPHABETA＆amp;我运行以下命令来获取满足sum to zero参数的资产子集。
AB <- subset(data.frame((ALPHABETA)), BETAdn+BETAdn >= -0.01 & BETAdn+BETAdn <= .01)
上面的代码返回以下内容：
AB <- structure(list(ALPHAdn = c(0, -2e-04, -2e-04, -0.002, 0.0129,
-0.0071, 7e-04, -0.0019, -2e-04, -2e-04, -2e-04, -0.0011, -2e-04,
-2e-04, -2e-04, 0.001, -2e-04, -2e-04, -2e-04, -0.0068, -2e-04,
-2e-04, -0.0016, -2e-04, -2e-04, -2e-04, -0.0029, 0.002, 3e-04,
0, -2e-04, -2e-04, -2e-04, -2e-04, -0.0072, 0.0028, -2e-04, -2e-04,
-2e-04, -2e-04, -0.0025, -0.003, -2e-04, -0.0026, -2e-04, -2e-04,
-2e-04, -0.002, -2e-04, -2e-04, -2e-04, -2e-04, -2e-04, -2e-04,
-2e-04, 0.0012, -9e-04, -2e-04, -2e-04, -2e-04, -2e-04, -2e-04,
-2e-04, -2e-04, 0.0106, -2e-04), BETAup = c(0, 0.7181, 0.2041,
2.0982, -3.6778, 0.2532, 0.0964, 0.4937, 0.6152, -0.597, -0.1563,
-0.0014, 0.1627, -0.1742, -0.0131, 0.8372, 0.2327, -0.0019, -0.4405,
1.3106, 0.44, -1.1557, 0.3253, -0.1607, 0.0171, 0.007, 1.0561,
0.4863, 0.2045, 0.4859, 0.3233, 0.7361, -0.0787, 0.273, 0.9931,
1.151, 0.1011, 0.1894, -0.2742, 0.0069, 0.1003, -0.2177, 0.0437,
0.9002, 0.4438, -0.0501, -0.031, 0.4498, 0.0115, 0.0031, 1e-04,
-0.0606, 0.0119, 1.4144, -0.0802, -0.004, 0.6779, -1.983, 0.818,
0.2103, 0.204, 0.4613, 0.058, 0.3685, -1.9407, 1.1002), BETAdn = c(0,
4e-04, 4e-04, 0.0012, 7e-04, 9e-04, -0.0034, -0.0011, 4e-04,
4e-04, 4e-04, -0.0027, 4e-04, 4e-04, 4e-04, 0.0033, 4e-04, 7e-04,
4e-04, 0.0048, 4e-04, 4e-04, -0.0021, 4e-04, 4e-04, 4e-04, -0.0047,
-0.002, 4e-04, 2e-04, -9e-04, 4e-04, 4e-04, 4e-04, 0.0015, 0.005,
4e-04, -1e-04, 0.0033, 4e-04, -0.0013, -0.0049, -0.001, 0.0048,
4e-04, 4e-04, 4e-04, 0.0048, 4e-04, 4e-04, 4e-04, 4e-04, 4e-04,
4e-04, 4e-04, 0.0046, -0.0023, 4e-04, 4e-04, 4e-04, 4e-04, 4e-04,
4e-04, 4e-04, 0.0042, 4e-04), ALPHAup = c(0, -0.0107, -0.0029,
4e-04, 0.027, 0.0137, 2e-04, -0.0016, -0.0043, 0.0084, -3e-04,
-9e-04, 0.0032, -0.0018, 0.008, 0.0031, -8e-04, 0, 5e-04, 0.0061,
0.0083, -9e-04, -4e-04, 0.0025, -0.005, 0.0011, 2e-04, -0.006,
-0.0026, -4e-04, 0.0018, -0.0055, -3e-04, -0.0059, 0.0053, -0.0094,
-0.002, 0.0015, 0.0021, 1e-04, 0.0048, 0.0101, -0.0015, 0.0068,
0.0071, 0.0041, -2e-04, -0.0086, -2e-04, -6e-04, -2e-04, 0.0015,
-0.0017, -0.005, -0.004, -0.0015, 0.0023, 0.0288, 0.0011, -0.0074,
-0.0041, -0.0092, -0.0011, -0.0066, 0.0137, -0.0124), UPpval = c(NaN,
0.7812, 0.5756, 0.039, 0.0162, 0.1637, 0.5131, 0.5469, 0.2043,
0.5301, 0.1703, 0.8713, 0.3905, 0.1556, 0.4194, 0.1435, 0.0175,
0.1676, 0.0891, 8e-04, 0.1421, 0.1439, 0.4381, 0.5572, 0.6719,
0.3788, 0, 0.4341, 0.404, 0.1174, 0.6871, 0.0417, 0.1056, 0.9757,
0.0019, 0.0102, 0.9733, 0.1693, 0.4, 0.1847, 0.6531, 0.385, 0.8424,
7e-04, 0.2976, 0.6284, 0.2896, 0.926, 0.8516, 0.5919, 0.5761,
0.837, 0.5985, 1e-04, 0.4427, 0.2718, 0.0776, 0.5231, 0.1577,
0.4862, 0.9704, 0.9688, 0.9509, 0.8703, 0.2898, 0.1635), DNpval = c(NaN,
0.5758, 0.324, 0.9876, 0.302, 0.433, 0.9699, 0.9317, 0.6168,
0.3043, 0.9366, 0.8997, 0.7444, 0.7153, 0.5165, 0.8601, 0.7079,
0.01, 0.9436, 0.6197, 0.5857, 0.9662, 0.9648, 0.184, 0.7238,
0.544, 0.9857, 0.3622, 0.3005, 0.9561, 0.9477, 0.3751, 0.8279,
0.4774, 0.6118, 0.1958, 0.4617, 0.7703, 0.6927, 0.6274, 0.809,
0.2253, 0.6894, 0.466, 0.7238, 0.6158, 0.8311, 0.3922, 0.4567,
0.6448, 0.6685, 0.3738, 0.6351, 0.5341, 0.6739, 0.4371, 0.851,
0.2468, 0.9466, 0.1929, 0.4586, 0.4569, 0.3455, 0.4226, 0.6683,
0.197)), .Names = c("ALPHAdn", "BETAup", "BETAdn", "ALPHAup",
"UPpval", "DNpval"), row.names = c("DGS6MO", "FLAGSHIP.EXPRESS.INC",
"AVESIS.INC", "FRANKLIN.ELECTRONIC.PUBLSHS.INC", "THERAGENICS.CORP",
"AIRSENSORS.INC", "BANK.OF.EAST.TENNESSEE", "ADTEC.INC", "ORBIS.INC",
"MAGIC.YEARS.CHILD...LRNG.CTR.INC", "DIGITAL.OPTRONICS.CORP",
"SPECTRUM.INFORMATION.TECHS.INC", "ENTREE.CORP", "VITEL.FIBER.OPTICS.CORP",
"ALS.FORMAL.WEAR.INC", "AMERICAN.ECOLOGY.CORP", "AUTHENTICOLOR.INC",
"COMPARATOR.SYSTEMS.CORP", "DATATAB.INC", "DOUGLAS...LOMASON.CO",
"GENERAL.BUILDERS.CORP", "EQUIPMENT.COMPANY.AMER", "SAN.CARLOS.MILLING.INC",
"FORTUNE.NATIONAL.CORP", "KIMCO.ENERGY.CORP", "KREISLER.MANUFACTURING.CORP",
"RUDDICK.CORP", "NORDSON.CORP", "TANDYCRAFTS.INC", "HUNT.MANUFACTURING.CO",
"MOUNTAIN.MEDICAL.EQUIP.INC", "ALTEX.INDUSTRIES.INC", "RAND.CAPITAL.CORP",
"S.F.E.TECHNOLOGIES", "UNO.RESTAURANT.CORP", "HIGH.YIELD.PLUS.FD.INC",
"TELEPHONE.SPECIALISTS.INC", "PATRIOT.PREMIUM.FD.I", "SAGE.SOFTWARE.INC.INTERSOLV",
"CITIZENS.SECURITY.GROUP.INC", "GATEWAY.FED.CORP", "ASIAMERICA.EQUITIES.LTD",
"NORCEN.ENERGY.RESOURCES.LTD", "DUTY.FREE.INTERNATIONAL.INC",
"CATTLEGUARD.INC", "TIMBERLINE.MINERALS.INC", "SUMMIT.FINANCIAL.CORP",
"RATHBONE.KING...SEELEY.INC", "INTERNATIONAL.SPECIALTY.PRODUCTS",
"CHIPPEWA.RESOURCES.CORP", "ADVANCED.PROMOTION.TECHS.INC", "BEVERLY.HILLS.FAN.COMPANY",
"JAKO.INC", "NEVADA.GOLDFIELDS.CORP", "QUANTUM.VENTURES.GROUP.INC",
"VINDICATOR.OF.FLORIDA.INC", "T.J.INTERNATIONAL.INC", "KWIK.PRODUCTS.INTERNATIONAL.CORP",
"BETHLEHEM.RESOURCES.CORP", "TRIAN.EQUITIES.LTD", "CONSOLIDATED.MERCANTILE.CORP",
"UNITED.SECURITY.FINL.CORP.IL", "EQUIVEST.FINANCE.INC", "I.V.C.I.CORP",
"KINGS.ROAD.ENTERTAINMENT.INC", "SPORTING.LIFE.INC"), class = "data.frame")
问题是它返回66个资产但我只想要最好的4个满足参数。如何再次对其进行子集以获得所需的输出？
答案 0 :(得分：1)
您可以使用top_n中的dplyr来选择＆＃39; n`顶行
library(dplyr)
top_n(AB, 4, BETAdn)
或者使用order中的base R，然后将顶部＆＃39; n＆＃39;行
AB[order(-AB$BETAdn),][1:4,]