我正在努力解决一个非常简单的问题。我有两个数据帧组合成一个数据帧，同时保留其原始行名称。如果您对使用这两个数据框感兴趣：
> dput(cc)
structure(list(c = c(166.081273211195, 117.874627144804, 85.7050194973198,
122.960746859139, 144.149802403233, 90.8034500957001, 89.5265981283352,
77.8535718910714, 144.544005656701, 115.597165020403, 109.028391182666,
89.4045716355402, 77.7944830105746, 69.3378920684953, 79.9094499459695,
146.768077595585, 157.933946809176, 92.562729606313, 62.5081476457419,
90.1081848285295, 111.830482369239, 111.596975757741, 99.2311075024839,
145.204385556523, 99.0215341758211, 75.4302512245677, 92.0094563363458,
77.7314808830408)), .Names = "c", row.names = c("26", "27", "28",
"29", "35", "36", "37", "38", "39", "40", "46", "47", "48", "49",
"50", "51", "52", "56", "57", "58", "59", "60", "61", "62", "69",
"70", "71", "72"), class = "data.frame")
> dput(ccc)
structure(list(b = c(76.376257255471, 61.8314936138378, 62.769450181685,
73.6356164203567, 111.690756826382, 76.9294523843767, 61.3534699857719,
69.3647221333577, 83.9764878084258, 81.3800252294203, 69.5091780233591,
87.3595961209547, 78.5074999563006, 74.4479256924594, 81.5920316281566,
96.3417259554163, 75.4138056616399, 76.0553034201146, 95.1759950844736,
81.3252467041995, 86.306305649635, 70.5626459312969, 72.7797520793756,
119.49702877934, 123.268678343102, 88.0450051118928, 76.2139948860248,
98.1496728839206, 126.396927030103, 146.058540478643, 115.7341525964,
87.280600158726, 78.0274068331766, 122.817977752389, 142.491559175427,
152.895839114334, 94.4932174696818, 117.167042165763, 85.5340971715004,
101.480170738897, 117.759691799033, 128.998051359269, 98.3180491401911,
84.5915489017958, 87.4927520958843, 75.5366495973031, 118.088343275321,
121.375320935357, 94.7724147096235, 70.2266610201599, 123.158462686523,
76.87408931845, 94.4365460662552, 105.952134808703, 76.821070196668
)), .Names = "b", row.names = c("1", "2", "3", "4", "5", "6",
"7", "8", "9", "10", "11", "12", "13", "14", "15", "16", "17",
"18", "19", "20", "21", "22", "23", "24", "25", "30", "31", "32",
"33", "34", "41", "42", "43", "44", "45", "53", "54", "55", "63",
"64", "65", "66", "67", "68", "73", "74", "75", "76", "77", "78",
"79", "80", "81", "82", "83"), class = "data.frame")
最简单的组合方式是
> c(cc$c, ccc$b)
[1] 166.08127 117.87463 85.70502 122.96075 144.14980 90.80345 89.52660 77.85357 144.54401
[10] 115.59717 109.02839 89.40457 77.79448 69.33789 79.90945 146.76808 157.93395 92.56273
[19] 62.50815 90.10818 111.83048 111.59698 99.23111 145.20439 99.02153 75.43025 92.00946
[28] 77.73148 76.37626 61.83149 62.76945 73.63562 111.69076 76.92945 61.35347 69.36472
[37] 83.97649 81.38003 69.50918 87.35960 78.50750 74.44793 81.59203 96.34173 75.41381
[46] 76.05530 95.17600 81.32525 86.30631 70.56265 72.77975 119.49703 123.26868 88.04501
[55] 76.21399 98.14967 126.39693 146.05854 115.73415 87.28060 78.02741 122.81798 142.49156
[64] 152.89584 94.49322 117.16704 85.53410 101.48017 117.75969 128.99805 98.31805 84.59155
[73] 87.49275 75.53665 118.08834 121.37532 94.77241 70.22666 123.15846 76.87409 94.43655
[82] 105.95213 76.82107
但这样做会重新创建行号。是否有任何简单的功能可以在保持行名完整的同时进行组合？谢谢！
答案 0 :(得分：0)
这些代码行应解决问题
z0 <- as.numeric(c(rownames(cc), rownames(ccc)))
z <- data.frame(c(cc$c, ccc$b))
row.names(z) <- z0
data.frame(z[order(as.numeric(row.names(z))),])