我有一个看起来像这样的数据框:
var1 var2
1 927720_2005 927720_2006
2 927720_2006 927720_2007
3 841555_2005 841555_2006
4 88095_2005 88095_2006
5 1003464_2005 1003464_2006
6 1003464_2005 1003464_2006
7 1003464_2006 1003464_2007
8 1037388_2005 1037388_2006
9 1037388_2006 1037388_2007
列1003464_2005
中的观察值var1
是重复的,因此当我应用rownames(MyMatrix) <- df$var1
时,行名的一个观察值为1003464_2005
,另一个为1003464_2005.1
。我不介意,但是当我使用colnames(MyMatrix) <- df$var2
时,列名可以重复。
我要使数据如下:
var1 var2
1 927720_2005 927720_2006
2 927720_2006 927720_2007
3 841555_2005 841555_2006
4 88095_2005 88095_2006
5 1003464_2005 1003464_2006
6 1003464_2005.1 1003464_2006.1
7 1003464_2006 1003464_2007
8 1037388_2005 1037388_2006
9 1037388_2006 1037388_2007
如果我在var1
中有“ 3个重复项”,只需在1003464_2005.2
或1003464_2005.1.1
上添加另一个“计数器”。这样,我将不会在var1
列中重复任何内容,而将曾经“添加”到var1
列中的内容添加到var2
列中。
在此先感谢您的帮助!
数据:
df <- structure(list(var1 = structure(c(7L, 8L, 5L, 6L, 1L, 1L, 2L,
3L, 4L), .Label = c("1003464_2005", "1003464_2006", "1037388_2005",
"1037388_2006", "841555_2005", "88095_2005", "927720_2005", "927720_2006"
), class = "factor"), var2 = c("927720_2006", "927720_2007",
"841555_2006", "88095_2006", "1003464_2006", "1003464_2006",
"1003464_2007", "1037388_2006", "1037388_2007")), class = "data.frame", row.names = c(NA,
-9L))
行名重复时,我的矩阵如下所示:
structure(c(0.0000000000000000111365086910415, 0.0242390433922595,
0.294121286748089, 0.302965878225595, 0.259626633772708, 0.25760904856241,
0.248574305825551, 0.17848782814175, 0.191657814393258, 0.0242390433922595,
0.0000000000000000113968217215608, 0.310381807852827, 0.293653514681392,
0.245957439956465, 0.249142123526167, 0.251115609138352, 0.166302748882678,
0.176256028117321, 0.294121286748089, 0.310381807852827, -0.00000000000000000151197688178523,
0.355703128500295, 0.319662657194485, 0.317127296846476, 0.305644319511071,
0.255031411391534, 0.275597914790561, 0.302965878225595, 0.293653514681392,
0.355703128500295, 0.00000000000000000801369440490437, 0.309841957462355,
0.311910981514099, 0.317253692884325, 0.254335300246398, 0.265496031285385,
0.259626633772708, 0.245957439956465, 0.319662657194485, 0.309841957462355,
0.0000000000000000105380873106143, 0.0104634838149491, 0.0245937753301301,
0.221744045353809, 0.22476375867925, 0.25760904856241, 0.249142123526167,
0.317127296846476, 0.311910981514099, 0.0104634838149491, 0.00000000000000000986038424517971,
0.0257337720292454, 0.220483645448676, 0.224712591289328, 0.248574305825551,
0.251115609138352, 0.305644319511071, 0.317253692884325, 0.0245937753301301,
0.0257337720292454, 0.0000000000000000121630774340264, 0.213285559165696,
0.229922308724439, 0.17848782814175, 0.166302748882678, 0.255031411391534,
0.254335300246398, 0.221744045353809, 0.220483645448676, 0.213285559165696,
0.0000000000000000139766402734024, 0.0152113168185518, 0.191657814393258,
0.176256028117321, 0.275597914790561, 0.265496031285385, 0.22476375867925,
0.224712591289328, 0.229922308724439, 0.0152113168185518, 0.0000000000000000120926010568502
), .Dim = c(9L, 9L), .Dimnames = list(c("927720_2005", "927720_2006",
"841555_2005", "88095_2005", "1003464_2005", "1003464_2005",
"1003464_2006", "1037388_2005", "1037388_2006"), c("927720_2005",
"927720_2006", "841555_2005", "88095_2005", "1003464_2005", "1003464_2005",
"1003464_2006", "1037388_2005", "1037388_2006")))
答案 0 :(得分:4)
使用make.unique
可以轻松完成。循环浏览感兴趣的列,然后应用函数make.unique
。它期望该列为character
类。根据{{1}}
名称-字符向量
因此,如果它是?make.unique
,则将其转换为factor
character
在df[] <- lapply(df, function(x) make.unique(as.character(x)))
中,这可以类似地完成,但是使用dplyr
mutate_if