我有8个数据框，我想添加一个名为“park”的列，然后用来自dataframe名称的最后四个字符的值填充此列。以下是我的八个数据框中的两个：
water_land_by_ownname_apis <- structure(list(OWNERNAME = c("Forest Service (USFS)", "Fish and Wildlife Service (FWS)",
"State Department of Natural Resources", "Private Landowner",
"National Park Service (NPS)", "Unknown", "Private Institution",
"Native American Land"), WATER = c(696600, 9900, 1758600, 26100,
112636800, 1586688300, 0, 11354400), LAND = c(258642900, 997200,
41905800, 2536200, 165591900, 1075917600, 461700, 314052300)), class = "data.frame", .Names = c("OWNERNAME",
"WATER", "LAND"), data_types = c("C", "F", "F"), row.names = c("1",
"2", "3", "4", "5", "6", "7", "8"))
water_land_by_ownname_indu <- structure(list(OWNERNAME = c("The Nature Conservancy (TNC)",
"Other State Land", "Private Institution", "State Department of Transportation",
"State Department of Natural Resources", "Unknown", "National Park Service (NPS)",
"Private Landowner", "Joint Ownership", "Private Non-profit",
"Land Trust"), WATER = c(24300, 1018800, 5282100, 0, 12600, 19192500,
802800, 139500, 0, 0, 0), LAND = c(719100, 10045800, 12556800,
900, 2018700, 1446426000, 42484500, 5769900, 38700, 852300, 70200
)), class = "data.frame", .Names = c("OWNERNAME", "WATER", "LAND"
), data_types = c("C", "F", "F"), row.names = c("1", "2", "3",
"4", "5", "6", "7", "8", "9", "10", "11"))
看起来像这样......
> water_land_by_ownname_apis
OWNERNAME WATER LAND
1 Forest Service (USFS) 696600 258642900
2 Fish and Wildlife Service (FWS) 9900 997200
3 State Department of Natural Resources 1758600 41905800
4 Private Landowner 26100 2536200
5 National Park Service (NPS) 112636800 165591900
6 Unknown 1586688300 1075917600
7 Private Institution 0 461700
8 Native American Land 11354400 314052300
> water_land_by_ownname_indu
OWNERNAME WATER LAND
1 The Nature Conservancy (TNC) 24300 719100
2 Other State Land 1018800 10045800
3 Private Institution 5282100 12556800
4 State Department of Transportation 0 900
5 State Department of Natural Resources 12600 2018700
6 Unknown 19192500 1446426000
7 National Park Service (NPS) 802800 42484500
8 Private Landowner 139500 5769900
9 Joint Ownership 0 38700
10 Private Non-profit 0 852300
11 Land Trust 0 70200
对于每个数据帧，我想添加一列（'park'）并用数据框名称的最后四个字符填充此列。例如......
water_land_by_ownname_apis$park <- 'apis'
water_land_by_ownname_indu$park <- 'indu'
导致这个......
> water_land_by_ownname_apis
OWNERNAME WATER LAND park
1 Forest Service (USFS) 696600 258642900 apis
2 Fish and Wildlife Service (FWS) 9900 997200 apis
3 State Department of Natural Resources 1758600 41905800 apis
4 Private Landowner 26100 2536200 apis
5 National Park Service (NPS) 112636800 165591900 apis
6 Unknown 1586688300 1075917600 apis
7 Private Institution 0 461700 apis
8 Native American Land 11354400 314052300 apis
> water_land_by_ownname_indu
OWNERNAME WATER LAND park
1 The Nature Conservancy (TNC) 24300 719100 indu
2 Other State Land 1018800 10045800 indu
3 Private Institution 5282100 12556800 indu
4 State Department of Transportation 0 900 indu
5 State Department of Natural Resources 12600 2018700 indu
6 Unknown 19192500 1446426000 indu
7 National Park Service (NPS) 802800 42484500 indu
8 Private Landowner 139500 5769900 indu
9 Joint Ownership 0 38700 indu
10 Private Non-profit 0 852300 indu
11 Land Trust 0 70200 indu
然后，将它们绑在一起......
water_land_by_ownname <- rbind (water_land_by_ownname_apis, water_land_by_ownname_indu)
然后，从内存中删除先前的数据帧...
rm (water_land_by_ownname_apis,water_land_by_ownname_indu)
答案 0 :(得分：6)
您可以这样做：
do.call(rbind,lapply(ls(pattern='water.*'),
function(x) {
dat=get(x)
dat$park = sub('.*_(.*)$','\\1',x)
dat
}))
ls将提取具有特定模式的所有data.frames名称，这里我假设您data.frame以单词water开头。这将是名单中的商店名称，方便lapply使用。sub将提取名称的最后一部分do.call + rbind已应用于结果列表以获取唯一的大数据。框架使用您获得的2个data.frames：
OWNERNAME WATER LAND park
1 Forest Service (USFS) 696600 258642900 apis
2 Fish and Wildlife Service (FWS) 9900 997200 apis
3 State Department of Natural Resources 1758600 41905800 apis
4 Private Landowner 26100 2536200 apis
5 National Park Service (NPS) 112636800 165591900 apis
6 Unknown 1586688300 1075917600 apis
7 Private Institution 0 461700 apis
8 Native American Land 11354400 314052300 apis
12 The Nature Conservancy (TNC) 24300 719100 indu
21 Other State Land 1018800 10045800 indu
31 Private Institution 5282100 12556800 indu
41 State Department of Transportation 0 900 indu
51 State Department of Natural Resources 12600 2018700 indu
61 Unknown 19192500 1446426000 indu
71 National Park Service (NPS) 802800 42484500 indu
答案 1 :(得分：2)
我会将data.frames放在命名列表中并执行此任务：
rslt <- list(water_land_by_ownname_apis = water_land_by_ownname_apis,
water_land_by_ownname_indu = water_land_by_ownname_indu)
for (i in names(rslt)) {
col <- unlist(strsplit(i, "_"))[5]
rslt[[i]]$park <- col
}
do.call("rbind", rslt)
答案 2 :(得分：0)
使用Map和一些"[<-"欺骗的变体：
vars <- ls(pattern="water_.")
l <- mget(vars)
names(l) <- substr(vars,nchar(vars)-3,nchar(vars))
do.call(rbind,Map("[<-",l,TRUE,"park",names(l)))