2013-10-16 29 views
2

我在網上查找,但沒有找到一個簡單而乾淨的解決方案。只從數據幀中提取數字列

這是我的一塊DF的:

structure(list(ID = structure(c(12L, 12L, 12L, 12L, 12L, 12L, 
12L, 12L, 12L, 12L), .Label = c("B0F", "B12T", "B1T", "B21T", 
"B22F", "B26T", "B2F", "B33F", "B3F", "B4F", "B7F", "P1", "P21", 
"P24", "P25", "P27", "P28", "P29"), class = "factor"), Data = structure(c(9646, 
9836, 9938, 10043, 10134, 10203, 10302, 10354, 10421, 10528), class = "Date"), 
    T = c(11.3, 9.7, 9.8, 10.5, 9.9, 10, 10, 10.1, 10, 10), ph = c(6.8, 
    6.9, 7.1, 6.9, 7, 6.93, 7.01, 6.9, 7.01, 6.84), EC = c(1840L, 
    1060L, 940L, 760L, 820L, 1038L, 1035L, 839L, 767L, 433L)), .Names = c("ID", 
"Data", "T", "ph", "EC"), row.names = c(NA, 10L), class = "data.frame") 

而這些變量:

str(df) 
'data.frame': 10 obs. of 5 variables: 
$ ID : Factor w/ 18 levels "B0F","B12T","B1T",..: 12 12 12 12 12 12 12 12 12 12 
$ Data: Date, format: "1996-05-30" "1996-12-06" "1997-03-18" ... 
$ T : num 11.3 9.7 9.8 10.5 9.9 10 10 10.1 10 10 
$ ph : num 6.8 6.9 7.1 6.9 7 6.93 7.01 6.9 7.01 6.84 
$ EC : int 1840 1060 940 760 820 1038 1035 839 767 433 

我需要的是一個新的DF與數值列(因此T的原始值,pH和EC)。我知道這可以用一個簡單的列提取(new_df=df[,3:5])來完成,但我有很多的df應該在這個操作上完成。

感謝

回答

7

如何

new_df <- df[sapply(df,is.numeric)] 

+0

我通常不會對「如何」的答案投票,但我可能會回答完全相同的問題。 – A5C1D2H2I1M1N2O1R2T1

+0

是的,它的工作原理!我嘗試過安樂死,但顯然我錯過了一些東西。謝謝 – matteo