基于R中的条件删除数据帧的列 [英] Remove columns of dataframe based on conditions in R
问题描述
我必须删除我的数据帧中有超过4000列和180行的列。我想要设置为删除数据框中的列的条件是：
（i）删除列如果有少那么该列中的两个值/条目
（ii）如果列中没有两个连续（一个接一个）
值，则删除列。
（iii）删除所有值为NA的列。
我提供了要删除列的条件。这里的目的不仅仅是通过它的名称找到列，如如何删除数据表中的列？。
I说明如下：
ABCDE
0.018 NA NA NA NA
0.017 NA NA NA NA
0.019 NA NA NA NA
0.018 0.034 NA NA NA
0.018 NA NA NA NA
0.015 NA NA NA 0.037
0.016 NA NA NA 0.031
0.019 NA 0.4 NA 0.025
0.016 0.03 NA NA 0.035
0.018 NA NA NA 0.035
0.017 NA NA NA 0.043
0.023 NA NA NA 0.040
0.022 NA NA NA 0.042
所需数据框架：
AE
0.018 NA
0.017 NA
0.019 NA
0.018 NA
0.018 NA
0.015 0.037
0.016 0.031
0.019 0.025
0.016 0.035
0.018 0.035
0.017 0.043
0.023 0.040
0.022 0.042
如何在一个代码中包含这三个条件。我很感谢您在这方面的帮助。
可复制的示例
结构（列表（Month = c（Jan-2000，Feb-2000 Mar-2000，
2000年4月，May-2000，Jun-2000），AGLS.INVS ... LON..DEAD ... 13.08.15 = c（NA_real_ ，
NA_real_，NA_real_，NA_real_，NA_real_，NA_real_），ABACUS.GROUP.DEAD ... 18.02.09 = c（0.00829384766220866，
0.00332213653674028,0,0，NA，NA） ..IRS。= c（NA_real_，
NA_real_，NA_real_，NA_real_，NA_real_，NA_real_））.Names = c（Month，
AGLSJ.INVS ... LON..DEAD ... 13.08.15，ABACUS.GROUP.DEAD ... 18.02.09，
ABB.R..IRS。），class = c（data.table frame），row.names = c（NA，
-6L），.internal.selfref =< pointer：0x0000000001c90788>）
解决方案我觉得这是非常复杂的。条件2已经包括所有其余条件，如同一列中至少有两个非
NA值，显然整个列不是NA。如果列中至少有两个连续的值，那么显然该列包含多个值。所以，而不是3条件，这总共是一个条件（我不喜欢运行许多函数每列，而是运行diff每列 - vecotrize的整个事情） ：cond< - colSums（is.na（sapply（df，diff））））& nrow（df） - 1
这是因为如果列中没有连续的值，列将变为
NA。
然后，只需
df [，cond，drop = FALSE ]
＃AE
＃1 0.018 NA
＃2 0.017 NA
＃3 0.019 NA
＃4 0.018 NA
＃5 0.018 NA
＃6 0.015 0.037
＃7 0.016 0.031
＃8 0.019 0.025
＃9 0.016 0.035
＃10 0.018 0.035
＃11 0.017 0.043
＃ 12 0.023 0.040
＃13 0.022 0.042
根据您的编辑，您似乎有一个
data.table对象，您还有一个日期column，因此代码需要修改。cond< - df [，lapply （x）sum（is.na（diff（x）））） .N-1，.SDcols = -1]
df [，c（TRUE，cond），with = FALSE]
$ b b
一些说明：
- 我们希望忽略计算中的第一列，当
.SD（这意味着 S ub D ）时，.SDcols = -1 / strong> data.table 中的ata 是 -
.N只是行数（类似nrow（df） - 下一步是通过条件子集我们不必忘记抓取第一列，所以我们从
c（TRUE，... - 开始
data.table默认情况下使用非标准评估，因此，如果你想选择列，就像你在data.frame中一样，你需要指定with = FALSE
A更好的方法，只是通过使用：= NULL
cond <-c（FALSE，df [，lapply（.SD，function（x）sum（is.na（diff（x））））= .N-1，.SDcols = -1] ）
df [，which（cond）：= NULL]
I have to remove columns in my dataframe which has over 4000 columns and 180 rows.The conditions I want to set in to remove the column in the dataframe are: (i) Remove the column if there are less then two values/entries in that column (ii) Remove the column if there are no two consecutive(one after the other) values in the column. (iii) Remove the column having all values as NA. I have provided with conditions on which a column is to be deleted. The aim here is not just to find a column by its name like in "How do you delete a column in data.table?". I Illustrate as follows:
A B C D E
0.018 NA NA NA NA
0.017 NA NA NA NA
0.019 NA NA NA NA
0.018 0.034 NA NA NA
0.018 NA NA NA NA
0.015 NA NA NA 0.037
0.016 NA NA NA 0.031
0.019 NA 0.4 NA 0.025
0.016 0.03 NA NA 0.035
0.018 NA NA NA 0.035
0.017 NA NA NA 0.043
0.023 NA NA NA 0.040
0.022 NA NA NA 0.042
Desired dataframe:
A E
0.018 NA
0.017 NA
0.019 NA
0.018 NA
0.018 NA
0.015 0.037
0.016 0.031
0.019 0.025
0.016 0.035
0.018 0.035
0.017 0.043
0.023 0.040
0.022 0.042
How can I incoporate these three conditions in one code. I would appreciate your help in this regard. Reproducible example
structure(list(Month = c("Jan-2000", "Feb-2000", "Mar-2000",
"Apr-2000", "May-2000", "Jun-2000"), A.G.L.SJ.INVS...LON..DEAD...13.08.15 = c(NA_real_,
NA_real_, NA_real_, NA_real_, NA_real_, NA_real_), ABACUS.GROUP.DEAD...18.02.09 = c(0.00829384766220866,
0.00332213653674028, 0, 0, NA, NA), ABB.R..IRS. = c(NA_real_,
NA_real_, NA_real_, NA_real_, NA_real_, NA_real_)), .Names = c("Month",
"A.G.L.SJ.INVS...LON..DEAD...13.08.15", "ABACUS.GROUP.DEAD...18.02.09",
"ABB.R..IRS."), class = c("data.table", "data.frame"), row.names = c(NA,
-6L), .internal.selfref = <pointer: 0x0000000001c90788>)
I feel like this is all over-complicated. Condition 2 already includes all the rest of the conditions, as if there are at least two non-NA values in a column, obviously the whole column aren't NAs. And if there are at least two consecutive values in a column, then obviously this column contains more than one value. So instead of 3 conditions, this all sums up into a single condition (I prefer not to run many functions per column, rather after running diff per column- vecotrize the whole thing):
cond <- colSums(is.na(sapply(df, diff))) < nrow(df) - 1
This works because if there are no consecutive values in a column, the whole column will become NAs.
Then, just
df[, cond, drop = FALSE]
# A E
# 1 0.018 NA
# 2 0.017 NA
# 3 0.019 NA
# 4 0.018 NA
# 5 0.018 NA
# 6 0.015 0.037
# 7 0.016 0.031
# 8 0.019 0.025
# 9 0.016 0.035
# 10 0.018 0.035
# 11 0.017 0.043
# 12 0.023 0.040
# 13 0.022 0.042
Per your edit, it seems like you have a data.table object and you also have a Date column so the code would need some modifications.
cond <- df[, lapply(.SD, function(x) sum(is.na(diff(x)))) < .N - 1, .SDcols = -1]
df[, c(TRUE, cond), with = FALSE]
Some explanations:
- We want to ignore the first column in our calculations so we specify
.SDcols = -1when operating on our.SD(which means Sub Data indata.tableis) .Nis just the rows count (similar tonrow(df)- Next step is to subset by condition. We need not forget to grab the first column too so we start with
c(TRUE,... - Finally,
data.tableworks with non standard evaluation by default, hence, if you want to select column as if you would in adata.frameyou will need to specifywith = FALSE
A better way though, would be just to remove the column by reference using := NULL
cond <- c(FALSE, df[, lapply(.SD, function(x) sum(is.na(diff(x)))) == .N - 1, .SDcols = -1])
df[, which(cond) := NULL]
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