[英]subset of a data frame in R
I have 2 data frames df2
and DF
. 我有2个数据帧df2
和DF
。
> DF
date tickers
1 2000-01-01 B
2 2000-01-01 GOOG
3 2000-01-01 V
4 2000-01-01 YHOO
5 2000-01-02 XOM
> df2
date tickers quantities
1 2000-01-01 BB 11
2 2000-01-01 XOM 23
3 2000-01-01 GOOG 42
4 2000-01-01 YHOO 21
5 2000-01-01 V 2112
6 2000-01-01 B 13
7 2000-01-02 XOM 24
8 2000-01-02 BB 422
i need the values from df2
those are present in DF
. 我需要来自df2
的值,这些值存在于DF
。 That means i require the following output: 这意味着我需要以下输出:
3 2000-01-01 GOOG 42
4 2000-01-01 YHOO 21
5 2000-01-01 V 2112
6 2000-01-01 B 13
7 2000-01-02 XOM 24
So I used the following code: 所以我使用了以下代码:
> subset(df2,df2$date %in% DF$date & df2$tickers %in% DF$tickers)
date tickers quantities
2 2000-01-01 XOM 23
3 2000-01-01 GOOG 42
4 2000-01-01 YHOO 21
5 2000-01-01 V 2112
6 2000-01-01 B 13
7 2000-01-02 XOM 24
But the output contains one extra column.That is because the ticker
'xom' is present in 2 days in df2
. 但是输出包含一个额外的列。这是因为在df2
中2天内存在ticker
'xom'。 so both rows are selected. 所以两行都被选中了。 What modification is needed in my code? 我的代码需要进行哪些修改?
The dput is as follows: 输入如下:
> dput(DF)
structure(list(date = structure(c(1L, 1L, 1L, 1L, 2L), .Label = c("2000-01-01",
"2000-01-02"), class = "factor"), tickers = structure(c(4L, 5L,
6L, 8L, 7L), .Label = c("A", "AA", "AAPL", "B", "GOOG", "V",
"XOM", "YHOO", "Z"), class = "factor")), .Names = c("date", "tickers"
), row.names = c(NA, -5L), class = "data.frame")
> dput(df2)
structure(list(date = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 2L,
2L), .Label = c("2000-01-01", "2000-01-02"), class = "factor"),
tickers = structure(c(2L, 5L, 3L, 6L, 4L, 1L, 5L, 2L), .Label = c("B",
"BB", "GOOG", "V", "XOM", "YHOO"), class = "factor"), quantities = c(11,
23, 42, 21, 2112, 13, 24, 422)), .Names = c("date", "tickers",
"quantities"), row.names = c(NA, -8L), class = "data.frame")
Using sqldf
package: 使用sqldf
包:
require(sqldf)
sqldf("SELECT d2.date, d2.tickers, d2.quantities FROM df2 d2
JOIN DF d1 ON d1.date=d2.date AND d1.tickers=d2.tickers")
## date tickers quantities
## 1 2000-01-01 GOOG 42
## 2 2000-01-01 YHOO 21
## 3 2000-01-01 V 2112
## 4 2000-01-01 B 13
## 5 2000-01-02 XOM 24
This is not so different from my answer to this post of yours , but requires a little modification: 这与我对你的帖子的回答没有什么不同,但需要稍加修改:
df2[duplicated(rbind(DF, df2[,1:2]))[-seq_len(nrow(DF))], ]
# date tickers quantities
# 3 2000-01-01 GOOG 42
# 4 2000-01-01 YHOO 21
# 5 2000-01-01 V 2112
# 6 2000-01-01 B 13
# 7 2000-01-02 XOM 24
Note: This provides the output with the rows in the same order as it were in df2
. 注意:这为输出提供了与df2
相同的顺序。
Alternatively, as Ben suggests, using merge
: 或者,正如Ben建议的那样,使用merge
:
merge(df2, DF, by=c("date", "tickers"))
will give the same result as well (but not necessarily in the same order). 也会得到相同的结果(但不一定是相同的顺序)。
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