My data looks like this:
我的数据如下所示：
CHROM Mutant_SNP_2
3RD T
4RD C
5RD
6RD G
7RD A
8RD
I have a CSV dataframe. I want a count from column "Mutant_SNP_2" of how many rows have an entry and therefore don't want a count of any blanks " ". I am separating it out by column "CHROM". I am getting the right output in terms of layout using this code in dplyr: count(combined, Mutant_SNP_2, wt = CHROM, sort = FALSE) however it is only counting the blank rows rather than those with a value. Any idea much appreciated. The output I get:
我有一个CSV数据帧。我想从列“Mutant_SNP_2”中计算有多少行有一个条目，因此不希望计算任何空白“”。我将其列为“CHROM”列。我在dplyr中使用此代码得到了正确的布局输出：count（合并，Mutant_SNP_2，wt = CHROM，sort = FALSE）但是它只计算空白行而不是具有值的行。任何想法都非常感激。我得到的输出：
Mutant_SNP_2 CHROM.x n
(fctr) (fctr) (int)
1 gi|339957448|gb|AENI01001139.1| 23
2 gi|339957449|gb|AENI01001138.1| 9
3 gi|339957451|gb|AENI01001136.1| 97
4 gi|339957452|gb|AENI01001135.1| 116
5 gi|339957453|gb|AENI01001134.1| 175
6 gi|339957454|gb|AENI01001133.1| 2
7 gi|339957455|gb|AENI01001132.1| 78
8 gi|339957456|gb|AENI01001131.1| 51
9 gi|339957457|gb|AENI01001130.1| 2
10 gi|339957458|gb|AENI01001129.1| 52
.. ... ... ...
4 个解决方案
#1
3
You can try with function table, the line with TRUE will give you the number of not blank value, by CHROM value) :
您可以尝试使用功能表，使用TRUE的行将为您提供非空白值的数量，按CHROM值）：
table(df$Mutant_SNP_2!="", df$CHROM)
You can get the result directly with table(df$Mutant_SNP_2!="", df$CHROM)[2, ]
您可以直接用表格获得结果（df $ Mutant_SNP_2！=“”，df $ CHROM）[2，]
Example:
例：
set.seed(123)
df <- data.frame(CHROM=sample(letters[1:3], 10, replace=TRUE), Mutant_SNP_2=sample(c("", "not blank"), 10, replace=TRUE), stringsAsFactors=FALSE)
table(df$Mutant_SNP_2!="", df$CHROM)
# a b c
# FALSE 0 2 3
# TRUE 2 2 1
table(df$Mutant_SNP_2!="", df$CHROM)[2, ]
# a b c
# 2 2 1
#2
1
We could try summing the boolean vector df$Mutant_SNP_2 != "" grouped by CHROM. This works because TRUE's will be coerced to 1, while FALSE's to 0.
我们可以尝试将由CHROM分组的布尔向量df $ Mutant_SNP_2！=“”求和。这是有效的，因为TRUE将被强制为1，而FALSE将被强制为0。
library(dplyr)
df %>% group_by(CHROM) %>%
summarise(n = sum(Mutant_SNP_2 != ""))
CHROM n
(fctr) (int)
1 3RD 1
2 4RD 1
3 5RD 0
4 6RD 1
5 7RD 1
6 8RD 0
#3
1
Try this:
尝试这个：
library(data.table)
setDT(df)[ Mutant_SNP_2 != "", .(count = .N), by=CHROM]
Perhaps this?
也许这个？
setDT(df)[ ,.(count= length(unique(Mutant_SNP_2))), by=CHROM]
#4
0
We can ave from base R to do this
我们可以从基地R做到这一点
with(df1, as.numeric(ave(Mutant_SNP_2, CHROM,
FUN= function(x) sum(nzchar(x)))))
#[1] 1 1 0 1 1 0
#1
3
You can try with function table, the line with TRUE will give you the number of not blank value, by CHROM value) :
您可以尝试使用功能表，使用TRUE的行将为您提供非空白值的数量，按CHROM值）：
table(df$Mutant_SNP_2!="", df$CHROM)
You can get the result directly with table(df$Mutant_SNP_2!="", df$CHROM)[2, ]
您可以直接用表格获得结果（df $ Mutant_SNP_2！=“”，df $ CHROM）[2，]
Example:
例：
set.seed(123)
df <- data.frame(CHROM=sample(letters[1:3], 10, replace=TRUE), Mutant_SNP_2=sample(c("", "not blank"), 10, replace=TRUE), stringsAsFactors=FALSE)
table(df$Mutant_SNP_2!="", df$CHROM)
# a b c
# FALSE 0 2 3
# TRUE 2 2 1
table(df$Mutant_SNP_2!="", df$CHROM)[2, ]
# a b c
# 2 2 1
#2
1
We could try summing the boolean vector df$Mutant_SNP_2 != "" grouped by CHROM. This works because TRUE's will be coerced to 1, while FALSE's to 0.
我们可以尝试将由CHROM分组的布尔向量df $ Mutant_SNP_2！=“”求和。这是有效的，因为TRUE将被强制为1，而FALSE将被强制为0。
library(dplyr)
df %>% group_by(CHROM) %>%
summarise(n = sum(Mutant_SNP_2 != ""))
CHROM n
(fctr) (int)
1 3RD 1
2 4RD 1
3 5RD 0
4 6RD 1
5 7RD 1
6 8RD 0
#3
1
Try this:
尝试这个：
library(data.table)
setDT(df)[ Mutant_SNP_2 != "", .(count = .N), by=CHROM]
Perhaps this?
也许这个？
setDT(df)[ ,.(count= length(unique(Mutant_SNP_2))), by=CHROM]
#4
0
We can ave from base R to do this
我们可以从基地R做到这一点
with(df1, as.numeric(ave(Mutant_SNP_2, CHROM,
FUN= function(x) sum(nzchar(x)))))
#[1] 1 1 0 1 1 0