grep 不从数据帧返回值

问题描述 投票:0回答:1

在下面的代码中,我尝试从数据框 df_meth 和 df_exp 中选择以 A 开头的所有行。 df_meth 和 df_exp 都有超过 65000 行以 A、B、C 和 D 开头。以 A 开头的 df_meth 和 df_exp 的子集如下所示。

df_meth <- read.table('meth.tsv', header = TRUE, sep = "\t")
df_exp <- read.table('exp.tsv', header = TRUE, sep = "\t")
meth_A <- df_meth[, grep("^A", names(df_meth))]
exp_A <- df_exp[, grep("^A", names(df_exp))]

meth_A and exp_A are empty
> print(head(meth_A))
data frame with 0 columns and 6 rows
> print(head(exp_A))
data frame with 0 columns and 6 rows

df_meth:
subgenome Meth_value
A 0.920802534
A 0.472727273
A 0.127757353
A 0.025291829
A 0.030303030
A 0.165847666
A 0.072727273
A 0.001315789
A 0.446351931
A 0.415584416
A 0.846153846
A 0.013698630
A 0.095505618
A 0.195121951
A 0.102941176
A 0.150150150
A 0.450000000
A 0.025316456
A 0.646153846

df_exp:
subgenome TPM
A 0.000000
A 0.000000
A 25.687089
A 5.644342
A 20.459905
A 196.714492
A 4.629544
A 0.346602
A 90.000952
A 13.099918
A 0.329252
A 4.434103
r
1个回答
0
投票

meth_B <- subset(df_meth, substr(subgenome, 1, 1) == "B") exp_B <- subset(df_exp, substr(subgenome, 1, 1) == "B")

© www.soinside.com 2019 - 2024. All rights reserved.