Select rows based on value in specific column

Question

I have a file like the one below. I would like to print the rows for which the value in the last column is higher than 90. I am not sure how to specify the correct column.

29974   A1CF    NM_138932       9606    hsa-miR-4711-3p 3       25      32      -0.018  -0.095  -0.108  0.003   0.017   -0.448  99
29974   A1CF    NM_138933       9606    hsa-miR-4711-3p 3       25      32      -0.018  -0.095  -0.108  0.003   0.017   -0.448  99
29974   A1CF    NM_014576       9606    hsa-miR-4711-3p 3       25      32      -0.018  -0.095  -0.108  0.003   0.017   -0.448  99
29974   A1CF    NM_001198820    9606    hsa-miR-4711-3p 3       25      32      -0.018  -0.095  -0.108  0.003   0.017   -0.448  69
29974   A1CF    NM_001198819    9606    hsa-miR-4711-3p 3       25      32      -0.018  -0.095  -0.108  0.003   0.017   -0.448  89
29974   A1CF    NM_001198818    9606    hsa-miR-4711-3p 3       25      32      -0.018  -0.095  -0.108  0.003   0.017   -0.448  90

Answer 1

print(df[which(df[,ncol(df)]>90),])

df是数据框对象的名称。

Answer 2

Through awk.

$ awk '$NF>90' file
29974   A1CF    NM_138932       9606    hsa-miR-4711-3p 3       25      32      -0.018  -0.095  -0.108  0.003   0.017   -0.448  99
29974   A1CF    NM_138933       9606    hsa-miR-4711-3p 3       25      32      -0.018  -0.095  -0.108  0.003   0.017   -0.448  99
29974   A1CF    NM_014576       9606    hsa-miR-4711-3p 3       25      32      -0.018  -0.095  -0.108  0.003   0.017   -0.448  99

Awk process the input file, record by record ie, row by row. NF is a special variable in awk which stores the last column number. So $NF contains the value of last column. So $NF>90 will check for the value of last column is greater than 90 or not. If it's true then awk prints the corresponding row.