gnuplot读取unix时间戳

Question

我有一个包含以下格式数据的文件：

1351649601.045 421 2945
1351649601.684 1036 28591
1351649603.310 128 1620
1351649603.413 93 0
1351649603.375 132 1762
1351649603.318 218 1761
1351649603.434 412 176
1351649603.441 520 5780
1351649603.480 565 11070
1351649603.556 644 11535


gnuplot> set term png
Terminal type set to 'png'
gnuplot> set output "output.png"
gnuplot> plot "data.txt" using 1:2 with lines

但是，gnuplot无法正确读取日期。 你有什么想法？

解决了：

set term png xffffff
set output "output.png"
set size 17,17
set title "HTTP payload size and response time"
set style data fsteps
set xlabel "Date"
set timefmt "%s"
set format x "%m/%d/%Y %H:%M:%S"
set xdata time
set ylabel "Payload/Response time"
set grid
set key left
plot 'response' usi 1:2 with linespoints

Answer 1

对不起，似乎无法删除。 打算建议：

set xdata time
set timefmt "%s"
set format x "%m/%d/%Y %H:%M:%S"
plot "file.txt" using 1:2 with linespoints

但它会截断分数并从线上掉落点。