gnuplot缺少具有表达式评估的数据 [英] gnuplot missing data with expression evaluation

问题描述

我想在gnuplot中使用plot命令进行表达式评估，即

plot "-" using ($1):($2) with lines
1 10
2 20
3 ?
4 40
5 50
e

但是我希望它忽略丢失的数据?"以这种方式连接线路(并且不会在2到4之间断开).
我尝试了set datafile missing "?"， 但与在线帮助一致，它不连接线路.可以，但是我不能使用表达式求值:

plot "-" using 1:2 with lines
1 10
2 20
3 ?
4 40
5 50
e

有什么想法如何连接和使用表达式求值?

解决方案

两列数据

如果您设置数据文件Data.csv

1 10
2 20
3 ?
4 40
5 50

您可以使用以下连接线绘制数据

plot '<grep -v "?" Data.csv' u ($1):($2) w lp

多于两列的数据

对于两个以上的列，您可以使用 awk .
使用数据文件 Data.csv

1 10 1
2 20 2
3 ?  3
4 40 ?
5 50 5

您可以在每个图的数据文件上运行脚本，如下所示:

plot "<awk '{if($2 != \"?\") print}' Data.csv" u ($1):($2) w lp, \
     "<awk '{if($3 != \"?\") print}' Data.csv" u ($1):($3) w lp

可以在此处找到有关gnuplot中脚本编写的参考. awk 用户手册此处.

I want to use the plot command in gnuplot with expression evaluation, i.e.

plot "-" using ($1):($2) with lines
1 10
2 20
3 ?
4 40
5 50
e

But I want it to ignore the missing data "?" in such a way that it connects the line (and doesn't break it between 2 and 4).
I tried set datafile missing "?", but in agreement with the online-help it does not connect the lines. The following would, but I cannot use expression evaluation:

plot "-" using 1:2 with lines
1 10
2 20
3 ?
4 40
5 50
e

Any ideas how to connect the lines and use expression evaluation?

解决方案

Two column data

If you set up a data file Data.csv

1 10
2 20
3 ?
4 40
5 50

you can plot your data with connected lines using

plot '<grep -v "?" Data.csv' u ($1):($2) w lp

More than two column data

For more than two columns you can make use of awk.
With a data file Data.csv

1 10 1
2 20 2
3 ?  3
4 40 ?
5 50 5

you can run a script over the data file for each plot like so:

plot "<awk '{if($2 != \"?\") print}' Data.csv" u ($1):($2) w lp, \
     "<awk '{if($3 != \"?\") print}' Data.csv" u ($1):($3) w lp

A reference on scripting in gnuplot can be found here. The awk user manual here.

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