查找列和，除以列值，然后使用awk打印到新列。

Question

我的数据是这样的：

file Gibbs kcal rel pop pop2
RR2.out -1752.142111    -1099486.696073  0.000000 -0.0000 1.0000
RR1.out -1752.141887    -1099486.555511  0.140562 -0.2374 0.7891
RR4.out -1752.140564    -1099485.725315  0.970758 -1.6398 0.1947
RR3.out -1752.140319    -1099485.571575  1.124498 -1.8995 0.1502
RR5.out -1752.138532    -1099484.450215  2.245858 -3.7937 0.0227
RR6.out -1752.138493    -1099484.425742  2.270331 -3.8351 0.0218

我想找出第6栏的和，然后将第6栏中的每个值除以该和，然后在一个名为“加权”的新列中打印这些值。

使用

 echo "weighted" >> allRE7
 awk 'NR==FNR{sum+= $6; next}{printf("%0.4f\n", $6/sum)}' input input >> out
 paste input out >> final

给我

 file Gibbs kcal rel pop pop2   weighted
 RR2.out    -1752.142111    -1099486.696073  0.000000 -0.0000 1.0000    0.0000
 RR1.out    -1752.141887    -1099486.555511  0.140562 -0.2374 0.7891    0.4590
 RR4.out    -1752.140564    -1099485.725315  0.970758 -1.6398 0.1947    0.3622
 RR3.out    -1752.140319    -1099485.571575  1.124498 -1.8995 0.1502    0.0894
 RR5.out    -1752.138532    -1099484.450215  2.245858 -3.7937 0.0227    0.0689
 RR6.out    -1752.138493    -1099484.425742  2.270331 -3.8351 0.0218    0.0104
         0.0100

我不知道0.0100是从哪里来的。

Answer 1

问题是，awk代码被打印为标题行的加权结果。为消除这一现象，请取代：

awk 'NR==FNR{sum+= $6; next}{printf("%0.4f\n", $6/sum)}' input input >> out

通过以下方式：

awk 'NR==FNR{sum+= $6; next} FNR>1{printf("%0.4f\n", $6/sum)}' input input >> out

FNR>1条件确保只为数据行打印$6/sum。

改进

echo和paste命令是不必要的。尝试：

$ awk 'NR==FNR{sum+= $6; next} FNR==1{print $0,"weighted"; next} {printf("%s %0.4f\n",$0,$6/sum)}' input input
file Gibbs kcal rel pop pop2 weighted
RR2.out -1752.142111    -1099486.696073  0.000000 -0.0000 1.0000 0.4590
RR1.out -1752.141887    -1099486.555511  0.140562 -0.2374 0.7891 0.3622
RR4.out -1752.140564    -1099485.725315  0.970758 -1.6398 0.1947 0.0894
RR3.out -1752.140319    -1099485.571575  1.124498 -1.8995 0.1502 0.0689
RR5.out -1752.138532    -1099484.450215  2.245858 -3.7937 0.0227 0.0104
RR6.out -1752.138493    -1099484.425742  2.270331 -3.8351 0.0218 0.0100

上面的变体使用了一个三元运算符(hat技巧：Ed Morton)，：

$ awk 'NR==FNR{sum+= $6; next} {print $0, (FNR>1 ? sprintf("%0.4f",$6/sum) : "weighted")}' input input
file Gibbs kcal rel pop pop2 weighted
RR2.out -1752.142111    -1099486.696073  0.000000 -0.0000 1.0000 0.4590
RR1.out -1752.141887    -1099486.555511  0.140562 -0.2374 0.7891 0.3622
RR4.out -1752.140564    -1099485.725315  0.970758 -1.6398 0.1947 0.0894
RR3.out -1752.140319    -1099485.571575  1.124498 -1.8995 0.1502 0.0689
RR5.out -1752.138532    -1099484.450215  2.245858 -3.7937 0.0227 0.0104
RR6.out -1752.138493    -1099484.425742  2.270331 -3.8351 0.0218 0.0100

Answer 2

你也在计算标题线的平均值。

要省略标题行，您的awk脚本应该是：

awk 'FNR==1{next}NR==FNR{sum+= $6; next}{printf("%0.4f\n", $6/sum)}' input input >> out
paste input out >> final 

一个更干净的awk脚本(包括paste命令)是：

awk 'FNR==1{next}NR==FNR{sum+= $6; next}{printf("%s %0.4f\n", $0, $6/sum)}' input input