我想在一列中平均每100个点，然后将平均值放在中间 - 在50点。
我尝试使用此脚本计算移动平均线：
BEGIN {
P = 100;
}
{
x = $2;
i = NR % P;
MA += (x - Z[i]) / P;
Z[i] = x;
print $1,"\t",$2,"\t",MA;
}
但我需要知道如何将它放在中间？
输入样本：
Depth Velocity
1150.315 434.929
1150.468 434.929
1150.62 434.929
1150.772 434.929
1150.925 434.929
1151.077 434.929
1151.23 434.929
1151.382 434.929
1151.534 434.929
1151.687 434.929
1151.839 434.929
1151.992 434.929
1152.144 434.929
1152.296 434.929
1152.449 434.929
1152.601 434.929
1152.754 434.929
1152.906 434.929
1153.058 434.929
1153.211 434.929
1153.363 434.929
1153.516 434.929
1153.668 434.929
1153.82 434.929
1153.973 434.929
1154.125 434.929
1154.278 434.929
1154.43 434.929
1154.582 434.929
1154.735 434.929
1154.887 434.929
1155.04 434.929
1155.192 434.929
1155.344 434.929
1155.497 434.929
1155.649 434.517
1155.802 434.105
1155.954 433.693
1156.106 433.233
1156.259 432.773
1156.411 432.313
1156.564 431.853
1156.716 431.853
1156.868 431.853
1157.021 431.853
1157.173 431.853
1157.326 431.853
1157.478 431.853
1157.63 431.853
1157.783 431.853
1157.935 431.853
1158.088 431.853
1158.24 431.853
1158.392 431.853
1158.545 431.853
1158.697 431.853
1158.85 431.853
1159.002 431.853
1159.154 432.642
1159.307 433.431
1159.459 434.221
1159.612 437.791
1159.764 441.363
1159.916 444.933
1160.069 448.505
1160.221 448.037
1160.374 447.569
1160.526 447.101
1160.678 455.151
1160.831 463.208
1160.983 471.259
1161.136 473.544
1161.288 475.826
1161.44 478.111
1161.593 465.778
1161.745 453.435
...... .......
输出应该是深度和平滑速度 - 平均 - 并且将删除第一个和最后50个点。
答案 0 :(得分：3)
您可以尝试以下脚本：
awk -vn=100 -f a.awk file
其中a.awk是
BEGIN {
m=int((n+1)/2)
}
{L[NR]=$2; sum+=$2}
NR>=m {d[++i]=$1}
NR>n {sum-=L[NR-n]}
NR>=n{
a[++k]=sum/n
}
END {
for (j=1; j<=k; j++)
print d[j],a[j]
}
例如给定测试数据：
1 2
2 2
3 3
4 3
5 4
6 14
7 5
8 5
9 6
10 6
11 7
12 7
13 8
14 8
15 9
并且正在运行awk -vn=5 -f a.awk file
3 2.8
4 5.2
5 5.8
6 6.2
7 6.8
8 7.2
9 5.8
10 6.2
11 6.8
12 7.2
13 7.8
答案 1 :(得分：1)
没有输入就很难理解你想要达到的目标，下面是5点移动平均值和3点索引的例子，根据你的需要实现
awk '
BEGIN{
OFS = "\t"
No_of_row = 20
Average_Point = 5
index_Point = 3
print "Column1" OFS "Column2"
for(i=1;i<=No_of_row;i++)
{
C1[i]=i
C2[i]=i
print i OFS i
}
print RS "index" OFS "Average" OFS "Data_used" OFS "Sum" OFS "No_of_Point"
}
END{
for(i=1;i<=No_of_row;i++){
for(j=1;j<=Average_Point;j++){
flag = 0
if(C1[i+j] || j == 1 && C1[i])
{
add = j==1 ? C2[i] : C2[(i+j)-1]
sum += add
ind = j==index_Point ? C1[(i+j)-1] : ind
flag = 1
s = s ? s "+" add : add
}
}
if(flag == 1){
print ind,sum/Average_Point,s,sum,Average_Point
}
sum = ind = s = ""
}
}
' /dev/null
所得
Column1 Column2
1 1
2 2
3 3
4 4
5 5
6 6
7 7
8 8
9 9
10 10
11 11
12 12
13 13
14 14
15 15
16 16
17 17
18 18
19 19
20 20
index Average Data_used Sum No_of_Point
3 3 1+2+3+4+5 15 5
4 4 2+3+4+5+6 20 5
5 5 3+4+5+6+7 25 5
6 6 4+5+6+7+8 30 5
7 7 5+6+7+8+9 35 5
8 8 6+7+8+9+10 40 5
9 9 7+8+9+10+11 45 5
10 10 8+9+10+11+12 50 5
11 11 9+10+11+12+13 55 5
12 12 10+11+12+13+14 60 5
13 13 11+12+13+14+15 65 5
14 14 12+13+14+15+16 70 5
15 15 13+14+15+16+17 75 5
16 16 14+15+16+17+18 80 5
17 17 15+16+17+18+19 85 5
答案 2 :(得分：0)
你必须修改否。为适合您的文件的行，输入应包含两列作为索引，另一列将被平均，输出将少于100点，每侧50个。
# Moving average over the Second column of a data file
BEGIN{
OFS = "\t"
No_of_row = 10294
Average_Point = 100
index_Point = 50
}
{
for(i=1;i<=No_of_row;i++){
NR == i &&
C1[i]= $1
NR == i &&
C2[i]= $2 } }
END { for(i=1;i<=No_of_row - Average_Point;i++){
for(j=1;j<=Average_Point;j++){
if(C1[i+j] || j == 1 && C1[i])
{
add = j==1 ? C2[i] : C2[(i+j)-1]
sum += add
ind = j==index_Point ? C1[(i+j)-1] : ind
}
}
{
print ind,sum/Average_Point
}
sum = ind = ""
}
}