Applying a certain formula to a repeated sample data
Dear Jim,
I don't think my problem is clear the way I put.
I have been trying to manually apply the formula to some rows.
This is what I have done.
I cut and past some rows from 1-7 and save each with a different file as
shown below:
1 8590 12516
2 8641 98143
3 8705 98916
4 8750 89911
5 8685 104835
6 8629 121963
7 8676 77655
1 8577 81081
2 8593 83385
3 8642 112164
4 8708 103684
5 8622 83982
6 8593 75944
7 8600 97036
1 8650 104911
2 8730 114098
3 8731 99421
4 8715 85707
5 8717 81273
6 8739 106462
7 8684 110635
1 8713 105214
2 8771 92456
3 8759 109270
4 8762 99150
5 8730 77306
6 8780 86324
7 8804 90214
1 8797 99894
2 8863 95177
3 8873 95910
4 8827 108511
5 8806 115636
6 8869 85542
7 8854 111018
1 8571 93247
2 8533 85105
3 8553 114725
4 8561 122195
5 8532 100945
6 8560 108552
7 8634 108707
1 8646 117420
2 8633 113823
3 8680 82763
4 8765 121072
5 8756 89835
6 8750 104578
7 8790 88429
Each of them are then read as:
d1<-read.table("dat1",col.names=c("n","CR","WW"))
d2<-read.table("dat2",col.names=c("n","CR","WW"))
d3<-read.table("dat3",col.names=c("n","CR","WW"))
d4<-read.table("dat4",col.names=c("n","CR","WW"))
d5<-read.table("dat5",col.names=c("n","CR","WW"))
d6<-read.table("dat6",col.names=c("n","CR","WW"))
d7<-read.table("dat7",col.names=c("n","CR","WW"))
And my formula for percentage change applied as follows for column 2:
a1<-((d1$CR-mean(d1$CR))/mean(CR))*100
a2<-((d2$CR-mean(d2$CR))/mean(CR))*100
a3<-((d3$CR-mean(d3$CR))/mean(CR))*100
a4<-((d4$CR-mean(d4$CR))/mean(CR))*100
a5<-((d5$CR-mean(d5$CR))/mean(CR))*100
a6<-((d6$CR-mean(d6$CR))/mean(CR))*100
a7<-((d7$CR-mean(d7$CR))/mean(CR))*100
a1-a7 actually gives percentage change in the data.
Instead of doing this one after the other, can you please give an
indication on how I may apply this formula to the data frame with probably
a code.
Thank you again.
Best
Ogbos
On Wed, Nov 28, 2018 at 5:15 AM Ogbos Okike <giftedlife2014 at gmail.com>
wrote:
Dear Jim, I wish also to use the means calculated and apply a certain formula on the same data frame. In particular, I would like to subtract the means of each of these seven days from each of the seven days and and divide the outcome by the same means. If I represent m1 by the means of each seven days in column 1, and c1 is taken as column 1 data. My formula will be of the form: aa<-(c1-m1)/m1. I tried it on the first 7 rows and I have what I am looking for.: -0.0089986156 -0.0031149054 0.0042685741 0.0094600831 0.0019612367 -0.0044993078 0.0009229349 But doing it manually will take much time. Many thanks for going a step further to assist me. Warmest regards. Ogbos On Wed, Nov 28, 2018 at 4:31 AM Jim Lemon <drjimlemon at gmail.com> wrote:
Hi Ogbos, If we assume that you have a 3 column data frame named oodf, how about: oodf[,4]<-floor((cumsum(oodf[,1])-1)/28) col2means<-by(oodf[,2],oodf[,4],mean) col3means<-by(oodf[,3],oodf[,4],mean) Jim On Wed, Nov 28, 2018 at 2:06 PM Ogbos Okike <giftedlife2014 at gmail.com> wrote:
Dear List, I have three data-column data. The data is of the form: 1 8590 12516 2 8641 98143 3 8705 98916 4 8750 89911 5 8685 104835 6 8629 121963 7 8676 77655 1 8577 81081 2 8593 83385 3 8642 112164 4 8708 103684 5 8622 83982 6 8593 75944 7 8600 97036 1 8650 104911 2 8730 114098 3 8731 99421 4 8715 85707 5 8717 81273 6 8739 106462 7 8684 110635 1 8713 105214 2 8771 92456 3 8759 109270 4 8762 99150 5 8730 77306 6 8780 86324 7 8804 90214 1 8797 99894 2 8863 95177 3 8873 95910 4 8827 108511 5 8806 115636 6 8869 85542 7 8854 111018 1 8571 93247 2 8533 85105 3 8553 114725 4 8561 122195 5 8532 100945 6 8560 108552 7 8634 108707 1 8646 117420 2 8633 113823 3 8680 82763 4 8765 121072 5 8756 89835 6 8750 104578 7 8790 88429 I wish to calculate average of the second and third columns based on the first column for each repeated 7 days. The length of the data is 1442.
That
is 206 by 7. So I should arrive at 207 data points for each of the two columns after calculating the mean of each group 1-7. I have both tried factor/tapply and aggregate functions but seem not to
be
making progress.
Thank you very much for your idea.
Best wishes
Ogbos
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