将数据帧转换为带计数的矩阵

将数据帧转换为带计数的矩阵

本文介绍了将数据帧转换为带计数的矩阵的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我的数据文件的结构如下:

I have data files structured like this:

OTU1    PIA0    1120
OTU2    PIA1    2
OTU2    PIA3    6
OTU2    PIA4    10
OTU2    PIA5    1078
OTU2    PIN1    24
OTU2    PIN2    45
OTU2    PIN3    261
OTU2    PIN4    102
OTU3    PIA0    16
OTU3    PIA1    59
OTU3    PIA2    27
OTU3    PIA3    180
OTU3    PIA4    200
OTU3    PIA5    251
OTU3    PIN0    36
OTU3    PIN1    61
OTU3    PIN2    156
OTU3    PIN3    590
OTU3    PIN4    277
OTU4    PIA0    401
OTU4    PIN0    2

我想创建一个矩阵,该矩阵显示第二列的数据组合,并以第一列作为组合计数的参考(显示按第一列号-OTU1,OTU2,OTU3进行测量的次数. ,OTU4-,则第二列中的每个基准点在同一OTU中一起显示).它需要看起来像这样:

And I want to create a matrix that shows combination of data from the second column taking the first column as reference for the counts of combination (showing how many times, measured each one by the first column number -OTU1, OTU2, OTU3, OTU4- each datum from the second column appears together with each other in the same OTU). It needs to look like this:

    PIA0  PIA1  PIA2  PIA3  PIA4  PIA5  PIN0  PIN1  PIN2  PIN3  PIN4
PIA0  1     1     1     1     1     1     2     1     1     1     1
PIA1  1     0     1     2     2     2     1     2     2     2     2
PIA2  1     1     0     1     1     1     1     1     1     1     1
PIA3  1     2     1     0     2     2     1     2     2     2     2
PIA4  1     2     1     2     0     2     1     2     2     2     2
PIA5  1     2     1     2     2     0     1     2     2     2     2
PIN0  2     1     1     1     1     1     0     1     1     1     1
PIN1  1     2     1     2     2     2     1     0     2     2     2
PIN2  1     2     1     2     2     2     1     2     0     2     2
PIN3  1     2     1     2     2     2     1     2     2     0     2
PIN4  1     2     1     2     2     2     1     2     2     2     0

在具有相同名称的行和列之间共享的数据反映了该数据在OTU中单独出现的次数.

Data shared between a row and a column with the same name reflects the number of times this datum appears alone in an OTU.

有什么想法吗?

我已经阅读了有关R库'reshape2'和命令'acast'的信息,但是我只能更改其中包含所有数据的矩阵的形状,而不能根据需要进行组合计数.我也一直在考虑一个Biopython脚本,但是如果我对编程的一点了解的话,将它写下来将会太大而且很难.

I have read about R libraries 'reshape2' and command 'acast' here, but with that I can only change the shape of a matrix with all data in it, not make combination counts as desired. I have also been thinking about a Biopython script, but I think it would be too big and difficult to write it down with my little knowledge about programming.

目标是建立一个类似于示例中的矩阵,这样我就可以运行 CIRCOS在线使用这些数据进行编程.

The goal is to build a matrix like the one in the example so I can run CIRCOS online program with these data.

推荐答案

您可以使用dcast创建一个二进制矩阵,指示每个OTU中每个PI的存在,然后将其自身相乘以得到计数./p>

You can use dcast to create a binary matrix indicating the presence of each PI inside each OTU, and then multiply it by itself to have the counts.

d <- read.fwf( textConnection("
OTU1    PIA0    1120
OTU2    PIA1    2
OTU2    PIA3    6
OTU2    PIA4    10
OTU2    PIA5    1078
OTU2    PIN1    24
OTU2    PIN2    45
OTU2    PIN3    261
OTU2    PIN4    102
OTU3    PIA0    16
OTU3    PIA1    59
OTU3    PIA2    27
OTU3    PIA3    180
OTU3    PIA4    200
OTU3    PIA5    251
OTU3    PIN0    36
OTU3    PIN1    61
OTU3    PIN2    156
OTU3    PIN3    590
OTU3    PIN4    277
OTU4    PIA0    401
OTU4    PIN0    2"), widths=c(8,8,10), header=FALSE, skip=1 )

library(reshape2)
A <- as.matrix( dcast( V1 ~ V2, data=d, length )[,-1]>0 )
#          PIA0     PIA1     PIA2     PIA3     PIA4     PIA5     PIN0     PIN1     PIN2     PIN3     PIN4
# [1,]     TRUE    FALSE    FALSE    FALSE    FALSE    FALSE    FALSE    FALSE    FALSE    FALSE    FALSE
# [2,]    FALSE     TRUE    FALSE     TRUE     TRUE     TRUE    FALSE     TRUE     TRUE     TRUE     TRUE
# [3,]     TRUE     TRUE     TRUE     TRUE     TRUE     TRUE     TRUE     TRUE     TRUE     TRUE     TRUE
# [4,]     TRUE    FALSE    FALSE    FALSE    FALSE    FALSE     TRUE    FALSE    FALSE    FALSE    FALSE
t(A) %*% A
#              PIA0     PIA1     PIA2     PIA3     PIA4     PIA5     PIN0     PIN1     PIN2     PIN3     PIN4
# PIA0            3        1        1        1        1        1        2        1        1        1        1
# PIA1            1        2        1        2        2        2        1        2        2        2        2
# PIA2            1        1        1        1        1        1        1        1        1        1        1
# PIA3            1        2        1        2        2        2        1        2        2        2        2
# PIA4            1        2        1        2        2        2        1        2        2        2        2
# PIA5            1        2        1        2        2        2        1        2        2        2        2
# PIN0            2        1        1        1        1        1        2        1        1        1        1
# PIN1            1        2        1        2        2        2        1        2        2        2        2
# PIN2            1        2        1        2        2        2        1        2        2        2        2
# PIN3            1        2        1        2        2        2        1        2        2        2        2
# PIN4            1        2        1        2        2        2        1        2        2        2        2

这篇关于将数据帧转换为带计数的矩阵的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!

08-20 10:14