gpt4 book ai didi

python - 计算 python 列中具有相同值的行

转载 作者:太空宇宙 更新时间:2023-11-04 03:44:25 25 4
gpt4 key购买 nike

我正在尝试在 python 中重现 R aggregate() 函数,但没有连接。对于每一行,我只想计算给定列中具有相似值的行的出现次数。

我正在尝试从这里获取的一段代码中解决这个问题: http://timotheepoisot.fr/2011/12/01/the-aggregate-function-in-python/

我实现的修改由### 表示。我目前遇到的问题是第一列 [0] 包含字符串,代码似乎只适用于 float 。

import numpy as np
import scipy as sp

def MSD(vec):
return [np.mean(vec),np.std(vec)]
def aggregate(df,by=0,to=1,func=np.sum):
Dat = []
# ColBy = df.T[by]
ColBy = int(df.T[by][3:]) ### my attempt to read only the numbers in the first column's character strings
ColTo = df.T[to]
UniqueBy = np.sort(np.unique(ColBy))
for ub in UniqueBy:
uTo = ColTo[ColBy==ub]
Out = func(uTo)
# Dat.append(np.concatenate(([ub],Out)))
Dat.append([ub],Out) ### because I do not want to concatenate
return Dat

test_df = np.loadtxt('in_test.txt')
Agr = aggregate(test_df,0,3,MSD)
sp.savetxt("out_test.txt", Agr)

这是错误信息:

Traceback (most recent call last):
File "count_same_reads.py", line 30, in <module>
test_df = np.loadtxt('in_test.txt')
File "/usr/lib/python2.7/dist-packages/numpy/lib/npyio.py", line 796, in loadtxt
items = [conv(val) for (conv, val) in zip(converters, vals)]
ValueError: could not convert string to float: Tag19184

我的数据是制表符分隔的,主要包含字符串,除了第 3 列,我想在其中写入行的出现次数。

测试数据如下:

Tag19184    CTAAC   hffef   1   a   36  -   chr1    10006   0   36M 36
Tag19184 CTAAC hffef 1 a 36 - chr1 10012 0 36M 36
Tag19184 CTAAC hffef 1 a 36 - chr1 10018 0 36M 36
Tag19184 CTAAC hffef 1 a 36 - chr1 10024 0 36M 36
Tag19184 CTAAC hffef 1 a 36 - chr1 10030 0 36M 36
Tag19184 CTAAC hffef 1 a 36 - chr1 10036 0 36M 36
Tag19184 CTAAC hffef 1 a 36 - chr1 10042 0 36M 36
Tag20198 CTAAC hffef 1 a 36 - chr1 10048 0 36M 36
Tag20198 CTAAC hffef 1 a 36 - chr1 10054 0 36M 36
Tag45093 CTAAC hffef 1 a 36 - chr1 10060 0 36M 36

结果应该是这样的:

Tag19184    CTAAC   hffef   7   a   36  -   chr1    10006   0   36M 36
Tag19184 CTAAC hffef 7 a 36 - chr1 10012 0 36M 36
Tag19184 CTAAC hffef 7 a 36 - chr1 10018 0 36M 36
Tag19184 CTAAC hffef 7 a 36 - chr1 10024 0 36M 36
Tag19184 CTAAC hffef 7 a 36 - chr1 10030 0 36M 36
Tag19184 CTAAC hffef 7 a 36 - chr1 10036 0 36M 36
Tag19184 CTAAC hffef 7 a 36 - chr1 10042 0 36M 36
Tag20198 CTAAC hffef 2 a 36 - chr1 10048 0 36M 36
Tag20198 CTAAC hffef 2 a 36 - chr1 10054 0 36M 36
Tag45093 CTAAC hffef 1 a 36 - chr1 10060 0 36M 36

您可能会说,我还不太擅长 Python。欢迎任何建议。

[编辑] 附言。数据已按列 [0] 排序。

最佳答案

我会推荐pandas,尤其是在你的基因组数据的情况下,数据的大小可能非常大:

In [44]:
#you can read you data by pandas.read_csv()
import pandas as pd
print df
v0 v1 v2 v3 v4 v5 v6 v7 v8 v9 v10 v11
0 Tag19184 CTAAC hffef 1 a 36 - chr1 10006 0 36M 36
1 Tag19184 CTAAC hffef 1 a 36 - chr1 10012 0 36M 36
2 Tag19184 CTAAC hffef 1 a 36 - chr1 10018 0 36M 36
3 Tag19184 CTAAC hffef 1 a 36 - chr1 10024 0 36M 36
4 Tag19184 CTAAC hffef 1 a 36 - chr1 10030 0 36M 36
5 Tag19184 CTAAC hffef 1 a 36 - chr1 10036 0 36M 36
6 Tag19184 CTAAC hffef 1 a 36 - chr1 10042 0 36M 36
7 Tag20198 CTAAC hffef 1 a 36 - chr1 10048 0 36M 36
8 Tag20198 CTAAC hffef 1 a 36 - chr1 10054 0 36M 36
9 Tag45093 CTAAC hffef 1 a 36 - chr1 10060 0 36M 36
In [45]:
#if we want to group by the first 3 fields
df.groupby(['v0','v1','v2']).transform(sum).v3
Out[45]:
0 7
1 7
2 7
3 7
4 7
5 7
6 7
7 2
8 2
9 1
Name: v3, dtype: int64
In [46]:
#all it takes is just one line
df['v3']=df.groupby(['v0','v1','v2']).transform(sum).v3
print df
v0 v1 v2 v3 v4 v5 v6 v7 v8 v9 v10 v11
0 Tag19184 CTAAC hffef 7 a 36 - chr1 10006 0 36M 36
1 Tag19184 CTAAC hffef 7 a 36 - chr1 10012 0 36M 36
2 Tag19184 CTAAC hffef 7 a 36 - chr1 10018 0 36M 36
3 Tag19184 CTAAC hffef 7 a 36 - chr1 10024 0 36M 36
4 Tag19184 CTAAC hffef 7 a 36 - chr1 10030 0 36M 36
5 Tag19184 CTAAC hffef 7 a 36 - chr1 10036 0 36M 36
6 Tag19184 CTAAC hffef 7 a 36 - chr1 10042 0 36M 36
7 Tag20198 CTAAC hffef 2 a 36 - chr1 10048 0 36M 36
8 Tag20198 CTAAC hffef 2 a 36 - chr1 10054 0 36M 36
9 Tag45093 CTAAC hffef 1 a 36 - chr1 10060 0 36M 36

关于python - 计算 python 列中具有相同值的行,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/24581967/

25 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com