开发生物信息学APP,用前面的条形码识别DNA seq

Question

我正在为python类的介绍工作,我在编写脚本来读取文件时遇到了很多麻烦,然后在文件的序列开头识别条形码.

这就是我要打开我的文件:

#!/usr/bin/python 

import sys  

fname  = sys.argv[1]

handle = open(fname , "r")
# read the file # 
for line in handle:
        print line.strip()

handle.close()

它完美地打开我的文件并将内容打印到屏幕上.

我有的问题是添加到此以完成作业我收到错误消息,我不知道我做错了什么.

我将不胜感激任何帮助或建议.

作业和正确的预期结果详细说明:

创建一个名为〜/ assignments/assignment07/assignment07.py的可执行文件

python脚本应该采用2个命令行参数(按顺序):

(1)DNA条形码(2)含有DNA序列的文件的名称

您的脚本应该打印序列文件中与序列开头的给定条形码匹配的所有DNA序列,但丢弃条形码.不要打印条形码,只打印与条形码匹配的序列,并且不匹配不在序列前面的条形码.

#!/usr/bin/python 
import sys
barcode  = sys.argv[1]
filename = sys.argv[2]

bclen = len(bacode)



handle = open(fname, "r")

# read the file # 

for line in handle:

        print line.strip()


for line in filename:

        bc    = line[4:][:bclen]

        seq   = line[4:19][bclen:]



        if bc == barcode:

                seqslice = sequence[4:]

                #print "barcode %s is at beginning of sequence %s" % (barcode, seqslice)



handle.close()

Answer 1

这个脚本充满了一些常见的开始错误(不匹配的变量名称,不理解切片的使用),但这里有一个更正的版本,注释应该有所帮助:

与之合作 python script_name.py 123barcode filename.csv

#!/usr/bin/python 
import sys
barcode  = sys.argv[1]
filename = sys.argv[2]

bclen = len(barcode) #fixed typo so from bacode
handle = open(filename, "r") #changed from fname

# read the file # 
## Combined for loops, no reason for double loop here

for line in handle:
    print line.strip()

    bc    = line[:bclen]  #changed to just slice of beginning to barcode length
    seq   = line[bclen:]  #from end of barcode to end (only want 19 just add)

    print "BC = " + bc    #Added these print statements: when problems occur
    print "SEQ = " + seq  # always look to see what variable actually contain

    #I don't know what you wanted here but this prints the matching sequence
    if bc == barcode:     
        print "barcode %s is at beginning of sequence %s" % (barcode, seq)
handle.close()