如何从具有 python 文件列表的单个文件中打开多个文件以及如何对它们进行处理？

Question

i am having single file called bar.txt which is having list of files as follows,

bar.txt -

1.txt

2.txt

3.txt

each file inside bar.txt having some similar contents.

1.txt -

spec = sadasdsad

2.txt -

spec = dddddd

3.txt -

spec = ppppppppp

how can i open all files inside bar.txt and extract the data from all file and store in other file called foo.txt?

in foo.txt i want extracted data mentioned below,

foo.txt -

spec = sadasdsad

spec = dddddd

spec = ppppppppp

 outfile = open('bar.txt', "rw")
 outfile_1 = open('foo.txt', "w")
     for f in outfile:
        f=f.rstrip()
        lines = open(f,'rw')
        lines = re.findall(".*SPEC.*\\n",lines)
        outfile_1.write(lines)
 outfile.close()

Answer 1

I'd do this:

infile = open('bar.txt', "r")
outfile = open('foo.txt', "w")
line = infile.readline()

while line:
    f = line.rstrip()
    contents_file = open(f,'rw')
    contents = contents_file.read()
    outfile.write(contents)
    f = infile.readline()
 
outfile.close()

Answer 2

You're code is almost correct. I guess you messed everything up by using f variable for almost everything. So you assign multiple different things to one f variable. First its single line on outfile, then it's the same line striped, then it's another opened file, and finally you try to use the same f variable outside of it's scope (outside for loop). Try using different variables for all theses beings.

Also make sure you have correct indents (eg for loop indent is incorrect in your example), and not that regex findall works on string, not file-like object, so second argument of findall should be contentfile.read() .

infile = open('bar.txt', "r")
outfile = open('foo.txt', "w")
for f in infile:
    name=f.rstrip()
    contentfile = open(name,'rw')
    #all_matches= re.findall(<define your real pattern here>,contentfile.read())
    result = 0 #do something with your all_matches
    outfile.write(result)
    contentfile.close()
outfile.close()
infile.close()