分隔元组并将其保存到文件

Question

I currently have this code:

from collections import  OrderedDict
a = [(1), (7), (3), (1)]
b = [(2), (4), (7), (4)]
c = [(8), (1), (3), (3)]
d = (list(OrderedDict((sub[0], sub) for sub in sorted(zip(a, b, c))).values()))
print(d)

Output:

[(1, 4, 3), (3, 7, 3), (7, 4, 1)]

I am currently trying to save to 3 files. D://a.txt , D://b.txt and D://c.txt

In D://a.txt I want to save:

1
3
7

In D://b.txt I want to save:

4
7
4

And in D://c.txt I want to save:

3
3
1

I know how to save it:

with open("D:\\a.txt", "a") as myfile1:
   myfile1.write(num1)
   myfile1.write("\n")

Or with b.txt:

with open("D:\\b.txt", "a") as myfile2:
   myfile2.write(num2)
   myfile2.write("\n")

with num1 and num2 in this case to be used like:

for num1 in a:
   myfile1.write(num1)
   myfile1.write("\n")

My objective is to save the numbers in a.txt , b.txt and c.txt without any ( ) [ ] ,

Answer 1

>>> d
[(1, 4, 3), (3, 7, 3), (7, 4, 1)]
>>> with open('a.txt','w') as fa, open('b.txt','w') as fb, open('c.txt','w') as fc:
...     for a,b,c in d:
...         fa.write(str(a) + '\n')
...         fb.write(str(b) + '\n')
...         fc.write(str(c) + '\n')
...         
>>> !cat a.txt
1
3
7

Answer 2

Try this:

letters = ("a", "b", "c")
for i, nums in enumerate(zip(d)):
    with open("D:\\{}.txt".format(letters[i]), "a") as myfile:
        myfile.write("\n".join(str(num) for num in nums))

Answer 3

Assuming the number of tuples as well as the number of entries in each tuple in d can vary:

for i, nums in enumerate(zip(*d)):
    fname = '/tmp/%s.txt' % chr(ord('a') + i)
    with open(fname, 'w') as f:
        f.write('\n'.join('%s' % n for n in nums))

This way you get exactly what you want:

!cat /tmp/a.txt
1
3
7
!cat /tmp/b.txt
4
7
4
!cat /tmp/c.txt
3
3
1

Answer 4

I'm going to start assuming that the code that gets the list of tuples works and does not need any adjusting.

since d is a list of tuples, I'm going to assign 3 variables, one to each tuple to make things simpler for now.

a = d[0] #(1,4,3)
b = d[1] #(3,7,3)
c = d[2] #(7,4,1)

Now, it's time to loop through them, and assign the values to the files.

for x in a:
    with open("D:\\a.txt", "a") as file:
        file.write(x)
        file.write("\n")

for y in b:
    with open("D:\\b.txt", "a") as file:
        file.write(y)
        file.write("\n")

for z in c:
    with open("D:\\c.txt", "a") as file:
        file.write(z)
        file.write("\n")

Your output should be free of any type of brackets or parenthesis.

Answer 5

While I'm not confident that you're doing what you are trying to do, your definition of d could be rewritten in a more simple fashion, and then e could be created to pair up the entries in each resultant tuple.

a = [(1), (7), (3), (1)]
b = [(2), (4), (7), (4)]
c = [(8), (1), (3), (3)]
d = sorted(zip(a, b, c))[1:] # [(1, 4, 3), (3, 7, 3), (7, 4, 1)]
e = list(zip(*d)) # [(1, 3, 7), (4, 7, 4), (3, 3, 1)]

Then, you could write to each file as:

for name, values in zip(['a', 'b', 'c'], e):
    with open('D:\\{}.txt'.format(name), 'w') as myfile:
        myfile.write('\n'.join(map(str,values))