CODEWITHSUNDEEP
pythonlistfile
Gábor
Gábor

Reputation: 327

Create list of lists from list of numbers

I'm working currently on a basic project, where I want to visualize later some data logged by some sensors.

I have a file with numbers:

75
0
0
30
35
32
38
45
53
44
51
62
43
34
56
42
28
32
43
56
43
46
16
33
48
...

And I want to create a list of lists, by every sub list will contain 6 element from the original ones, like [75,0,0,30,35,32],[38,45,53,44,51,62],...].

f = open("log2.txt", "r")
content = f.readlines()
 
# Variable for storing the sum
idx = 0

line = []
db =  [] 

# Iterating through the content
for number in content:
    print(idx%6)
    if idx % 6 == 0:
        #newline
        db.append(line)
        line = []
        print("AAAAA")
    else:
        #continue line
        line.append(int(number))
    idx += idx+1

print(db)

But the result of db is only [[]]. Why? How can I do it well?

Upvotes: 1

Views: 316

Answers (4)

fshabashev
fshabashev

Reputation: 677

Taking inspiration from another answer here, we can provide a solution with lazy evaluation.

def chunks(lst, n):
    """Yield successive n-sized chunks from lst."""
    for i in range(0, len(lst), n):
        yield lst[i:i + n]


f = open("log2.txt", "r")
content = list(int(line) for line in f)

list_of_lists = list(chunks(content, 6))

Here we don't keep intermediate data in memory and this approach can be used to process very large files.

Upvotes: 2

Abhyuday Vaish
Abhyuday Vaish

Reputation: 2379

Method 1(List comprehension):

You can create a list comprehension looping through the lines of f:

f = open("log2.txt", "r")
content = f.readlines()
l = list(map(int, content)) # Converting each line to integer.
l = [l[i:i+6] for i in range(0,len(l), 6)] # Creating lists of 6 elements
print(l)

For eg, if a file contains:

1
2
3
4
5
6
7
8
9
10
11
12

Then the output is:

[[1, 2, 3, 4, 5, 6], [7, 8, 9, 10, 11, 12]]

Method 2(Improvements in your code):

f = open("log2.txt", "r")
content = f.readlines()
line = []
db =  [] 

# Iterating through the content
for idx, number in enumerate(content):
    if idx % 6 == 0:
        #newline
        line = [int(number)] # You need to add that number too!
        db.append(line)
    else:
        #continue line
        line.append(int(number))
print(db)

Output:

[[1, 2, 3, 4, 5, 6], [7, 8, 9, 10, 11, 12]]

What were you doing wrong?

  1. Using a variable to keep track of the index. Instead, you can use enumerate() to get both the value and the index. Then you won't need to update or initialize idx.
  2. You were not appending the number when idx % 6 == 0.

Upvotes: 1

ex4
ex4

Reputation: 2428

There is much more simple way to do that:

f = open("log2.txt", "r")
content = f.readlines()

# now your content looks like this: ['75\n', '0\n', '0\n',...]
content = [int(x.strip()) for x in content]

# now your content looks like this [75, 0, 0, 30, 35, ...]

result = []
while content:
   result.append(content[:6])
   content = content[6:]

# now your result looks like this: [[75, 0, 0, 30, 35, 32], [38, 45, 53, 44, 51, 62],...]
# and your content variable is empty list

Upvotes: 1

Rahil Kadakia
Rahil Kadakia

Reputation: 135

f = open("log2.txt", "r", encoding='utf-8')
content = f.read()

content = content.split()
int_content = []
for i in content:
    int_content.append(int(i))

print(int_content)

final = []
# Iterating through the content
rng = 0
temp = []
for number in int_content:
    if rng != 6:
        temp.append(number)
        print(temp)
        range += 1
        continue
    else:
        rng = 0
        final.append(temp)
        temp = []

print(final)

Upvotes: 1

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