CODEWITHSUNDEEP
python
RJK
RJK

Reputation: 242

Convert list into table - python

I have two arrays. column_names hold the column titles. values hold all the values.

I understand if I do this:

column_names = ["a", "b", "c"]
values = [1, 2, 3]
for n, v in zip(column_names, values):
    print("{} = {}".format(n, v))

I get

a = 1
b = 2
c = 3

How do I code it so if I pass:

column_names = ["a", "b", "c"]
values = [1, 2, 3, 4, 5, 6, 7, 8, 9]

I would get

a = 1, 4, 7
b = 2, 5, 8
c = 3, 6, 9

Thank you!

Upvotes: 3

Views: 51249

Answers (5)

PoweredBy90sAi
PoweredBy90sAi

Reputation: 1233

With pandas and numpy it is easy and the result will be a much more useful table. Pandas excels at arranging tabular data. So lets take advantage of it: install pandas with:

pip install pandas --user

#pandas comes with numpy
import numpy as np
import pandas as pd

# this makes a normal python list for integers 1-9
input = list(range(1,10))

#lets convert that to numpy array as np.array
num = np.array(input)

#currently its shape is single dimensional, lets change that to a two dimensional matrix that turns it into the clean breaks you want
reshaped = num.reshape(3,3)

#now construct a beautiful table
pd.DataFrame(reshaped, columns=['a','b','c'])

#ouput is
   a  b  c
0  1  2  3
1  4  5  6
2  7  8  9

Upvotes: 5

RoadRunner
RoadRunner

Reputation: 26335

You can also use slicing and a collections.defaultdict to collect your values:

from collections import defaultdict

column_names = ["a", "b", "c"]
values = [1, 2, 3, 4, 5, 6, 7, 8, 9]

column_len = len(column_names)

d = defaultdict(list)
for i in range(0, len(values), column_len):
    seq = values[i:i+column_len]

    for idx, number in enumerate(seq):
        d[column_names[idx]].append(number)

for k, v in d.items():
    print('%s = %s' % (k, ', '.join(map(str, v))))

Which Outputs:

a = 1, 4, 7
b = 2, 5, 8
c = 3, 6, 9

This can be imporoved if you create zipped lists with itertools.cycle, avoiding the slicing all together:

from collections import defaultdict
from itertools import cycle

column_names = ["a", "b", "c"]
values = [1, 2, 3, 4, 5, 6, 7, 8, 9]

column_names = cycle(column_names)

d = defaultdict(list)
for column, val in zip(column_names, values):
    d[column].append(val)

for k, v in d.items():
    print('%s = %s' % (k, ', '.join(map(str, v))))

Upvotes: 0

chickity china chinese chicken
chickity china chinese chicken

Reputation: 8057

itertools.cycle seems appropriate in this case. Here's another version for future readers:

import itertools

column_names = ["a", "b", "c"]
values = [1, 2, 3, 4, 5, 6, 7, 8, 9]

L = zip(itertools.cycle(column_names), values)

for g, v in itertools.groupby(sorted(L), lambda x: x[0]):
    print("{} = {}".format(g, [i[1] for i in v]))

gives:

a = [1, 4, 7]
b = [2, 5, 8]
c = [3, 6, 9]

Upvotes: 1

AmphotericLewisAcid
AmphotericLewisAcid

Reputation: 2299

This has two sub-steps that you want to do.

First, you want to divide your list into chunks, and then you want to assign those chunks to a dictionary.

To split the list into chunks, we can create a function:

def chunk(values, chunk_size):
    assert len(values)%chunk_size == 0 # Our chunk size has to evenly fit in our list
    steps = len(values)/chunk_size
    chunky_list = []
    for i in range(0,steps):
        position = 0 + i
        sub_list = []
        while position < len(values):
            sub_list.append(values[position])
            position += chunk_size
        chunky_list.append(sub_list)
    return chunky_list

At this point we will have: [[1,4,7],[2,5,8],[3,6,9]]

From here, creating the dict is really easy. First, we zip the two lists together:

zip(column_names, chunk(3))

And take advantage of the fact that Python knows how to convert a list of tuples into a dictionary:

dict(zip(column_names, chunk(3)))

Upvotes: 0

Sunitha
Sunitha

Reputation: 12025

You can do it as follows

>>> for n, v in zip(column_names, zip(*[values[i:i+3] for i in range(0,len(values),3)])):
...     print("{} = {}".format(n, ', '.join(map(str, v))))
... 
a = 1, 4, 7
b = 2, 5, 8
c = 3, 6, 9

Alternatively, you can use grouper defined in itertools

>>> def grouper(iterable, n, fillvalue=None):
...     "Collect data into fixed-length chunks or blocks"
...     # grouper('ABCDEFG', 3, 'x') --> ABC DEF Gxx"
...     args = [iter(iterable)] * n
...     return zip_longest(*args, fillvalue=fillvalue)
... 
>>> from itertools import zip_longest
>>> for n, v in zip(column_names, zip(*grouper(values, 3))):
...     print("{} = {}".format(n, ', '.join(map(str, v))))
... 
a = 1, 4, 7
b = 2, 5, 8
c = 3, 6, 9

Upvotes: 2

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