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!
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
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
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]
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)))
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))))
