List Comprehension Nested Loop, Multiple Operations - python

Given a list like so:
[[1,2,3],[4,5,6],[7,8,9]]
I'm trying to get my output to look like this, with using only a list comprehension:
1 2 3
4 5 6
7 8 9
Right now I have this:
[print(x,end="") for row in myList for x in row]
This only outputs:
1 2 3 4 5 6 7 8 9
Is there a way to have it break to a new line after it processes each inner list?


You could do like the below.
>>> l = [[1,2,3],[4,5,6],[7,8,9]]
>>> for i in l:
print(' '.join(map(str, i)))
1 2 3
4 5 6
7 8 9


You can do as follows:
print("\n".join(" ".join(map(str, x)) for x in mylist))
Gives:
1 2 3
4 5 6
7 8 9


>>> l = [[1,2,3],[4,5,6],[7,8,9]]
>>> for line in l:
print(*line)
1 2 3
4 5 6
7 8 9
A good explanation of the star in this answer. tl;dr version: if line is [1, 2, 3], print(*line) is equivalent to print(1, 2, 3), which is distinct from print(line) which is print([1, 2, 3]).


I agree with Ashwini Chaudhary that using list comprehension to print is probably never the "right thing to do".
Marcin's answer is probably the best one-liner and Andrew's wins for readability.
For the sake of answering the question that was asked...
>>> from __future__ import print_function # python2.x only
>>> list = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
>>> [print(*x, sep=' ') for x in list]
1 2 3
4 5 6
7 8 9
[None, None, None]


Try this:
print("\n".join([' '.join([str(val) for val in list]) for list in my_list]))
In [1]: my_list = [[1,2,3],[4,5,6],[7,8,9]]
In [2]: print("\n".join([' '.join([str(val) for val in list]) for list in my_list]))
1 2 3
4 5 6
7 8 9

Related

How I can get the loop result as list or string

I write this code
for i in range(1,6):
for m in range(i):
print(i)
And I get the output like this:
1
2
2
3
3
3
4
4
4
4
5
5
5
5
5
How I can get the output as an iterator (list or tuple) like this
1[1]
2[2,2]
or simple string like this (In each category of numbers, we return to the line)
1
22
333

You can do this for iterable format output:
for i in range(1,6):
print(i, [i]*i)
output ->
1 [1]
2 [2, 2]
3 [3, 3, 3]
4 [4, 4, 4, 4]
5 [5, 5, 5, 5, 5]
or this for string format output:
for i in range(1,6):
print(str(i)*i)
output ->
1
22
333
4444
55555

How to change a list [1,2,3] into 1 2 3 with spaces in between?

I created a list
x = list(range(0,5))
which gives the result [0,1,2,3,4] but I want to convert it as 0 1 2 3 4
How do I do that?

myList = [1, 2, 3]
print(' '.join(myList))

Try using * to unpack the list
print(*[1,2,3]) # -> 1 2 3

Slicing lists containing lists [duplicate]

This question already has answers here:
Slicing list of lists in Python
(6 answers)
Closed 2 years ago.
I have a list containing 3 lists:
V1 = [0,0,0,0,1,1,1,1,2,2,2,2,3,3,3,3]
V2 = [4,4,4,4,5,5,5,5,6,6,6,6,7,7,7,7]
V3 = [8,8,8,8,9,9,9,9,10,10,10,10,11,11,11,11]
V = [V1,V2,V3]
I would like to slice V so that the indexes 4:7 of V1,V2 and V3 are selected and put into list Z. I tried doing it like this:
Z = V[:,4:7]
But I get the error:
TypeError: list indices must be integers or slices, not tuple
How do I correct this?

You need to use numpy to get an array that is sliceable in multi-dimension
V = np.array([V1, V2, V3])
Z = V[:, 4:7]
# V
[[ 0 0 0 0 1 1 1 1 2 2 2 2 3 3 3 3]
[ 4 4 4 4 5 5 5 5 6 6 6 6 7 7 7 7]
[ 8 8 8 8 9 9 9 9 10 10 10 10 11 11 11 11]]
# Z
[[1 1 1]
[5 5 5]
[9 9 9]]

Your code will work if it was numpy.array
V1 = [0,0,0,0,1,1,1,1,2,2,2,2,3,3,3,3]
V2 = [4,4,4,4,5,5,5,5,6,6,6,6,7,7,7,7]
V3 = [8,8,8,8,9,9,9,9,10,10,10,10,11,11,11,11]
V = [V1,V2,V3]
import numpy as np
spam = np.array(V)
print(spam[:,4:7])
output
[[1 1 1]
[5 5 5]
[9 9 9]]
without numpy you can use list comprehension:
eggs = [item[4:7] for item in V]
print(eggs)
output
[[1, 1, 1], [5, 5, 5], [9, 9, 9]]

You cannot simply slice a sublist, but you can iterate over each sublist and slice them one by one:
V1 = [0,0,0,0,1,1,1,1,2,2,2,2,3,3,3,3]
V2 = [4,4,4,4,5,5,5,5,6,6,6,6,7,7,7,7]
V3 = [8,8,8,8,9,9,9,9,10,10,10,10,11,11,11,11]
V = [V1, V2, V3]
Z = []
for sublist in V:
Z.append(sublist[4:7])
Or use numpy as suggested in other answers.

Split string with double enter in Python

I have something like that:
1
2
3
4
5
6
And I want it to have :
[[1,2,3],[4,5,6]]
How can I do it? Thanks a lot

With list comprehension
import os
src = """1
2
3
4
5
6"""
print [[int(x) for x in sub.split(os.linesep)] for sub in src.split(os.linesep*2)]
#special case if this is for windows and src is a string
print [[int(x) for x in sub.split('\n')] for sub in src.split('\n\n')]
would return
[[1, 2, 3], [4, 5, 6]]

Try this:
input = '''1
2
3
4
5
6'''
def parse(string):
out = []
groups = string.split('\n\n') # Split by empty line
for group in groups:
out.append([item.strip() for item in group.split('\n')])
return out
print(parse(input))

get rid of commas in a list python

I have a list [1,2,4,7,5,2] where I want to get rid of the commas to make it [1 2 4 7 5 2]
how would I go about this?
np.random.randint(0,4,12) will print out like [0 3 4 1 3 4 2 1 2 4 3 4] and thats the kind of thing I want :)

You could do this:
out = '[' + ' '.join(str(item) for item in numbers)+ ']'
print out

In [7]: data = [1,2,4,7,5,2]
In [11]: '[{}]'.format(' '.join(map(str, data)))
Out[11]: '[1 2 4 7 5 2]'
or,
In [14]: str(data).replace(',','')
Out[14]: '[1 2 4 7 5 2]'

If you want a numpy ndarray, use:
np.array([1, 2, 4, 7, 5, 2])

Categories

Resources