Aligning stdout print in python

Question

I have the following program that uses a deque that I copied from the web.

from collections import Counter,deque
import re
import time
import my_ds

num = 100000

def append(c):
    for i in range(num):
        c.append(i)

def appendleft(c):
    if isinstance(c, deque):
        for i in range(num):
            c.appendleft(i)
    else:
        for i in range(num):
            c.insert(0,i)

def pop(c):
    for i in range(num):
        c.pop()

def popleft(c):
    if isinstance(c,deque):
        for i in range(num):
            c.popleft()
    else:
        for i in range(num):
            c.pop(0)


for container in [deque, list]:
    for operation in [append, appendleft, pop, popleft]:
        c = container(range(num))
        start = time.time()
        operation(c)
        elapsed = time.time() - start
        print('Completed {0}/{1}       in {2} seconds: {3} ops/sec'.format(container.__name__,operation.__name__, elapsed, num/elapsed))

The output looks like below.

Completed deque/append       in 0.011004447937011719 seconds: 9087234.595718866 ops/sec
Completed deque/appendleft       in 0.00800323486328125 seconds: 12494947.56911344 ops/sec
Completed deque/pop       in 0.00800323486328125 seconds: 12494947.56911344 ops/sec
Completed deque/popleft       in 0.009003400802612305 seconds: 11106914.175250906 ops/sec
Completed list/append       in 0.011004447937011719 seconds: 9087234.595718866 ops/sec
Completed list/appendleft       in 8.727489709854126 seconds: 11458.048456601553 ops/sec
Completed list/pop       in 0.01900768280029297 seconds: 5261030.555416185 ops/sec
Completed list/popleft       in 1.781712532043457 seconds: 56125.776858800775 ops/sec

I am looking for a nice way to align the numbers seconds and ops/second.How is this done in python

viraptor · Accepted Answer · 2016-04-27 00:47:02Z

2

If you need it in more places than just this one, you could also use the tabulate module which will format it nicely for you and doesn't require guessing the lengths upfront.

answered Apr 27, 2016 at 0:47

viraptor

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3 Comments

liv2hak Over a year ago

I tried this approach. But I am getting an error - `Type Error: 'float' object is not iterable.

viraptor Over a year ago

@liv2hak You'd have to post the actual code you're using.

liv2hak Over a year ago

Thanks :) Please see stackoverflow.com/questions/36878944/…

Hun · Accepted Answer · 2016-04-27 00:18:17Z

You need to align Completed deque/append part as well for the rest of columns to be aligned. Try this:

print('Completed {0}/{1:<11} in {2:<20} seconds: {3:<18} ops/sec'.format(container.__name__,operation.__name__, elapsed, num/elapsed))

Completed deque/append      in 0.009192228317260742 seconds: 10878755.025288548 ops/sec
Completed deque/appendleft  in 0.008057117462158203 seconds: 12411386.636681067 ops/sec
Completed deque/pop         in 0.009001970291137695 seconds: 11108679.18531663  ops/sec
Completed deque/popleft     in 0.008355855941772461 seconds: 11967654.863469055 ops/sec
Completed  list/append      in 0.009819984436035156 seconds: 10183315.528794795 ops/sec
Completed  list/appendleft  in 5.856244802474976    seconds: 17075.78890106128  ops/sec
Completed  list/pop         in 0.012813091278076172 seconds: 7804517.8817312345 ops/sec
Completed  list/popleft     in 1.437035083770752    seconds: 69587.72345181856  ops/sec

Carlos Mermingas · Accepted Answer · 2016-04-27 00:12:24Z

0

You can use the < and > alignment operators of the format() string method. Look at the documentation here:

https://docs.python.org/3.5/library/string.html#format-string-syntax

For example:

>>> a='deque'
>>> b='append'
>>> c=0.000102301230102
>>> d=0.1242344213
>>> print('Completed {0}/{1}       in {2:<15} seconds: {3:<15} ops/sec'.format(a, b, c, d))
Completed deque/append       in 0.000102301230102 seconds: 0.1242344213    ops/sec

answered Apr 27, 2016 at 0:12

Carlos Mermingas

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Comments

microwaver · Accepted Answer · 2019-01-13 17:01:13Z

0

You can use f-strings with format spec. Basically It's just like str.format. For example, you can use

a = 2
f'length:{a:<5}cm'

and it will output

length:2    cm

but the point is you can set the alignment length programmatically like this

a = 2
f'length:{a:<{a+3}}cm'

and I get result

length:2    cm

f-string is new in Python 3.6 and I'm using Python 3.7.1

https://docs.python.org/3/reference/lexical_analysis.html#formatted-string-literals

answered Jan 13, 2019 at 17:01

microwaver

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