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

Answer 1

如果您需要的位置不仅限于此，还可以使用制表模块，该模块将为您很好地设置格式，而无需事先猜测长度。

Answer 2

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

Answer 3

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

Answer 4

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