Why doesn't round() work as expected on exponential numbers?

Question

Formatting the number works and cut unwanted decimals. Is there a reason the round() function give the same value passed by parameter?

number = 5.7821323121209e19
print(round(number,2)) # 5.7821323121209e19.
# Works.
print('%.3g' % number) # 5.78e19
print('{:.3g}'.format(number))  # 5.78e19
print(f'{number:.3g}')   # 5.78e19

"Doesn't work"? What does it give you, and what did you expect? — khelwood, Aug 28 '20 at 17:05

score 2 · Accepted Answer · answered Aug 28 '20 at 17:13

2

Everything works correctly. 5.7821323121209e19 is actually 5.7821323121209 * (10 ** 19) so it is a number with only 0 after the comma.

Therefore there is nothing to round from mathematical point of view.

answered Aug 28 '20 at 17:13

Magiczne

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score 2 · Answer 2 · answered Aug 28 '20 at 17:16

2

Looks like you want -17 instead of 2:

>>> round(5.7821323121208e19, -17)
5.78e+19

answered Aug 28 '20 at 17:16

superb rain

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score 1 · Answer 3 · answered Aug 28 '20 at 17:13

1

round(number, 2) rounds to 2 decimal places (i.e. hundredths). 5.7821323121209e19 is much too big for that to have any effect. It is 57821323121209000000.

answered Aug 28 '20 at 17:13

khelwood

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Why doesn't round() work as expected on exponential numbers?

3 Answers3