CODEWITHSUNDEEP
python
YEp d
YEp d

Reputation: 189

How to make python output more decimal places of a recurring decimal?

```python
import decimal
x=[]
#functions etc. x is changed
#now returns is a list full of integers
print(sum(x)/len(x))
#This happens to give 0.6999
print(decimal.Decimal(sum(x)/len(x))
# This happens to give 0.6998999999996034 blah blah blah
```

The decimal module gives too many decimal places and round(decimal.Decimal(x),5) gives 0.69990

I would like it to output 0.69999(5 d.p.) but it outputs 0.6999 or 0.69990

Upvotes: 0

Views: 686

Answers (3)

iyunbo
iyunbo

Reputation: 106

you already have the right answer: round(decimal.Decimal(x),5) which gives 0.69990. In your case, you should not expect 0.69999, it is not even close to 0.699899999999. The python built-in function round is good to use, but sometimes it can be surprising, see more details here

To summary:

import decimal
x=0.699899999
print("x =", x)
g = round(decimal.Decimal(x),5)
print("rounded:", g)

output:

x = 0.699899999
rounded: 0.69990

Upvotes: 1

Blckknght
Blckknght

Reputation: 104752

The number 0.699899999999... is mathematically equivalent to 0.6999000..., assuming the 9s repeat farther than we want to go before we round things off.

This is the same as the question of why 0.9999... is equal to 1 (for which you can find many explanations online, I'm fond of this one myself), just shifted several decimal places to the right. All the 9s at the end turn into a 1. That one gets added to the last 8 and turn it into a 9. That's all followed by infinite zeros, which Python omits by default.

Upvotes: 0

Akshay Nailwal
Akshay Nailwal

Reputation: 206

You can try something like this

x=0.6999932
g = float("{0:.5f}".format(x))
print(g)

#output be like
0.69999

Upvotes: 1

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