如何使python输出一个重复小数点的更多小数位？

Question

```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 dp) but it outputs 0.6999 or 0.69990

Answer 1

You can try something like this

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

#output be like
0.69999

Answer 2

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

Answer 3

The number 0.699899999999... is mathematically equivalent to 0.6999000... , assuming the 9 s 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 9 s 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.