How to set the number of decimal places in scientific notation?

Question

I am working on python 2.7. I want to fix the number of decimal places in a number written in scientific notation like 1.32e6, but instead of using for example "%3f". I want to write something like:

n=3
"%.nf"

where n is the number of decimal places, but it can be changed from my application.

Answer 1

n = 4
print("{0:.{1}f}".format(1.987213,n))
n = 5
print("{0:.{1}f}".format(1.987213,n))

---

Output

1.9872
1.98721

Answer 2

i = 123456789
n = 3
print('{:.{n}e}'.format(i, n=n))

Output:

'1.235e+08'

Answer 3

You can try this,

>>> a = 10e6
>>> n = 2
>>> '%.{}f'.format(n) % a
'10000000.00'

Or cast to float,

>>> a = 10e6
>>> n = 2
>>> float('%.{}f'.format(n) % a)
10000000.0

If you want to return to scientific notation,

>>> a = 103.023232
>>> n = 3
>>> '%.{}E'.format(n) % a
'1.030E+02'

Answer 4

The old printf -style formatting operator supports this, though you'll probably want to use the newer format method.

>>> n = 3
>>> "%.*f" % (n, 3.14159)
'3.142'

When you use * as the precision, the value to use precedes the floating-point value in the tuple on the RHS.

Answer 5

Thank you every body for the answers. I tried the method of Ajay Dabas and that worked for me perfectly. This example is going to be used in a major (but easy) application. The complete code with the improved solution is:

enter code here entrada_n= "Ingrese n:" enter code here n= input (entrada_n)

enter code here entrada_numero="Ingrese numero:" enter code here numero=input(entrada_numero)

enter code here print("{0:.{1}f}".format(numero,n)) enter code here print("{0:.{1}e}".format(numero,n))