What I want is:
if 1700 / 1000 = 1.7 = int(1) # I want this to be True lst.append("T") My original code was:
if 1700 / 1000 == int(1) # This is False and I want it to be True lst.append("T") The if statement is False because the answer is 1.7 and not 1. I would like this to be True. So I want the 1.7 to round down to 1 using int so that the if statement will be True.
// always provide integer value. If you want to get always integer then use it otherwise follow int(a/b)
if 1700 // 1000 == int(1): # I want this to be True lst.append("T") 
You either need to put the int on the other side:
if int(1700 / 1000) == 1 lst.append("T") i.e. round the 1700/1000 to an integer before comparing it, or use //, which is integer division and discards the fraction part:
if 1700 // 1000 == 1 lst.append("T") You can either use the // operator (integer division) or use the floor function from the math module:
>>> from math import floor >>> floor(1.7/1) 1 >>> floor(1.7/1) == int(1) True >>> 1.7 // 1 1.0 >>> 1.7 // 1 == 1 True You can try
if int(1700/1000) == 1 # I want this to be True lst.append("T") int(1700/1000) will convert the 1.7 into 1 by ignoring the decimal portion of the number
Everything brilliant is simple
if int(1700 / 1000) == int(1): lst.append("T") You should take a look at the floor and ceil functions. The floor function rounds to the last number, whereas ceil to the next. In your case, you need to do it like this:
import math if math.floor(1700 / 1000) == int(1): print("TRUE") else: print("FALSE") 
You could round up or down in Python by 'floor' and 'ceil' methods in the 'math' library.
from math import floor, ceil print (floor(1.7), ' , ', ceil(1.7)) The result will be
1 , 2 This will work in either Python 2.x or Python 3.x
