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Why python print() prints a rounded value rather than the exact value for non-representable float

The value 0.1 is not representable as a 64 bits floats.
The exact value is roughly equals to 0.10000000000000000555
https://www.exploringbinary.com/why-0-point-1-does-not-exist-in-floating-point/
You can highlight this behavior with this simple code:
timestep = 0.1
iterations = 1_000_000
total = 0
for _ in range(iterations):
total += timestep
print(total - timestep * iterations) # output is not zero but 1.3328826753422618e-06
I totally understand why 0.1 is not representable as an exact value as a float 64, but what I don't get is why when I do print(0.1), it outputs 0.1 and not the underlying value as a float 64.
Of course, the underlying value has many more digits on a base 10 system so there should be some rounding involved, but I am looking for the specification for all values and how to control that.
I had the issue with some application storing data in database:
the python app (using str(0.1)) would show 0.1
another database client UI would show 0.10000000000000000555, which would throw off the end user
P-S: I had other issues with other values
Regards,

First, you are right, floats (single, double, whatever) have an exact value.
For 64 bits IEEE-754 double, the nearest representable value to 0.1 would be exactly 0.1000000000000000055511151231257827021181583404541015625, quite long as you can see. But representable floating point values all have a finite number of decimal digits, because the base (2) is a divisor of some power of 10.
For a REPL language like python, it is essential to have this property:
the printed representation of the float shall be reinterpreted as the same value
A consequence is that
every two different float shall have different printed representation
For obtaining those properties, there are several possbilities:
print the exact value. That can be many digits, and for the vast majority of humans, just noise.
print enough digits so that every two different float have a different representation. For double precision, that's 17 digits in the worse case. So a naive implementation for representing floating point values would be to always print 17 significant digits.
print the shortest representation that would be reinterpreted unchanged.
Python, and many other languages have chosen the 3rd solution, because it is considered annoying to print 0.10000000000000001 when user have entered 0.1. Human users generally choose the shorter representation and printed representation is for human consumption. The shorter, the better.
The bad property is that it could give the false impression that those floating point values are storing exact decimal values like 1/10. That's a knowledge that is evangelized here and in many places now.

Extract significant digit from 0.0007

I want to extract significant digit of 7 from XX=0.0007
The code is as follows
XX=0.0007
enX1=XX//10**np.floor(np.log10(XX));
But XX becomes 6not 7. Can anyone help me?

In some sense, you were lucky to start out with the value 0.0007. As it turns out, that value is one of the (many!) decimal values that cannot be represented exactly in a floating point format.
A floating point number gets usually stored in the common IEEE-754 format as powers of 2. Just like a whole number such as 175 is stored as the sum of bits with increasing powers-of-two values (165 = 128 + 32 + 4 + 1), fractions are stored as a sum of 1/power-of-two numbers. That means that a value of 1/2, 1/4, and 1/65536 can be stored exactly (and sums thereof, such as 3/4), but your 0.0007 can not. Its closest value is actually 0.0000699999999999999992887633748495. ("Closest" in the sense that adding just one more one-bit at the end will make it slightly larger than 0.0007, and the difference is ever so slightly larger than this lower one.)
In your calculation, you use the double divide slash //, which instructs Python to do an integer division and discard the fractional part. So while the intermediate calculation is correct and you get something like 6.99999..., this gets truncated and you end up with 6.
If you use a single slash, the result will keep its (exact!) decimals but Python will represent it as 7.0000, give or take a few zeroes. By default, Python displays only a small number of decimals.
Note that this still "is" not the exact value 7. The calculation starts out with an imprecise number, and although there may be some intermediate rounding here and there, there is only a small chance you end up with a precise integer. Again, not for all decimals, but for a large number of them. Other fractional values may be stored fractionally larger than the value you enter – 0.0004, for examplea – but the underlying 'problem' of accuracy is also present there. It's just not as visible as with yours.
If you want a nearest integer result, use a single divide slash for the exact calculation, followed by round to force the number to the nearest integer anyway.
a To be precise, as somewhere about 0.000400000000000000019168694409544. After your routine, Python will display it as 4 but internally it's still just a bit larger than that.

Floor division // vs int() rounded off

I am a new user to Python 3.6.0. I am trying to divide 2 numbers that produces a big output. However, using
return ans1 // ans2
produces 55347740058143507128 while using
return int(ans1 / ans2)
produces 55347740058143506432.
Which is more accurate and why is that so?

The first one is more accurate since it gives the exact integer result.
The second represents the intermediate result as a float. Floats have limited resolution (53 bits of mantissa) whereas the result needs 66 bits to be represented exactly. This results in a loss of accuracy.
If we looks at the hex representation of both results:
>>> hex(55347740058143507128)
'0x3001aac56864d42b8L'
>>> hex(55347740058143506432)
'0x3001aac56864d4000L'
we can see that the least-significant bits of the result that didn't fit in a 53-bit mantissa all got set to zero.
One way to see the rounding directly, without any complications brought about by division is:
>>> int(float(55347740058143507128))
55347740058143506432L

The flooring integer division is more accurate, in that sense.
The problem with this construction int(ans1 / ans2), is that the result is temporarily a float (before, obviously, converting it to an integer), introducing rounding to the nearest float (the amount of rounding depends on the magnitude of the number). This can even be seen by just trying to round-trip that value through a float:
print(int(float(55347740058143507128)))
Which prints 55347740058143506432. So, because plain / results in a float, that limits its accuracy.

Python: Why does have 2^-n work for n>52 and not 1+2^-n-1?

I'm pretty new to python, and I've made a table which calculates T=1+2^-n-1 and C=2^n, which both give the same values from n=40 to n=52, but for n=52 to n=61 I get 0.0 for T, whereas C gives me progressively smaller decimals each time - why is this?
I think I understand why T becomes 0.0, because of python using binary floating point and because of the machine epsilon value - but I'm slightly confused as to why C doesn't also become 0.0.
import numpy as np
import math
t=np.zeros(21)
c=np.zeros(21)
for n in range(40,61):
m=n-40
t[m]=1+2**(-n)-1
c[m]=2**(-n)
print (n,t[m],c[m])

The "floating" in floating point means that values are represented by storing a fixed number of leading digits and a scale factor, rather than assuming a fixed scale (which would be fixed point).
2**-53 only takes one (binary) digit to represent (not including the scale), but 1+2**-53 would take 54 to represent exactly. Python floats only have 53 binary digits of precision; 2**-53 can be represented exactly, but 1+2**-53 gets rounded to exactly 1, and subtracting 1 from that gives exactly 0. Thus, we have
>>> 2**-53
1.1102230246251565e-16
>>> 1+(2**-53)-1
0.0
Postscript: you might wonder why 2**-53 displays as a value not equal to the exact mathematical value when I said it was exact. That's due to the float->string conversion logic, which only keeps enough decimal digits to reconstruct the original float (instead of printing a bunch of digits at the end that are usually just noise).

The difference between both is indeed due to floating-point representation. Indeed, if you perform 1 + X where X is a very very small number, then the floating-point representation sets its exponent value to 0 and the precision is ensured by the mantissa, which is 52-bit on a 64-bit computer. Therefore, 1 + 2^(-X) if X > 52 is equal to 1. However, even 2^-100 can be represented in double-precision floating-point, so you can see C decrease for a larger number of samples.

Safest way to convert float to integer in python?

Python's math module contain handy functions like floor & ceil. These functions take a floating point number and return the nearest integer below or above it. However these functions return the answer as a floating point number. For example:
import math
f=math.floor(2.3)
Now f returns:
2.0
What is the safest way to get an integer out of this float, without running the risk of rounding errors (for example if the float is the equivalent of 1.99999) or perhaps I should use another function altogether?

All integers that can be represented by floating point numbers have an exact representation. So you can safely use int on the result. Inexact representations occur only if you are trying to represent a rational number with a denominator that is not a power of two.
That this works is not trivial at all! It's a property of the IEEE floating point representation that int∘floor = ⌊⋅⌋ if the magnitude of the numbers in question is small enough, but different representations are possible where int(floor(2.3)) might be 1.
To quote from Wikipedia,
Any integer with absolute value less than or equal to 224 can be exactly represented in the single precision format, and any integer with absolute value less than or equal to 253 can be exactly represented in the double precision format.

Use int(your non integer number) will nail it.
print int(2.3) # "2"
print int(math.sqrt(5)) # "2"

You could use the round function. If you use no second parameter (# of significant digits) then I think you will get the behavior you want.
IDLE output.
>>> round(2.99999999999)
3
>>> round(2.6)
3
>>> round(2.5)
3
>>> round(2.4)
2

Combining two of the previous results, we have:
int(round(some_float))
This converts a float to an integer fairly dependably.

That this works is not trivial at all! It's a property of the IEEE floating point representation that int∘floor = ⌊⋅⌋ if the magnitude of the numbers in question is small enough, but different representations are possible where int(floor(2.3)) might be 1.
This post explains why it works in that range.
In a double, you can represent 32bit integers without any problems. There cannot be any rounding issues. More precisely, doubles can represent all integers between and including 253 and -253.
Short explanation: A double can store up to 53 binary digits. When you require more, the number is padded with zeroes on the right.
It follows that 53 ones is the largest number that can be stored without padding. Naturally, all (integer) numbers requiring less digits can be stored accurately.
Adding one to 111(omitted)111 (53 ones) yields 100...000, (53 zeroes). As we know, we can store 53 digits, that makes the rightmost zero padding.
This is where 253 comes from.
More detail: We need to consider how IEEE-754 floating point works.
1 bit 11 / 8 52 / 23 # bits double/single precision
[ sign | exponent | mantissa ]
The number is then calculated as follows (excluding special cases that are irrelevant here):
-1sign × 1.mantissa ×2exponent - bias
where bias = 2exponent - 1 - 1, i.e. 1023 and 127 for double/single precision respectively.
Knowing that multiplying by 2X simply shifts all bits X places to the left, it's easy to see that any integer must have all bits in the mantissa that end up right of the decimal point to zero.
Any integer except zero has the following form in binary:
1x...x where the x-es represent the bits to the right of the MSB (most significant bit).
Because we excluded zero, there will always be a MSB that is one—which is why it's not stored. To store the integer, we must bring it into the aforementioned form: -1sign × 1.mantissa ×2exponent - bias.
That's saying the same as shifting the bits over the decimal point until there's only the MSB towards the left of the MSB. All the bits right of the decimal point are then stored in the mantissa.
From this, we can see that we can store at most 52 binary digits apart from the MSB.
It follows that the highest number where all bits are explicitly stored is
111(omitted)111. that's 53 ones (52 + implicit 1) in the case of doubles.
For this, we need to set the exponent, such that the decimal point will be shifted 52 places. If we were to increase the exponent by one, we cannot know the digit right to the left after the decimal point.
111(omitted)111x.
By convention, it's 0. Setting the entire mantissa to zero, we receive the following number:
100(omitted)00x. = 100(omitted)000.
That's a 1 followed by 53 zeroes, 52 stored and 1 added due to the exponent.
It represents 253, which marks the boundary (both negative and positive) between which we can accurately represent all integers. If we wanted to add one to 253, we would have to set the implicit zero (denoted by the x) to one, but that's impossible.

If you need to convert a string float to an int you can use this method.
Example: '38.0' to 38
In order to convert this to an int you can cast it as a float then an int. This will also work for float strings or integer strings.
>>> int(float('38.0'))
38
>>> int(float('38'))
38
Note: This will strip any numbers after the decimal.
>>> int(float('38.2'))
38

math.floor will always return an integer number and thus int(math.floor(some_float)) will never introduce rounding errors.
The rounding error might already be introduced in math.floor(some_large_float), though, or even when storing a large number in a float in the first place. (Large numbers may lose precision when stored in floats.)

Another code sample to convert a real/float to an integer using variables.
"vel" is a real/float number and converted to the next highest INTEGER, "newvel".
import arcpy.math, os, sys, arcpy.da
.
.
with arcpy.da.SearchCursor(densifybkp,[floseg,vel,Length]) as cursor:
for row in cursor:
curvel = float(row[1])
newvel = int(math.ceil(curvel))

Since you're asking for the 'safest' way, I'll provide another answer other than the top answer.
An easy way to make sure you don't lose any precision is to check if the values would be equal after you convert them.
if int(some_value) == some_value:
some_value = int(some_value)
If the float is 1.0 for example, 1.0 is equal to 1. So the conversion to int will execute. And if the float is 1.1, int(1.1) equates to 1, and 1.1 != 1. So the value will remain a float and you won't lose any precision.