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Computing pi by using definite integral of a circle

I am new to coding, and my assignment requires that I approximate pi by using the definite integral of area of a disk (circle) with radius 1. I have created the following code, via Python. This gives me the correct answer, however pi is truncated to six digits. Is there a way for me to expand it to 7 digits, per assignment parameters? Thanks so much!
Code:
GlowScript 2.7 VPython
x=-1
dx=0.00001
A=0
while x<1:
A=A+sqrt(1-x**2)*dx
x=x+dx
tpi=2*A
print(tpi)

I think, you're using Python version 2.7? Try to format the float when you're printing the value:
print('%.7f' % tpi)
This should print your value as a float rounded to 7 decimal places. To read more about that, have a look at String Formatting Operations

What you see in the console does not always match the value of a variable. The console might cut of some digits (a logger usually does this in order to save space on the screen) or the python interpreter might round. To combat this you can use the following code to extend the number of digits
# coding: utf-8
import math
x=-1
dx=0.00001
A=0
while x<1:
A=A+math.sqrt(1-x**2)*dx
x=x+dx
tpi=2*A
print("{0:.30g}".format(tpi)) #this will give you 30 digits including the digits before the . !

Python : arbitrary precision with floats

I tried to compute math.exp(9500) but encountered an OverflowError: math range error (it's roughly 6.3e4125). From this question it seems like it's due to a too large float, the accepted answer says "(...) is slightly outside of the range of a double, so it causes an overflow".
I know that Python can deal with arbitrarily large integers (long type), is there a way to deal with arbitrarily large floats in the same manner ?
Edit : my original question was about using integers for calculating exp(n) but as Eric Duminil said, the simplest way to do that would be 3**n which doesn't provide any useful result. I know realize this question might be similar to this one.

I don't think it's possible to approximate exp() with integers. If you use 3**n instead of 2.71828182845905**n, your calculations will be completely useless.
One possible solution would be to use Sympy. According to the documentation:
There is essentially no upper precision limit
>>> from sympy import *
>>> exp(9500)
exp(9500)
>>> exp(9500).evalf()
6.27448493490172e+4125
You can also specify the desired precision:
>>> exp(9500).evalf(1000)
6.274484934901720177929867046175406311474380389941415760684209191232450360090766458256588885184199320756050569665785657269735313171886975309933254563488343491718198237894473901620914303565550450204805537225888529509352754121292701357622411614860860409639719786022989336837263283678476008817556351031696366815467221836948040042378034720460820127399855873232167818091083005170669482845098735176209372328114732133251096196535355946589133977397512846130629857604295369747597459602137604440011394793443041829253598478244189078131130488653468669559814695095974271938947640276013215753183113041899037415404445478806695965167014404297848725756879184380559837391976534521522360723388582608454995349380217499779247330557664230806254642768796486899322646423713763772064068933790640394967085887914192401473425799354391464743910233873602389444180426155866237536459654917521713769608318128404177877383203786348495822099924812081683286880293701785567962687838594752986160305764297117036426951203418854463404773701882e+4125
With exp(9500).evalf(5000), you even get the integer closest to exp(9500).

Here's another way to calculate the result with Python:
exp(9500)
is too big.
But log10(exp(9500)) isn't. You cannot calculate it this way in Python, but log10(exp(9500)) is log(exp(9500))/ln(10), which is 9500/ln(10):
>>> from math import log
>>> 9500/log(10)
4125.797578080892
>>> int(9500/log(10))
4125
>>> 10**(9500/log(10) % 1)
6.274484934896202
This way, you can calculate that exp(9500) is 6.27448493 * 10**4125 in plain Python, without any library!

try long type.
int type has been remove from python since 3.0 version.

Arithmetic error: Python is incorrectly dividing variables

I'm getting something that doesn't seem to be making a lot of sense. I was practicing my coding by making a little program that would give me the probability of getting certain cards within a certain timeframe of a card game. In order to calculate the chances, I would need to create a method that would perform division, and report the chances as a fraction and as a decimal. so I designed this:
from fractions import Fraction
def time_odds(card_count,turns,deck_size=60):
chance_of_occurence = float(card_count)/float(deck_size)
opening_hand_odds = 7*chance_of_occurence
turn_odds = (7 + turns)*chance_of_occurence
print ("Chance of it being in the opening hand: %s or %s" %(opening_hand_odds,Fraction(opening_hand_odds)))
print ("Chance of it being acquired by turn %s : %s or %s" %(turns,turn_odds,Fraction(turn_odds) ))
and then I used it like so:
time_odds(3,5)
but for whatever reason I got this as the answer:
"Chance of it being in the opening hand: 0.35000000000000003 or
6305039478318695/18014398509481984"
"Chance of it being acquired by turn 5 : 0.6000000000000001 or
1351079888211149/2251799813685248"
so it's like, almost right, except the decimal is just slightly off, giving like a 0.0000000000003 difference or a 0.000000000000000000001 difference.
Python doesn't do this when I just make it do division like this:
print (7*3/60)
This gives me just 0.35, which is correct. The only difference that I can observe, is that I get the slightly incorrect values when I am dividing with variables rather than just numbers.
I've looked around a little for an answer, and most incorrect division problems have to do with integer division (or I think it can be called floor division) , but I didn't manage to find anything addressing this.
I've had a similar problem with python doing this when I was dividing really big numbers. What's going on?
Why is this so? what can I do to correct it?

The extra digits you're seeing are floating point precision errors. As you do more and more operations with floating point numbers, the errors have a chance of compounding.
The reason you don't see them when you try to replicate the computation by hand is that your replication performs the operations in a different order. If you compute 7 * 3 / 60, the mutiplication happens first (with no error), and the division introduces a small enough error that Python's float type hides it for you in its repr (because 0.35 unambiguously refers to the same float value as the computation). If you do 7 * (3 / 60), the division happens first (introducing error) and then the multiplication increases the size of the error to the point that it can't be hidden (because 0.35000000000000003 is a different float value than 0.35).
To avoid printing out the the extra digits that are probably error, you may want to explicitly specify a precision to use when turning your numbers into strings. For instance, rather than using the %s format code (which calls str on the value), you could use %.3f, which will round off your number after three decimal places.
There's another issue with your Fractions. You're creating the Fraction directly from the floating point value, which already has the error calculated in. That's why you're seeing the fraction print out with a very large numerator and denominator (it's exactly representing the same number as the inaccurate float). If you instead pass integer numerator and denominator values to the Fraction constructor, it will take care of simplifying the fraction for you without any floating point inaccuracy:
print("Chance of it being in the opening hand: %.3f or %s"
% (opening_hand_odds, Fraction(7*card_count, deck_size)))
This should print out the numbers as 0.350 and 7/20. You can of course choose whatever number of decimal places you want.
Completely separate from the floating point errors, the calculation isn't actually getting the probability right. The formula you're using may be a good enough one for doing in your head while playing a game, but it's not completely accurate. If you're using a computer to crunch the numbers for you, you might as well get it right.
The probability of drawing at least one of N specific cards from a deck of size M after D draws is:
1 - (comb(M-N, D) / comb(M, D))
Where comb is the binary coefficient or "combination" function (often spoken as "N choose R" and written "nCr" in mathematics). Python doesn't have an implementation of that function in the standard library, but there are a lot of add on modules you may already have installed that provide one or you can pretty easily write your own. See this earlier question for more specifics.
For your example parameters, the correct odds are '5397/17110' or 0.315.

Numpy to weak to calculate a precise mean value

This question is very similar to this post - but not exactly
I have some data in a .csv file. The data has precision to the 4th digit (#.####).
Calculating the mean in Excel or SAS gives a result with precision to 5th digit (#.#####) but using numpy gives:
import numpy as np
data = np.recfromcsv(path2file, delimiter=';', names=['measurements'], dtype=np.float64)
rawD = data['measurements']
print np.average(rawD)
gives a number like this
#.#####999999999994
Clearly something is wrong..
using
from math import fsum
print fsum(rawD.ravel())/rawD.size
gives
#.#####
Is there anything in the np.average that I set wrong _______?
BONUS info:
I'm only working with 200 data points in the array
UPDATE
I thought I should make my case more clear.
I have numbers like 4.2730 in my csv (giving a 4 decimal precision - even though the 4th always is zero [not part of the subject so don't mind that])
Calculating an average/mean by numpy gives me this
4.2516499999999994
Which gives a print by
>>>print "%.4f" % np.average(rawD)
4.2516
During the same thing in Excel or SAS gives me this:
4.2517
Which I actually believe as being the true average value because it finds it to be 4.25165.
This code also illustrate it:
answer = 0
for number in rawD:
answer += int(number*1000)
print answer/2
425165
So how do I tell np.average() to calculate this value ___?
I'm a bit surprised that numpy did this to me... I thought that I only needed to worry if I was dealing with 16 digits numbers. Didn't expect a round off on the 4 decimal place would be influenced by this..
I know I could use
fsum(rawD.ravel())/rawD.size
But I also have other things (like std) I want to calculate with the same precision
UPDATE 2
I thought I could make a temp solution by
>>>print "%.4f" % np.float64("%.5f" % np.mean(rawD))
4.2416
Which did not solve the case. Then I tried
>>>print "%.4f" % float("4.24165")
4.2416
AHA! There is a bug in the formatter: Issue 5118
To be honest I don't care if python stores 4.24165 as 4.241649999... It's still a round off error - NO MATTER WHAT.
If the interpeter can figure out how to display the number
>>>print float("4.24165")
4.24165
Then should the formatter as well and deal with that number when rounding..
It still doesn't change the fact that I have a round off problem (now both with the formatter and numpy)
In case you need some numbers to help me out then I have made this modified .csv file:
Download it from here
(I'm aware that this file does not have the number of digits I explained earlier and that the average gives ..9988 at the end instead of ..9994 - it's modified)
Guess my qeustion boils down to how do I get a string output like the one excel gives me if I use =average()
and have it round off correctly if I choose to show only 4 digits
I know that this might seem strange for some.. But I have my reasons for wanting to reproduce the behavior of Excel.
Any help would be appreciated, thank you.

To get exact decimal numbers, you need to use decimal arithmetic instead of binary. Python provides the decimal module for this.
If you want to continue to use numpy for the calculations and simply round the result, you can still do this with decimal. You do it in two steps, rounding to a large number of digits to eliminate the accumulated error, then rounding to the desired precision. The quantize method is used for rounding.
from decimal import Decimal,ROUND_HALF_UP
ten_places = Decimal('0.0000000001')
four_places = Decimal('0.0001')
mean = 4.2516499999999994
print Decimal(mean).quantize(ten_places).quantize(four_places, rounding=ROUND_HALF_UP)
4.2517

The result value of average is a double. When you print out a double, by default all digits are printed. What you see here is the result of limited digital precision, which is not a problem of numpy, but a general computing problem. When you care of the presentation of your float value, use "%.4f" % avg_val. There is also a package for rational numbers, to avoid representing fractions as real numbers, but I guess that's not what you're looking for.
For your second statement, summarizing all the values by hand and then dividing it, I suppose you're using python 2.7 and all your input values are integer. In that way, you would have an integer division, which truncates everything after the dot, resulting in another integer value.

Why do Python's math.ceil() and math.floor() operations return floats instead of integers?

Can someone explain this (straight from the docs- emphasis mine):
math.ceil(x) Return the ceiling of x as a float, the smallest integer value greater than or equal to x.
math.floor(x) Return the floor of x as a float, the largest integer value less than or equal to x.
Why would .ceil and .floor return floats when they are by definition supposed to calculate integers?
EDIT:
Well this got some very good arguments as to why they should return floats, and I was just getting used to the idea, when #jcollado pointed out that they in fact do return ints in Python 3...

As pointed out by other answers, in python they return floats probably because of historical reasons to prevent overflow problems. However, they return integers in python 3.
>>> import math
>>> type(math.floor(3.1))
<class 'int'>
>>> type(math.ceil(3.1))
<class 'int'>
You can find more information in PEP 3141.

The range of floating point numbers usually exceeds the range of integers. By returning a floating point value, the functions can return a sensible value for input values that lie outside the representable range of integers.
Consider: If floor() returned an integer, what should floor(1.0e30) return?
Now, while Python's integers are now arbitrary precision, it wasn't always this way. The standard library functions are thin wrappers around the equivalent C library functions.

Because python's math library is a thin wrapper around the C math library which returns floats.

The source of your confusion is evident in your comment:
The whole point of ceil/floor operations is to convert floats to integers!
The point of the ceil and floor operations is to round floating-point data to integral values. Not to do a type conversion. Users who need to get integer values can do an explicit conversion following the operation.
Note that it would not be possible to implement a round to integral value as trivially if all you had available were a ceil or float operation that returned an integer. You would need to first check that the input is within the representable integer range, then call the function; you would need to handle NaN and infinities in a separate code path.
Additionally, you must have versions of ceil and floor which return floating-point numbers if you want to conform to IEEE 754.

Before Python 2.4, an integer couldn't hold the full range of truncated real numbers.
http://docs.python.org/whatsnew/2.4.html#pep-237-unifying-long-integers-and-integers

Because the range for floats is greater than that of integers -- returning an integer could overflow

This is a very interesting question! As a float requires some bits to store the exponent (=bits_for_exponent) any floating point number greater than 2**(float_size - bits_for_exponent) will always be an integral value! At the other extreme a float with a negative exponent will give one of 1, 0 or -1. This makes the discussion of integer range versus float range moot because these functions will simply return the original number whenever the number is outside the range of the integer type. The python functions are wrappers of the C function and so this is really a deficiency of the C functions where they should have returned an integer and forced the programer to do the range/NaN/Inf check before calling ceil/floor.
Thus the logical answer is the only time these functions are useful they would return a value within integer range and so the fact they return a float is a mistake and you are very smart for realizing this!

Maybe because other languages do this as well, so it is generally-accepted behavior. (For good reasons, as shown in the other answers)

This totally caught me off guard recently. This is because I've programmed in C since the 1970's and I'm only now learning the fine details of Python. Like this curious behavior of math.floor().
The math library of Python is how you access the C standard math library. And the C standard math library is a collection of floating point numerical functions, like sin(), and cos(), sqrt(). The floor() function in the context of numerical calculations has ALWAYS returned a float. For 50 YEARS now. It's part of the standards for numerical computation. For those of us familiar with the math library of C, we don't understand it to be just "math functions". We understand it to be a collection of floating-point algorithms. It would be better named something like NFPAL - Numerical Floating Point Algorithms Libary. :)
Those of us that understand the history instantly see the python math module as just a wrapper for the long-established C floating-point library. So we expect without a second thought, that math.floor() is the same function as the C standard library floor() which takes a float argument and returns a float value.
The use of floor() as a numerical math concept goes back to 1798 per the Wikipedia page on the subject: https://en.wikipedia.org/wiki/Floor_and_ceiling_functions#Notation
It never has been a computer science covert floating-point to integer storage format function even though logically it's a similar concept.
The floor() function in this context has always been a floating-point numerical calculation as all(most) the functions in the math library. Floating-point goes beyond what integers can do. They include the special values of +inf, -inf, and Nan (not a number) which are all well defined as to how they propagate through floating-point numerical calculations. Floor() has always CORRECTLY preserved values like Nan and +inf and -inf in numerical calculations. If Floor returns an int, it totally breaks the entire concept of what the numerical floor() function was meant to do. math.floor(float("nan")) must return "nan" if it is to be a true floating-point numerical floor() function.
When I recently saw a Python education video telling us to use:
i = math.floor(12.34/3)
to get an integer I laughed to myself at how clueless the instructor was. But before writing a snarkish comment, I did some testing and to my shock, I found the numerical algorithms library in Python was returning an int. And even stranger, what I thought was the obvious answer to getting an int from a divide, was to use:
i = 12.34 // 3
Why not use the built-in integer divide to get the integer you are looking for! From my C background, it was the obvious right answer. But low and behold, integer divide in Python returns a FLOAT in this case! Wow! What a strange upside-down world Python can be.
A better answer in Python is that if you really NEED an int type, you should just be explicit and ask for int in python:
i = int(12.34/3)
Keeping in mind however that floor() rounds towards negative infinity and int() rounds towards zero so they give different answers for negative numbers. So if negative values are possible, you must use the function that gives the results you need for your application.
Python however is a different beast for good reasons. It's trying to address a different problem set than C. The static typing of Python is great for fast prototyping and development, but it can create some very complex and hard to find bugs when code that was tested with one type of objects, like floats, fails in subtle and hard to find ways when passed an int argument. And because of this, a lot of interesting choices were made for Python that put the need to minimize surprise errors above other historic norms.
Changing the divide to always return a float (or some form of non int) was a move in the right direction for this. And in this same light, it's logical to make // be a floor(a/b) function, and not an "int divide".
Making float divide by zero a fatal error instead of returning float("inf") is likewise wise because, in MOST python code, a divide by zero is not a numerical calculation but a programming bug where the math is wrong or there is an off by one error. It's more important for average Python code to catch that bug when it happens, instead of propagating a hidden error in the form of an "inf" which causes a blow-up miles away from the actual bug.
And as long as the rest of the language is doing a good job of casting ints to floats when needed, such as in divide, or math.sqrt(), it's logical to have math.floor() return an int, because if it is needed as a float later, it will be converted correctly back to a float. And if the programmer needed an int, well then the function gave them what they needed. math.floor(a/b) and a//b should act the same way, but the fact that they don't I guess is just a matter of history not yet adjusted for consistency. And maybe too hard to "fix" due to backward compatibility issues. And maybe not that important???
In Python, if you want to write hard-core numerical algorithms, the correct answer is to use NumPy and SciPy, not the built-in Python math module.
import numpy as np
nan = np.float64(0.0) / 0.0 # gives a warning and returns float64 nan
nan = np.floor(nan) # returns float64 nan
Python is different, for good reasons, and it takes a bit of time to understand it. And we can see in this case, the OP, who didn't understand the history of the numerical floor() function, needed and expected it to return an int from their thinking about mathematical integers and reals. Now Python is doing what our mathematical (vs computer science) training implies. Which makes it more likely to do what a beginner expects it to do while still covering all the more complex needs of advanced numerical algorithms with NumPy and SciPy. I'm constantly impressed with how Python has evolved, even if at times I'm totally caught off guard.