This is the first of several lessons on the data types used by Python. Computer programs can process instructions that work with many different kinds of data, and the instructions need to be very precise. If you add a word to this sentence, 'add' means something very different from when you add 2 and 3. A computer language has to have a set of rules defining what operations can be applied to different kinds of data, and for this to work, there also has to be a set of rules defining exactly what data can be used with each operation. For example, if you want to calculate grossProfit = salesIncome - costs, a program has to know that these quantities are variables containing numbers rather than just strings of letters. They must have a numerical data type rather than string data type.
If you are not clear about the meaning in computer science of variables and of data types, it may help to brush up on the lesson Introduction_to_Programming/Variables.
Python has several data types to represent numbers. This lesson introduces two: integers, and floating point numbers, or 'floats'. We'll discuss floats later in the lesson. An integer, commonly abbreviated to int, is a whole number (positive, negative, or zero). So
>>> isinstance(7, int) True >>> isinstance(0, int) True >>> isinstance(-11, int) True >>> isinstance(2, int) True >>> isinstance(5, int) True >>> isinstance(3.14159, int) False >>> isinstance(0.0001, int) False >>> isinstance(11.11111, int) False >>> isinstance(2.0, int) False
>>> 2+2 4 >>> 4-2 2 >>> 6+1 7 >>> 6+7-3 10 >>> 2*2 4 >>> 2*2*2 8 >>> -2 -2 >>> 8/2 4.0 >>> 4*4/2 8.0 >>> 4-4*2 -4 >>> 2-4 -2 >>> 10+10/2 15.0
You can do more mathematical operations than the previously demonstrated ones. We can perform a floor division by using two forward slashes (
>>> 4 // 2 2 >>> 1 // 8 0 >>> 5 // 5 1 >>> 100 // 5 20 >>> 4 // 3 1
>>> 5 % 4 1 >>> 1 % 4 1 >>> 4 % 4 0 >>> 2 % 4 2 >>> 2 % 1 0 >>> 20 % 2 0 >>> 20 % 3 2 >>> -20 % 3 1
>>> divmod(7,3) (2, 1) >>> (q,r) = divmod(7,3) >>> q; r 2 1
You can also find the power of a number by using two asterisk symbols (
>>> 4 ** 2 16 >>> 4 ** 4 256 >>> 1 ** 11278923689 1 >>> 2 ** 4 16 >>> 10 ** 2 100 >>> 1024 ** 2 1048576 >>> 10 ** 6 1000000 >>> 25 ** (-1/2) 0.2 >>> 4 * - 3 ** 2 -36 >>> 4 * (- 3) ** 2 36 >>> 8 / 4 ** 2 0.5
The operator of exponentiation
If unsure of precedence, you can always use parentheses to force the desired result:
>>> (4 * (- 3)) ** 2 144
There is no limit for the length of integer literals apart from what can be stored in available memory.
Almost everyone is familiar with ten based numbers. While base 10 is useful for everyday tasks, it isn't ideal for use in the computer world. Three other numeral systems are commonly used in computer science; binary, octal, and hexadecimal. We'll lightly cover Python's use of these in this section. The binary system is essential as all information is represented in binary form in computer hardware. Octal and hexadecimal are convenient for condensing binary numbers to a form that is more easily read by humans, while (unlike decimal) being simple to translate to or from binary. If you have difficulty with this part of the lesson, it may help to brush up on the lesson Numeral_systems in the course Introduction_to_Computers.
Most people have heard of binary and it is often associated with computers. Actually, modern binary made its way into the world far before electricity was widely in use. The binary system is 2 based, which means that only two numbers are used. Of course, these numbers are 0 and 1. So unlike the decimal's To use binary numbers in python, prepend
>>> 0B11 3 >>> 0B1 + 0B1 2 >>> 0B11 + 0B1 4 >>> 0B10001 + 0B1 18 >>> 0B10001 - 0B1 16 >>> bin(2345) '0b100100101001' >>> 0b_111_0101_0011 1875
The value returned by
>>> 0o3 3 >>> 0o12 10 >>> 0o12 + 0o10 18 >>> 0o12 - 0o03 7 >>> 0o100 64 >>> 0o777 511 >>> 0o777 - 0o111 438 >>> oct(1_234_987) '0o4554053' >>> 0o_1234_9876 File "<stdin>", line 1 0o_1234_9876 ^ SyntaxError: invalid token >>> 0o_1234_0765 2736629
>>> 0xF 15 >>> 0xF0 240 >>> 0xFF - 0xF 240 >>> 0xF + 0xA 25 >>> 0x2 + 0x2 4 >>> 0x12 - 0xA 8 >>> 0xFF / 0xF 17.0 >>> 0xF * 0xF 225 >>> hex(1_234_987) '0x12d82b' >>> 0x_12_D82B 1234987
All integers may be tested or modified by the bitwise operators:
These operators are called 'bitwise' because they operate on individual bits within the integer.
>>> bin (0b1010101 & 0b1111) '0b101' >>> bin (0b1010101 & 0b111000) '0b10000' >>> hex (0xFF00FF & 0xFF00) '0x0'
In the first example both input operands
0b1010101 0b 1111 ^ ^
>>> bin (0b1010101 | 0b1110) '0b1011111' >>> bin (0b1010101 | 0b1100) '0b1011101' >>> hex (0xFF00FF | 0x3F0) '0xff03ff'
In the first example both input operands
0b1010101 0b 1110 ^ ^^^^^
>> bin (0b1010101 ^ 0b1110) '0b1011011' >>> bin (0b1010101 ^ 0b1100) '0b1011001' >>> hex (0xFF00FF ^ 0x3F0) '0xff030f'
In the first example both input operands
0b1010101 0b 1110 ^ ^^ ^^
>> bin(0b10101 << 2) '0b1010100' >>> bin(0b10101 << 5) '0b1010100000' >>> hex(0xFF00FF << 8) '0xff00ff00' >>> (0xFF00FF << 8) == (0xFF00FF * 2**8) True
In the first example the output is the input shifted left 2 bits:
0b 10101 0b1010100 ^^
>> bin(0b10101 >> 2) '0b101' >>> bin(0b10101 >> 5) '0b0' >>> hex(0xFF00FF >> 8) '0xff00' >>> (0xFF00FF >> 8) == (0xFF00FF // 2**8) True
In the first example the output is the input shifted right 2 bits:
0b10101 0b 101
twoBits = operand & 0x3
The bitwise operators above perform as expected on all integers of (almost) unlimited length:
>>> hex( ( 0x1234_FEDC << 120 ) | ( 0x_CDE_90AB << 60 ) ) '0x1234fedc00000000cde90ab000000000000000' >>> hex( ( 0x1234_FEDC << 200 ) ^ ( 0x_CDE_90AB << 207 ) ) '0x67d7cab5c00000000000000000000000000000000000000000000000000'
6. The behavior of the invert (~) operator shows that negative numbers are treated as their 2's complement value:
>>> a = 0b1100101100101 ; bin(~a) '-0b1100101100110'
For a true 1's complement bitwise invert here is one way to do it:
>>> a = 0b1100101100101 ; b = a ^ ( (1 << a.bit_length) - 1 ); bin(b) '0b11010011010' >>> c = a + b; bin(c) '0b1111111111111' # to test the operation, all bits of c should be set. >>> (c+1) == ( 1 << (c.bit_length) ) True # they are.
And another way to do it:
from decimal import * a = 0b11100100011001110001010111 # a is int b = bin(a) # b is string print ('a =', b) formerPrecision = getcontext.prec getcontext.prec = a.bit_length d = Decimal.logical_invert( Decimal( b[2:] ) ) # d is string getcontext.prec = formerPrecision print ('d =', d) e = int(str(d),2) # e is int print ('e =', bin(e)) ( (a + e) == ( ( 1 << a.bit_length ) - 1 ) ) and print ('successful inversion')
When you execute the above code, you see the following results:
a = 0b11100100011001110001010111 d = 11011100110001110101000 e = 0b11011100110001110101000 successful inversion
The Decimal.logical_invert performs a 1's complement inversion.
Although integers are great for many situations, they lack one problem, integers are whole numbers. This means that they are not real numbers. A real number is a value that represents a quantity along a continuous line, which means that it can have fractions in decimal forms.
>>> isinstance(4.5, float) True >>> isinstance(1.25, float) True >>> isinstance(0.75, float) True >>> isinstance(3.14159, float) True >>> isinstance(2.71828, float) True >>> isinstance(1.0, float) True >>> isinstance(271828, float) False >>> isinstance(0, float) False >>> isinstance(0.0, float) True
The basic arithmetic operations used for integers will also work for floats. (Bitwise operators will not work with floats.)
>>> 4.0 + 2.0 6.0 >>> -1.0 + 4.5 3.5 >>> 1.75 - 1.5 0.25 >>> 4.13 - 1.1 3.03 >>> 4.5 // 1.0 4.0 >>> 4.5 / 1.0 4.5 >>> 4.5 % 1.0 0.5 >>> 7.75 * 0.25 1.9375 >>> 0.5 * 0.5 0.25 >>> 1.5 ** 2.0 2.25
A floating point literal can be either pointfloat or exponentfloat.
34.45 ; 34. ; .45 ; 0. ; -.00 ; -33. ;
e9 ; e-0 ; e+1 ; E2 ; E-3 ; E+4.
The exponent is interpreted as follows:
decinteger exponent, for example: 0e0 ; -3e1 ; 15E-6; or
pointfloat exponent, for example: .5E+2 ; -3.00e-5 ; 123_456.75E-5 ;
The Precision of Floats
Before you start calculating with floats you should understand that the precision of floats has limits, due to Python and the architecture of a computer. Some examples of errors due to finite precision are displayed below.
>>> 1.13 - 1.1 0.029999999999999805 >>> 0.001 / 11.11 9.000900090009002e-05 >>> 1 + .0000000000000001 1.0 >>> -5.5 % 3.2 0.9000000000000004 >>> float(1_234_567_890_123_456) 1234567890123456.0 >>> float(12_345_678_901_234_567) 1.2345678901234568e+16
In the second example,
In the third example, the sum of the addends
The fourth example gives the correct result if rewritten:
>>> ((-5.5*10 ) % (3.2*10)) / 10.0 0.9
When working with Python floats, we need to be aware that there will probably be a margin of error.
The Python "Decimal" module provides support for fast correctly-rounded decimal floating point arithmetic. The module offers several advantages over the float datatype, including:
The usual start to using decimals is importing the module, viewing the current context with getcontext and, if necessary, setting new values for precision, rounding, or enabled traps:
>>> from decimal import * >>> getcontext Context(prec=28, rounding=ROUND_HALF_EVEN, Emin=-999999, Emax=999999, capitals=1, clamp=0, flags=[Inexact, FloatOperation, Rounded], traps=[InvalidOperation, DivisionByZero, Overflow]) >>> setcontext(ExtendedContext) >>> getcontext Context(prec=9, rounding=ROUND_HALF_EVEN, Emin=-999999, Emax=999999, capitals=1, clamp=0, flags=, traps=) >>> setcontext(BasicContext) >>> getcontext Context(prec=9, rounding=ROUND_HALF_UP, Emin=-999999, Emax=999999, capitals=1, clamp=0, flags=, traps=[Clamped, InvalidOperation, DivisionByZero, Overflow, Underflow]) >>> c = getcontext >>> c.flags[Inexact] = True >>> c.flags[FloatOperation] = True >>> c.flags[Rounded] = True >>> getcontext Context(prec=9, rounding=ROUND_HALF_UP, Emin=-999999, Emax=999999, capitals=1, clamp=0, flags=[Inexact, FloatOperation, Rounded], traps=[Clamped, InvalidOperation, DivisionByZero, Overflow, Underflow]) >>> getcontext.prec = 75 # set desired precision >>> getcontext Context(prec=75, rounding=ROUND_HALF_UP, Emin=-999999, Emax=999999, capitals=1, clamp=0, flags=[Inexact, FloatOperation, Rounded], traps=[Clamped, InvalidOperation, DivisionByZero, Overflow, Underflow])
We are now ready to use the decimal module.
>>> Decimal(3.14) # input to decimal is float Decimal('3.140000000000000124344978758017532527446746826171875') # exact value of float 3.14. >>> Decimal('3.14') # input to decimal is string Decimal('3.14') # exact value of 3.14 in decimal floating point arithmetic >>> (2 ** 0.5)**2 2.0000000000000004 # result of binary floating point operation. We expect 2. >>> (Decimal('2') ** Decimal('0.5')) ** Decimal('2') Decimal('1.99999999999999999999999999999999999999999999999999999999999999999999999999') # result of decimal floating point operation with string input. We expect 2. >>> (2.12345678 ** (1/2.345)) ** 2.345 2.1234567800000006 # result of floating point operation. We expect 2.12345678. >>> (Decimal('2.12345678') ** (Decimal('1')/Decimal('2.345'))) ** Decimal('2.345') Decimal('2.12345677999999999999999999999999999999999999999999999999999999999999999999') # result of decimal floating point operation with string input . We expect 2.12345678. >>> getcontext.rounding=ROUND_UP >>> (Decimal('2.12345678') ** (Decimal('1')/Decimal('2.345'))) ** Decimal('2.345') Decimal('2.12345678000000000000000000000000000000000000000000000000000000000000000003') # result of decimal floating point operation with string input . We expect 2.12345678.
Some mathematical functions are also available to Decimal:
>>> getcontext.prec = 30 >>> Decimal(2).sqrt Decimal('1.41421356237309504880168872421') >>> (Decimal(2).sqrt)**2 Decimal('2.00000000000000000000000000001') # We expect 2. >>> Decimal(1).exp Decimal('2.71828182845904523536028747135') # value of 'e', base of natural logs. >>> Decimal( Decimal(1).exp ).ln Decimal('0.999999999999999999999999999999') # We expect 1.
>>> a = 899_999_999_999_999.1 ; a - (a - .1) 0.125 >>> 1.13 - 1.1 0.029999999999999805
Simple tests indicate that the error inherent in floating point operations is about
This raises the question "How much precision do we need?"
If you must have an answer correct to 30 places of decimals, Python's integer math comes to the rescue. Suppose your calculation is:
and you want 30 significant digits after the decimal point. Your calculation becomes:
The correct result
In Python and most languages, a Boolean can be either
>>> 1 == 1 True >>> 1 == 0 False >>> bool(0) False >>> bool(1) True >>> bool(10001219830) True >>> bool(-1908) True >>> bool("Hello!") True >>> bool("") False >>> bool(" ") True >>> bool(None) False >>> bool(0.000000000000000000000000000000000) False >>> bool("0.000000000000000000000000000000000") True >>> bool(0.0) False >>> bool() False >>> bool([1, 2, 3]) True >>> bool False >>> bool(True) True >>> bool(False) False >>> bool(1==1) True >>> bool(1==0) False
You can also use three operators to alter a Boolean statement:
>>> not False True >>> not True False >>> True and True True >>> True and False False >>> True or False True >>> False or False False >>> not(False or False) True >>> not(False and False) True >>> not(False and True) True
A complex number is represented as
>>> 1+2j (1+2j) >>> -1+5.5j (-1+5.5j) >>> 0+5.5j 5.5j >>> 2j 2j >>> 1+0j (1+0j) >>> complex(3,-2) (3-2j)
Note also that j cannot be used on its own without b. If you try to use j on its own, Python will look for a variable
>>> a = 5 + 3j >>> a - j Traceback (most recent call last): File "<stdin>", line 1, in <module> NameError: name 'j' is not defined >>> a - 1j (5+2j) >>> j = -3j >>> a - j (5+6j) >>> a - 1j (5+2j)
The last result illustrates that even when the variable j has a numerical value, 1j (where, as above, can be any number) is always interpreted as the imaginary number j, not the variable j.
>>> (1+3j)+(2-5j) (3-2j) >>> (1+3j)-(2-5j) (-1+8j) >>> (1+3j)*(2-5j) (17+1j) >>> a = complex(3,-5) ; b = 1 ; b += 2j ; a ; b (3-5j) (1+2j) >>> a + b ; a - b (4-3j) (2-7j) >>> a * b ; a / b (13+1j) (-1.4-2.2j) >>> a + 4 ; b - 2j ; a * 3.1 ; b / 2 (7-5j) (1+0j) (9.3-15.5j) (0.5+1j) >>> b ; b /= 5 ; b (1+2j) (0.2+0.4j) >>> a = complex(3,-5j) ; a (8-0j)
Look closely at the last example. It does not produce an error, but is it what you want?
>>> (1+2j).real 1.0 >>> (1+2j).imag 2.0 >>> var = 5+3j >>> var.real 5.0 >>> var.imag 3.0
Since integers and floats can't be mixed together in some situations, you'll need to be able to convert them from one type to another. Luckily, it's very easy to preform a conversion. To convert a data type to an integer, you'll need to use the
>>> int(1.5) 1 >>> int(10.0) 10 >>> int(True) 1 >>> int(False) 0 >>> int('0xFF', base=16) ; int('0xF1F0', 16) ; int('0b110100111', 0) ; int('11100100011',2) 255 61936 423 1827
>>> int("100") 100
>>> float(102) 102.0 >>> float(932) 932.0 >>> float(True) 1.0 >>> float(False) 0.0 >>> float("101.42") 101.42 >>> float("4") 4.0
>>> bool(1) True >>> bool(0) False >>> bool(0.0) False >>> bool(0.01) True >>> bool(14) True >>> bool(14+3j) True >>> bool(3j) True >>> bool(0j) False >>> bool("") False >>> bool("Hello") True >>> bool("True") True >>> bool("False") True
>>> complex(True) (1+0j) >>> complex(False) 0j >>> complex(3, 1) (3+1j) >>> complex(1, 22/7) (1+3.142857142857143j) >>> complex(0, 1.5) 1.5j >>> complex(7, 8) (7+8j) >>> complex("1") (1+0j) >>> complex("1+4j") (1+4j) >>> complex("9.75j") 9.75j
8. Python's built-in functions:
9. Python's documentation:
"3.1.1. Numbers", "Numeric Types", "Integer literals", "Floating point literals", "Imaginary literals", "Operator precedence", "Why are floating-point calculations so inaccurate?", "15. Floating Point Arithmetic: Issues and Limitations"
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