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I wrote this code and it's alright with positive numbers, but when I tried negative numbers it crashes. Can you give any hints on how to make it work with negative numbers as well? It needs to be recursive, not iterative, and to calculate the sum of the digits of an integer.
def sum_digits(n):
if n != 0:
return (n % 10 + sum_digits(n // 10))
else:
return 0
if __name__=='__main__':
print(sum_digits(123))
Input: 123
Output: 6
On the assumption that the 'sum' of the three digits of a negative number is the same as that of the absolute value of that number, this will work:
def sum_digits(n):
if n < 0:
return sum_digits(-n)
elif n != 0:
return (n % 10 + sum_digits(n // 10))
else:
return 0
That said, your actual problem here is that Python's handling of modulo for a negative number is different than you expect:
>>> -123 % 10
7
Why is that? It's because of the use of trunc() in the division. This page has a good explanation, but the short answer is that when you divide -123 by 10, in order to figure out the remainder, Python truncates in a different direction than you'd expect. (For good, if obscure, reasons.) Thus, in the above, instead of getting the expected 3 you get 7 (which is 10, your modulus, minus 3, the leftover).
Similarly, it's handling of integer division is different:
>>> -123 // 10
-13
>>> 123 // 10
12
This is un-intuitively correct because it is rounding 'down' rather than 'towards zero'. So a -12.3 rounds 'down' to -13.
These reasons are why the easiest solution to your particular problem is to simply take the absolute value prior to doing your actual calculation.
Separate your function into two functions: one, a recursive function that must always be called with a non-negative number, and two, a function that checks its argument can calls the recursive function with an appropriate argument.
def sum_digits(n):
return _recursive_sum_digits(abs(n))
def _recursive_sum_digits(n):
if n != 0:
return (n % 10 + sum_digits(n // 10))
else:
return 0
Since _recursive_sum_digits can assume its argument is non-negative, you can dispense with checking its sign on every recursive call, and guarantee that n // 10 will eventually produce 0.
If you want to just sum the digits that come after the negative sign, remove the sign by taking the absolute value of the number. If you're considering the first digit of the negative number to be a negative digit, then manually add that number in after performing this function on the rest of the digits.
Here is your hint. This is happening because the modulo operator always yields a result with the same sign as its second operand (or zero). Look at these examples:
>>> 13 % 10
3
>>> -13 % 10
7
In your specific case, a solution is to first get the absolute value of the number, and then you can go on with you approach:
def sum_digits(n):
n = abs(n)
if n != 0:
return (n % 10 + sum_digits(n // 10))
else:
return 0
This question already has answers here:
Negative integer division surprising result
(5 answers)
Closed 7 years ago.
I am new to Python, and I am learning operators right now.
I understood that:
The / operator is used for floating point division and
// for integer division.
Example:
7//3 = 2
And 7//-3=-3. Why is the answer -3?
I am stuck here.
// is not integer division, but floor division:
7/-3 -> -2.33333...
7//-3 -> floor(7/-3) -> floor(-2.33333...) -> -3
PEP 238 on Changing the Division Operator:
The // operator will be available to request floor division unambiguously.
See also Why Python's Integer Division Floors (thanks to #eugene y) - Basically 7//-3 is -7//3, so you want to be able to write:
-7 = 3 * q + r
With q an integer (negative, positive or nul) and r an integer such that 0 <= r < 3. This only works if q = -3:
-7 = 3 * (-3) + 2
// is the operator for floor division.
This means that after the division is completed the "floor" function is applied (the value retrieved from the division is rounded down to the nearest integer regardless of whether the decimal part is greater or less than .5)
As for your example, be careful to note that for negative answers the floor division operator will still be rounding down (e.g. -5/2 --> -2.5 --> -3).
This question already has answers here:
How does the modulo (%) operator work on negative numbers in Python?
(12 answers)
Closed last month.
Been looking through other answers and I still don't understand the modulo for negative numbers in python
For example the answer by df
x == (x/y)*y + (x%y)
so it makes sense that (-2)%5 = -2 - (-2/5)*5 = 3
Doesn't this (-2 - (-2/5)*5) =0 or am I just crazy?
Modulus operation with negatives values - weird thing?
Same with this
negative numbers modulo in python
Where did he get -2 from?
Lastly if the sign is dependent on the dividend why don't negative dividends have the same output as their positive counterparts?
For instance the output of
print([8%5,-8%5,4%5,-4%5])
is
[3, 2, 4, 1]
In Python, modulo is calculated according to two rules:
(a // b) * b + (a % b) == a, and
a % b has the same sign as b.
Combine this with the fact that integer division rounds down (towards −∞), and the resulting behavior is explained.
If you do -8 // 5, you get -1.6 rounded down, which is -2. Multiply that by 5 and you get -10; 2 is the number that you'd have to add to that to get -8. Therefore, -8 % 5 is 2.
In Python, a // b is defined as floor(a/b), as opposed to most other languages where integer division is defined as trunc(a/b). There is a corresponding difference in the interpretation of a % b = a - (a // b) * b.
The reason for this is that Python's definition of the % operator (and divmod) is generally more useful than that of other languages. For example:
def time_of_day(seconds_since_epoch):
minutes, seconds = divmod(seconds_since_epoch, 60)
hours, minutes = divmod(minutes, 60)
days, hours = divmod(hours, 24)
return '%02d:%02d:%02d' % (hours, minutes, seconds)
With this function, time_of_day(12345) returns '03:25:45', as you would expect.
But what time is it 12345 seconds before the epoch? With Python's definition of divmod, time_of_day(-12345) correctly returns '20:34:15'.
What if we redefine divmod to use the C definition of / and %?
def divmod(a, b):
q = int(a / b) # I'm using 3.x
r = a - b * q
return (q, r)
Now, time_of_day(-12345) returns '-3:-25:-45', which isn't a valid time of day. If the standard Python divmod function were implemented this way, you'd have to write special-case code to handle negative inputs. But with floor-style division, like my first example, it Just Works.
The rationale behind this is really the mathematical definition of least residue. Python respects this definition, whereas in most other programming language the modulus operator is really more like a 'reaminder after division' operator. To compute the least residue of -5 % 11, simply add 11 to -5 until you obtain a positive integer in the range [0,10], and the result is 6.
When you divide ints (-2/5)*5 does not evaluate to -2, as it would in the algebra you're used to. Try breaking it down into two steps, first evaluating the part in the parentheses.
(-2/5) * 5 = (-1) * 5
(-1) * 5 = -5
The reason for step 1 is that you're doing int division, which in python 2.x returns the equivalent of the float division result rounded down to the nearest integer.
In python 3 and higher, 2/5 will return a float, see PEP 238.
Check out this BetterExplained article and look # David's comment (No. 6) to get what the others are talking about.
Since we're working w/ integers, we do int division which, in Python, floors the answer as opposed to C. For more on this read Guido's article.
As for your question:
>>> 8 % 5 #B'coz (5*1) + *3* = 8
3
>>> -8 % 5 #B'coz (5*-2) + *2* = -8
2
Hope that helped. It confused me in the beginning too (it still does)! :)
Say -a % b needs to be computed. for ex.
r= 11 % 10
find the next number after 11 that is perfectly divisible by 10 i.e on dividing that next number after 11 gives the remainder 0.
In the above case its 20 which on dividing by 10 gives 0.
Hence, 20-11 = 9 is the number that needs to be added to 11.
The concept if 60 marbles needs to be equally divided among 8 people, actually what you get after dividing 60/8 is 7.5 since you can'nt halve the marbles, the next value after 60 that is perfectly divisible by 8 is 64. Hence 4 more marbles needs to be added to the lot so that everybody share the same joy of marbles.
This is how Python does it when negatives numbers are divided using modulus operator.
How could I go about finding the division remainder of a number in Python?
For example:
If the number is 26 and divided number is 7, then the division remainder is 5.
(since 7+7+7=21 and 26-21=5.)
For simple divisibility testing, see How do you check whether a number is divisible by another number?.
you are looking for the modulo operator:
a % b
for example:
>>> 26 % 7
5
Of course, maybe they wanted you to implement it yourself, which wouldn't be too difficult either.
The remainder of a division can be discovered using the operator %:
>>> 26%7
5
In case you need both the quotient and the modulo, there's the builtin divmod function:
>>> seconds= 137
>>> minutes, seconds= divmod(seconds, 60)
26 % 7 (you will get remainder)
26 / 7 (you will get divisor, can be float value)
26 // 7 (you will get divisor, only integer value)
If you want to get quotient and remainder in one line of code (more general usecase), use:
quotient, remainder = divmod(dividend, divisor)
#or
divmod(26, 7)
From Python 3.7, there is a new math.remainder() function:
from math import remainder
print(remainder(26,7))
Output:
-2.0 # not 5
Note, as above, it's not the same as %.
Quoting the documentation:
math.remainder(x, y)
Return the IEEE 754-style remainder of x with
respect to y. For finite x and finite nonzero y, this is the
difference x - n*y, where n is the closest integer to the exact value
of the quotient x / y. If x / y is exactly halfway between two
consecutive integers, the nearest even integer is used for n. The
remainder r = remainder(x, y) thus always satisfies abs(r) <= 0.5 *
abs(y).
Special cases follow IEEE 754: in particular, remainder(x, math.inf)
is x for any finite x, and remainder(x, 0) and remainder(math.inf, x)
raise ValueError for any non-NaN x. If the result of the remainder
operation is zero, that zero will have the same sign as x.
On platforms using IEEE 754 binary floating-point, the result of this
operation is always exactly representable: no rounding error is
introduced.
Issue29962 describes the rationale for creating the new function.
If you want to avoid modulo, you can also use a combination of the four basic operations :)
26 - (26 // 7 * 7) = 5
Use the % instead of the / when you divide. This will return the remainder for you. So in your case
26 % 7 = 5
We can solve this by using modulus operator (%)
26 % 7 = 5;
but
26 / 7 = 3 because it will give quotient but % operator will give remainder.
Modulo would be the correct answer, but if you're doing it manually this should work.
num = input("Enter a number: ")
div = input("Enter a divisor: ")
while num >= div:
num -= div
print num
You can find remainder using modulo operator
Example
a=14
b=10
print(a%b)
It will print 4
If you want the remainder of your division problem, just use the actual remainder rules, just like in mathematics. Granted this won't give you a decimal output.
valone = 8
valtwo = 3
x = valone / valtwo
r = valone - (valtwo * x)
print "Answer: %s with a remainder of %s" % (x, r)
If you want to make this in a calculator format, just substitute valone = 8
with valone = int(input("Value One")). Do the same with valtwo = 3, but different vairables obviously.
Here's an integer version of remainder in Python, which should give the same results as C's "%" operator:
def remainder(n, d):
return (-1 if n < 0 else 1) * (abs(n) % abs(d))
Expected results:
remainder(123, 10) == 3
remainder(123, -10) == 3
remainder(-123, 10) == -3
remainder(-123, -10) == -3
you can define a function and call it remainder with 2 values like rem(number1,number2) that returns number1%number2
then create a while and set it to true then print out two inputs for your function holding number 1 and 2 then print(rem(number1,number2)
Exactly how does the % operator work in Python, particularly when negative numbers are involved?
For example, why does -5 % 4 evaluate to 3, rather than, say, -1?
Unlike C or C++, Python's modulo operator (%) always return a number having the same sign as the denominator (divisor). Your expression yields 3 because
(-5) / 4 = -1.25 --> floor(-1.25) = -2
(-5) % 4 = (-2 × 4 + 3) % 4 = 3.
It is chosen over the C behavior because a nonnegative result is often more useful. An example is to compute week days. If today is Tuesday (day #2), what is the week day N days before? In Python we can compute with
return (2 - N) % 7
but in C, if N ≥ 3, we get a negative number which is an invalid number, and we need to manually fix it up by adding 7:
int result = (2 - N) % 7;
return result < 0 ? result + 7 : result;
(See http://en.wikipedia.org/wiki/Modulo_operator for how the sign of result is determined for different languages.)
Here's an explanation from Guido van Rossum:
http://python-history.blogspot.com/2010/08/why-pythons-integer-division-floors.html
Essentially, it's so that a/b = q with remainder r preserves the relationships b*q + r = a and 0 <= r < b.
In python, modulo operator works like this.
>>> mod = n - math.floor(n/base) * base
so the result is (for your case):
mod = -5 - floor(-1.25) * 4
mod = -5 - (-2*4)
mod = 3
whereas other languages such as C, JAVA, JavaScript use truncation instead of floor.
>>> mod = n - int(n/base) * base
which results in:
mod = -5 - int(-1.25) * 4
mod = -5 - (-1*4)
mod = -1
If you need more information about rounding in python, read this.
Other answers, especially the selected one have clearly answered this question quite well. But I would like to present a graphical approach that might be easier to understand as well, along with python code to perform normal mathematical modulo in python.
Python Modulo for Dummies
Modulo function is a directional function that describes how much we have to move further or behind after the mathematical jumps that we take during division over our X-axis of infinite numbers.
So let's say you were doing 7%3
So in forward direction, your answer would be +1, but in backward direction-
your answer would be -2. Both of which are correct mathematically.
Similarly, you would have 2 moduli for negative numbers as well. For eg: -7%3, can result both in -1 or +2 as shown -
Forward direction
Backward direction
In mathematics, we choose inward jumps, i.e. forward direction for a positive number and backward direction for negative numbers.
But in Python, we have a forward direction for all positive modulo operations. Hence, your confusion -
>>> -5 % 4
3
>>> 5 % 4
1
Here is the python code for inward jump type modulo in python:
def newMod(a,b):
res = a%b
return res if not res else res-b if a<0 else res
which would give -
>>> newMod(-5,4)
-1
>>> newMod(5,4)
1
Many people would oppose the inward jump method, but my personal opinion is, that this one is better!!
As pointed out, Python modulo makes a well-reasoned exception to the conventions of other languages.
This gives negative numbers a seamless behavior, especially when used in combination with the // integer-divide operator, as % modulo often is (as in math.divmod):
for n in range(-8,8):
print n, n//4, n%4
Produces:
-8 -2 0
-7 -2 1
-6 -2 2
-5 -2 3
-4 -1 0
-3 -1 1
-2 -1 2
-1 -1 3
0 0 0
1 0 1
2 0 2
3 0 3
4 1 0
5 1 1
6 1 2
7 1 3
Python % always outputs zero or positive*
Python // always rounds toward negative infinity
* ... as long as the right operand is positive. On the other hand 11 % -10 == -9
There is no one best way to handle integer division and mods with negative numbers. It would be nice if a/b was the same magnitude and opposite sign of (-a)/b. It would be nice if a % b was indeed a modulo b. Since we really want a == (a/b)*b + a%b, the first two are incompatible.
Which one to keep is a difficult question, and there are arguments for both sides. C and C++ round integer division towards zero (so a/b == -((-a)/b)), and apparently Python doesn't.
You can use:
result = numpy.fmod(x,y)
it will keep the sign , see numpy fmod() documentation.
It's also worth to mention that also the division in python is different from C:
Consider
>>> x = -10
>>> y = 37
in C you expect the result
0
what is x/y in python?
>>> print x/y
-1
and % is modulo - not the remainder! While x%y in C yields
-10
python yields.
>>> print x%y
27
You can get both as in C
The division:
>>> from math import trunc
>>> d = trunc(float(x)/y)
>>> print d
0
And the remainder (using the division from above):
>>> r = x - d*y
>>> print r
-10
This calculation is maybe not the fastest but it's working for any sign combinations of x and y to achieve the same results as in C plus it avoids conditional statements.
It's what modulo is used for. If you do a modulo through a series of numbers, it will give a cycle of values, say:
ans = num % 3
num
ans
3
0
2
2
1
1
0
0
-1
2
-2
1
-3
0
I also thought it was a strange behavior of Python. It turns out that I was not solving the division well (on paper); I was giving a value of 0 to the quotient and a value of -5 to the remainder. Terrible... I forgot the geometric representation of integers numbers. By recalling the geometry of integers given by the number line, one can get the correct values for the quotient and the remainder, and check that Python's behavior is fine. (Although I assume that you have already resolved your concern a long time ago).
#Deekshant has explained it well using visualisation. Another way to understand %(modulo) is ask a simple question.
What is nearest smaller number to dividend that can be divisible by divisor on X-axis ?
Let's have a look at few examples.
5 % 3
5 is Dividend, 3 is divisor. If you ask above question 3 is nearest smallest number that is divisible by divisor. ans would be 5 - 3 = 2. For positive Dividend, nearest smallest number would be always right side of dividend.
-5 % 3
Nearest smallest number that is divisible by 3 is -6 so ans would be -5 - (-6) = 1
-5 %4
Nearest smallest number that is divisible by 4 is -8 so ans would be -5 - (-8) = 3
Python answers every modulo expression with this method. Hope you can understand next how expression would be going to execute.
I attempted to write a general answer covering all input cases, because many people ask about various special cases (not just the one in OP, but also especially about negative values on the right-hand side) and it's really all the same question.
What does a % b actually give us in Python, explained in words?
Assuming that a and b are either float and/or int values, finite (not math.inf, math.nan etc.) and that b is not zero....
The result c is the unique number with the sign of b, such that a - c is divisible by b and abs(c) < abs(b). It will be an int when a and b are both int, and a float (even if it is exactly equal to an integer) when either a or b is an int.
For example:
>>> -9 % -5
-4
>>> 9 % 5
4
>>> -9 % 5
1
>>> 9 % -5
-1
The sign preservation also works for floating-point numbers; even when a is divisible by b, it is possible to get distinct 0.0 and -0.0 results (recalling that zero is signed in floating-point), and the sign will match b.
Proof of concept:
import math
def sign(x):
return math.copysign(1, x)
def test(a: [int, float], b: [int, float]):
c = a % b
if isinstance(a, int) and isinstance(b, int):
assert isinstance(c, int)
assert c * b >= 0 # same sign or c == 0
else:
assert isinstance(c, float)
assert sign(c) == sign(b)
assert abs(c) < abs(b)
assert math.isclose((a - c) / b, round((a - c) / b))
It's a little hard to phrase this in a way that covers all possible sign and type combinations and accounts for floating-point imprecision, but I'm pretty confident in the above. One specific gotcha for floats is that, because of that floating-point imprecision, the result for a % b might sometimes appear to give b rather than 0. In fact, it simply gives a value very close to b, because the result of the division wasn't quite exact:
>>> # On my version of Python
>>> 3.5 % 0.1
0.09999999999999981
>>> # On some other versions, it might appear as 0.1,
>>> # because of the internal rules for formatting floats for display
What if abs(a) < abs(b)?
A lot of people seem to think this is a special case, or for some reason have difficulty understanding what happens. But there is nothing special here.
For example: consider -1 % 3. How much, as a positive quantity (because 3 is positive), do we have to subtract from -1, in order to get a result divisible by 3? -1 is not divisible by 3; -1 - 1 is -2, which is also not divisible; but -1 - 2 is -3, which is divisible by 3 (dividing in exactly -1 times). By subtracting 2, we get back to divisibility; thus 2 is our predicted answer - and it checks out:
>>> -1 % 3
2
What about with b equal to zero?
It will raise ZeroDivisionError, regardless of whether b is integer zero, floating-point positive zero, or floating-point negative zero. In particular, it will not result in a NaN value.
What about special float values?
As one might expect, nan and signed infinity values for a produce a nan result, as long as b is not zero (which overrides everything else). nan values for b result in nan as well. NaN cannot be signed, so the sign of b is irrelevant in these cases.
Also as one might expect, inf % inf gives nan, regardless of the signs. If we are sharing out an infinite amount of as to an infinite amount of bs, there's no way to say "which infinity is bigger" or by how much.
The only slightly confusing cases are when b is a signed infinity value:
>>> 0 % inf
0.0
>>> 0 % -inf
-0.0
>>> 1 % inf
1.0
>>> 1 % -inf
-inf
As always, the result takes the sign of b. 0 is divisible by anything (except NaN), including infinity. But nothing else divides evenly into infinity. If a has the same sign as b, the result is simply a (as a floating-point value); if the signs differ, it will be b. Why? Well, consider -1 % inf. There isn't a finite value we can subtract from -1, in order to get to 0 (the unique value that we can divide into infinity). So we have to keep going, to infinity. The same logic applies to 1 % -inf, with all the signs reversed.
What about other types?
It's up to the type. For example, the Decimal type overloads the operator so that the result takes the sign of the numerator, even though it functionally represents the same kind of value that a float does. And, of course, strings use it for something completely different.
Why not always give a positive result, or take the sign of a?
The behaviour is motivated by integer division. While % happens to work with floating-point numbers, it's specifically designed to handle integer inputs, and the results for floats fall in line to be consistent with that.
After making the choice for a // b to give a floored division result, the % behaviour preserves a useful invariant:
>>> def check_consistency(a, b):
... assert (a // b) * b + (a % b) == a
...
>>> for a in range(-10, 11):
... for b in range(-10, 11):
... if b != 0:
... check_consistency(a, b) # no assertion is raised
...
In other words: adding the modulus value back, corrects the error created by doing an integer division.
(This, of course, lets us go back to the first section, and say that a % b simply computes a - ((a // b) * b). But that just kicks the can down the road; we still need to explain what // is doing for signed values, especially for floats.)
One practical application for this is when converting pixel coordinates to tile coordinates. // tells us which tile contains the pixel coordinate, and then % tells us the offset within that tile. Say we have 16x16 tiles: then the tile with x-coordinate 0 contains pixels with x-coordinates 0..15 inclusive, tile 1 corresponds to pixel coordinate values 16..31, and so on. If the pixel coordinate is, say, 100, we can easily calculate that it is in tile 100 // 16 == 6, and offset 100 % 16 == 4 pixels from the left edge of that tile.
We don't have to change anything in order to handle tiles on the other side of the origin. The tile at coordinate -1 needs to account for the next 16 pixel coordinates to the left of 0 - i.e., -16..-1 inclusive. And indeed, we find that e.g. -13 // 16 == -1 (so the coordinate is in that tile), and -13 % 16 == 3 (that's how far it is from the left edge of the tile).
By setting the tile width to be positive, we defined that the within-tile coordinates progress left-to-right. Therefore, knowing that a point is within a specific tile, we always want a positive result for that offset calculation. Python's % operator gives us that, on both sides of the y-axis.
What if I want it to work another way?
math.fmod will take the sign of the numerator. It will also return a floating-point result, even for two integer inputs, and raises an exception for signed-infinity a values with non-nan b values:
>>> math.fmod(-13, 16)
-13.0
>>> math.fmod(13, -16)
13.0
>>> math.fmod(1, -inf) # not -inf
1.0
>>> math.fmod(inf, 1.0) # not nan
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
ValueError: math domain error
It otherwise handles special cases the same way - a zero value for b raises an exception; otherwise any nan present causes a nan result.
If this also doesn't suit your needs, then carefully define the exact desired behaviour for every possible corner case, figure out where they differ from the built-in options, and make a wrapper function.