#!/usr/bin/env python
# -*- coding: utf-8 -*-

"""Humanizing functions for numbers."""

import math
import re
from fractions import Fraction

from . import compat
from .i18n import gettext as _
from .i18n import ngettext
from .i18n import ngettext_noop as NS_
from .i18n import pgettext as P_
from .i18n import thousands_separator


def ordinal(value):
    """Converts an integer to its ordinal as a string.

    For example, 1 is "1st", 2 is "2nd", 3 is "3rd", etc. Works for any integer or
    anything `int()` will turn into an integer. Anything other value will have nothing
    done to it.

    Examples:
        ```pycon
        >>> ordinal(1)
        '1st'
        >>> ordinal(1002)
        '1002nd'
        >>> ordinal(103)
        '103rd'
        >>> ordinal(4)
        '4th'
        >>> ordinal(12)
        '12th'
        >>> ordinal(101)
        '101st'
        >>> ordinal(111)
        '111th'
        >>> ordinal("something else")
        'something else'
        >>> ordinal(None) is None
        True

        ```
    Args:
        value (int, str, float): Integer to convert.

    Returns:
        str: Ordinal string.
    """
    try:
        value = int(value)
    except (TypeError, ValueError):
        return value
    t = (
        P_("0", "th"),
        P_("1", "st"),
        P_("2", "nd"),
        P_("3", "rd"),
        P_("4", "th"),
        P_("5", "th"),
        P_("6", "th"),
        P_("7", "th"),
        P_("8", "th"),
        P_("9", "th"),
    )
    if value % 100 in (11, 12, 13):  # special case
        return "%d%s" % (value, t[0])
    return "%d%s" % (value, t[value % 10])


def intcomma(value, ndigits=None):
    """Converts an integer to a string containing commas every three digits.

    For example, 3000 becomes "3,000" and 45000 becomes "45,000". To maintain some
    compatibility with Django's `intcomma`, this function also accepts floats.

    Examples:
        ```pycon
        >>> intcomma(100)
        '100'
        >>> intcomma("1000")
        '1,000'
        >>> intcomma(1_000_000)
        '1,000,000'
        >>> intcomma(1_234_567.25)
        '1,234,567.25'
        >>> intcomma(1234.5454545, 2)
        '1,234.55'
        >>> intcomma(14308.40, 1)
        '14,308.4'
        >>> intcomma(None) is None
        True

        ```
    Args:
        value (int, float, str): Integer or float to convert.
        ndigits (int, None): Digits of precision for rounding after the decimal point.

    Returns:
        str: string containing commas every three digits.
    """
    sep = thousands_separator()
    try:
        if isinstance(value, compat.string_types):
            float(value.replace(sep, ""))
        else:
            float(value)
    except (TypeError, ValueError):
        return value

    if ndigits:
        orig = "{0:.{1}f}".format(value, ndigits)
    else:
        orig = str(value)

    new = re.sub(r"^(-?\d+)(\d{3})", r"\g<1>%s\g<2>" % sep, orig)
    if orig == new:
        return new
    else:
        return intcomma(new)


powers = [10 ** x for x in (3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 100)]
human_powers = (
    NS_("thousand", "thousand"),
    NS_("million", "million"),
    NS_("billion", "billion"),
    NS_("trillion", "trillion"),
    NS_("quadrillion", "quadrillion"),
    NS_("quintillion", "quintillion"),
    NS_("sextillion", "sextillion"),
    NS_("septillion", "septillion"),
    NS_("octillion", "octillion"),
    NS_("nonillion", "nonillion"),
    NS_("decillion", "decillion"),
    NS_("googol", "googol"),
)


def intword(value, format="%.1f"):
    """Converts a large integer to a friendly text representation.

    Works best for numbers over 1 million. For example, 1_000_000 becomes "1.0 million",
    1200000 becomes "1.2 million" and "1_200_000_000" becomes "1.2 billion". Supports up
    to decillion (33 digits) and googol (100 digits).

    Examples:
        ```pycon
        >>> intword("100")
        '100'
        >>> intword("12400")
        '12.4 thousand'
        >>> intword("1000000")
        '1.0 million'
        >>> intword(1_200_000_000)
        '1.2 billion'
        >>> intword(8100000000000000000000000000000000)
        '8.1 decillion'
        >>> intword(None) is None
        True
        >>> intword("1234000", "%0.3f")
        '1.234 million'

        ```
    Args:
        value (int, float, str): Integer to convert.
        format (str): To change the number of decimal or general format of the number
            portion.

    Returns:
        str: Friendly text representation as a string, unless the value passed could not
        be coaxed into an `int`.
    """
    try:
        value = int(value)
    except (TypeError, ValueError):
        return value

    if value < powers[0]:
        return str(value)
    for ordinal, power in enumerate(powers[1:], 1):
        if value < power:
            chopped = value / float(powers[ordinal - 1])
            if float(format % chopped) == float(10 ** 3):
                chopped = value / float(powers[ordinal])
                singular, plural = human_powers[ordinal]
                return (
                    " ".join([format, ngettext(singular, plural, math.ceil(chopped))])
                ) % chopped
            else:
                singular, plural = human_powers[ordinal - 1]
                return (
                    " ".join([format, ngettext(singular, plural, math.ceil(chopped))])
                ) % chopped
    return str(value)


def apnumber(value):
    """Converts an integer to Associated Press style.

    Examples:
      ```pycon
      >>> apnumber(0)
      'zero'
      >>> apnumber(5)
      'five'
      >>> apnumber(10)
      '10'
      >>> apnumber("7")
      'seven'
      >>> apnumber("foo")
      'foo'
      >>> apnumber(None) is None
      True

      ```
    Args:
        value (int, float, str): Integer to convert.

    Returns:
        str: For numbers 0-9, the number spelled out. Otherwise, the number. This always
        returns a string unless the value was not `int`-able, unlike the Django filter.
    """
    try:
        value = int(value)
    except (TypeError, ValueError):
        return value
    if not 0 <= value < 10:
        return str(value)
    return (
        _("zero"),
        _("one"),
        _("two"),
        _("three"),
        _("four"),
        _("five"),
        _("six"),
        _("seven"),
        _("eight"),
        _("nine"),
    )[value]


def fractional(value):
    """Convert to fractional number.

    There will be some cases where one might not want to show ugly decimal places for
    floats and decimals.

    This function returns a human-readable fractional number in form of fractions and
    mixed fractions.

    Pass in a string, or a number or a float, and this function returns:

    * a string representation of a fraction
    * or a whole number
    * or a mixed fraction

    Examples:
        ```pycon
        >>> fractional(0.3)
        '3/10'
        >>> fractional(1.3)
        '1 3/10'
        >>> fractional(float(1/3))
        '1/3'
        >>> fractional(1)
        '1'
        >>> fractional("ten")
        'ten'
        >>> fractional(None) is None
        True

        ```
    Args:
        value (int, float, str): Integer to convert.

    Returns:
        str: Fractional number as a string.
    """
    try:
        number = float(value)
    except (TypeError, ValueError):
        return value
    whole_number = int(number)
    frac = Fraction(number - whole_number).limit_denominator(1000)
    numerator = frac._numerator
    denominator = frac._denominator
    if whole_number and not numerator and denominator == 1:
        # this means that an integer was passed in
        # (or variants of that integer like 1.0000)
        return "%.0f" % whole_number
    elif not whole_number:
        return "{:.0f}/{:.0f}".format(numerator, denominator)
    else:
        return "{:.0f} {:.0f}/{:.0f}".format(whole_number, numerator, denominator)


def scientific(value, precision=2):
    """Return number in string scientific notation z.wq x 10ⁿ.

    Examples:
        ```pycon
        >>> scientific(float(0.3))
        '3.00 x 10⁻¹'
        >>> scientific(int(500))
        '5.00 x 10²'
        >>> scientific(-1000)
        '1.00 x 10⁻³'
        >>> scientific(1000, 1)
        '1.0 x 10³'
        >>> scientific(1000, 3)
        '1.000 x 10³'
        >>> scientific("99")
        '9.90 x 10¹'
        >>> scientific("foo")
        'foo'
        >>> scientific(None) is None
        True

        ```

    Args:
        value (int, float, str): Input number.
        precision (int): Number of decimal for first part of the number.

    Returns:
        str: Number in scientific notation z.wq x 10ⁿ.
    """
    exponents = {
        "0": "⁰",
        "1": "¹",
        "2": "²",
        "3": "³",
        "4": "⁴",
        "5": "⁵",
        "6": "⁶",
        "7": "⁷",
        "8": "⁸",
        "9": "⁹",
        "+": "⁺",
        "-": "⁻",
    }
    negative = False
    try:
        if "-" in str(value):
            value = str(value).replace("-", "")
            negative = True

        if isinstance(value, compat.string_types):
            value = float(value)

        fmt = "{:.%se}" % str(int(precision))
        n = fmt.format(value)

    except (ValueError, TypeError):
        return value

    part1, part2 = n.split("e")
    if "-0" in part2:
        part2 = part2.replace("-0", "-")

    if "+0" in part2:
        part2 = part2.replace("+0", "")

    new_part2 = []
    if negative:
        new_part2.append(exponents["-"])

    for char in part2:
        new_part2.append(exponents[char])

    final_str = part1 + " x 10" + "".join(new_part2)

    return final_str


def clamp(value, format="{:}", floor=None, ceil=None, floor_token="<", ceil_token=">"):
    """Returns number with the specified format, clamped between floor and ceil.

    If the number is larger than ceil or smaller than floor, then the respective limit
    will be returned, formatted and prepended with a token specifying as such.

    Examples:
        ```pycon
        >>> clamp(123.456)
        '123.456'
        >>> clamp(0.0001, floor=0.01)
        '<0.01'
        >>> clamp(0.99, format="{:.0%}", ceil=0.99)
        '99%'
        >>> clamp(0.999, format="{:.0%}", ceil=0.99)
        '>99%'
        >>> clamp(1, format=intword, floor=1e6, floor_token="under ")
        'under 1.0 million'
        >>> clamp(None) is None
        True

        ```

    Args:
        value (int, float): Input number.
        format (str OR callable): Can either be a formatting string, or a callable
        function than receives value and returns a string.
        floor (int, float): Smallest value before clamping.
        ceil (int, float): Largest value before clamping.
        floor_token (str): If value is smaller than floor, token will be prepended
        to output.
        ceil_token (str): If value is larger than ceil, token will be prepended
        to output.

    Returns:
        str: Formatted number. The output is clamped between the indicated floor and
        ceil. If the number if larger than ceil or smaller than floor, the output will
        be prepended with a token indicating as such.

    """
    if value is None:
        return None

    if floor is not None and value < floor:
        value = floor
        token = floor_token
    elif ceil is not None and value > ceil:
        value = ceil
        token = ceil_token
    else:
        token = ""

    if isinstance(format, compat.string_types):
        return token + format.format(value)
    elif callable(format):
        return token + format(value)
    else:
        raise ValueError(
            "Invalid format. Must be either a valid formatting string, or a function "
            "that accepts value and returns a string."
        )