# SOME DESCRIPTIVE TITLE. # Copyright (C) 2001-2025, Python Software Foundation # This file is distributed under the same license as the Python package. # FIRST AUTHOR , YEAR. # # Translators: # Dmytro Kazanzhy, 2022 # #, fuzzy msgid "" msgstr "" "Project-Id-Version: Python 3.13\n" "Report-Msgid-Bugs-To: \n" "POT-Creation-Date: 2025-04-11 14:19+0000\n" "PO-Revision-Date: 2021-06-28 01:50+0000\n" "Last-Translator: Dmytro Kazanzhy, 2022\n" "Language-Team: Ukrainian (https://app.transifex.com/python-doc/teams/5390/" "uk/)\n" "MIME-Version: 1.0\n" "Content-Type: text/plain; charset=UTF-8\n" "Content-Transfer-Encoding: 8bit\n" "Language: uk\n" "Plural-Forms: nplurals=4; plural=(n % 1 == 0 && n % 10 == 1 && n % 100 != " "11 ? 0 : n % 1 == 0 && n % 10 >= 2 && n % 10 <= 4 && (n % 100 < 12 || n % " "100 > 14) ? 1 : n % 1 == 0 && (n % 10 ==0 || (n % 10 >=5 && n % 10 <=9) || " "(n % 100 >=11 && n % 100 <=14 )) ? 2: 3);\n" msgid "Floating-Point Arithmetic: Issues and Limitations" msgstr "" msgid "" "Floating-point numbers are represented in computer hardware as base 2 " "(binary) fractions. For example, the **decimal** fraction ``0.625`` has " "value 6/10 + 2/100 + 5/1000, and in the same way the **binary** fraction " "``0.101`` has value 1/2 + 0/4 + 1/8. These two fractions have identical " "values, the only real difference being that the first is written in base 10 " "fractional notation, and the second in base 2." msgstr "" msgid "" "Unfortunately, most decimal fractions cannot be represented exactly as " "binary fractions. A consequence is that, in general, the decimal floating-" "point numbers you enter are only approximated by the binary floating-point " "numbers actually stored in the machine." msgstr "" "На жаль, більшість десяткових дробів не можна представити точно як двійкові " "дроби. Наслідком цього є те, що загалом десяткові числа з плаваючою комою, " "які ви вводите, є лише наближеними до двійкових чисел з плаваючою комою, які " "фактично зберігаються в машині." msgid "" "The problem is easier to understand at first in base 10. Consider the " "fraction 1/3. You can approximate that as a base 10 fraction::" msgstr "" "Задачу легше зрозуміти спочатку з основою 10. Розглянемо дріб 1/3. Ви можете " "наблизити це як дріб з основою 10:" msgid "0.3" msgstr "" msgid "or, better, ::" msgstr "або, краще, ::" msgid "0.33" msgstr "" msgid "0.333" msgstr "" msgid "" "and so on. No matter how many digits you're willing to write down, the " "result will never be exactly 1/3, but will be an increasingly better " "approximation of 1/3." msgstr "" "і так далі. Незалежно від того, скільки цифр ви готові записати, результат " "ніколи не буде рівно 1/3, а буде дедалі кращим наближенням 1/3." msgid "" "In the same way, no matter how many base 2 digits you're willing to use, the " "decimal value 0.1 cannot be represented exactly as a base 2 fraction. In " "base 2, 1/10 is the infinitely repeating fraction ::" msgstr "" "Таким же чином, незалежно від того, скільки цифр з основою 2 ви бажаєте " "використовувати, десяткове значення 0,1 не може бути представлено точно як " "дріб з основою 2. За основою 2 1/10 — нескінченно повторюваний дріб ::" msgid "0.0001100110011001100110011001100110011001100110011..." msgstr "" msgid "" "Stop at any finite number of bits, and you get an approximation. On most " "machines today, floats are approximated using a binary fraction with the " "numerator using the first 53 bits starting with the most significant bit and " "with the denominator as a power of two. In the case of 1/10, the binary " "fraction is ``3602879701896397 / 2 ** 55`` which is close to but not exactly " "equal to the true value of 1/10." msgstr "" "Зупиніться на будь-якій кінцевій кількості бітів, і ви отримаєте наближення. " "На більшості сучасних машин числа з плаваючою точкою наближаються за " "допомогою двійкового дробу з чисельником, використовуючи перші 53 біти, " "починаючи зі старшого біта, а знаменник – ступінь двійки. У випадку 1/10 " "двійковий дріб дорівнює ``3602879701896397 / 2 ** 55``, що близько до " "справжнього значення 1/10, але не зовсім дорівнює йому." msgid "" "Many users are not aware of the approximation because of the way values are " "displayed. Python only prints a decimal approximation to the true decimal " "value of the binary approximation stored by the machine. On most machines, " "if Python were to print the true decimal value of the binary approximation " "stored for 0.1, it would have to display::" msgstr "" msgid "" ">>> 0.1\n" "0.1000000000000000055511151231257827021181583404541015625" msgstr "" msgid "" "That is more digits than most people find useful, so Python keeps the number " "of digits manageable by displaying a rounded value instead:" msgstr "" msgid "" ">>> 1 / 10\n" "0.1" msgstr "" msgid "" "Just remember, even though the printed result looks like the exact value of " "1/10, the actual stored value is the nearest representable binary fraction." msgstr "" "Просто пам’ятайте, навіть якщо надрукований результат виглядає як точне " "значення 1/10, фактичне збережене значення є найближчим двійковим дрібом, " "який можна представити." msgid "" "Interestingly, there are many different decimal numbers that share the same " "nearest approximate binary fraction. For example, the numbers ``0.1`` and " "``0.10000000000000001`` and " "``0.1000000000000000055511151231257827021181583404541015625`` are all " "approximated by ``3602879701896397 / 2 ** 55``. Since all of these decimal " "values share the same approximation, any one of them could be displayed " "while still preserving the invariant ``eval(repr(x)) == x``." msgstr "" "Цікаво, що існує багато різних десяткових чисел, які мають однаковий " "найближчий наближений двійковий дріб. Наприклад, числа ``0,1`` і " "``0,100000000000000001`` і " "``0,100000000000000005511151231257827021181583404541015625`` наближено до " "``360287970751596`` Оскільки всі ці десяткові значення мають однакове " "наближення, будь-яке з них може бути відображено, зберігаючи інваріант " "``eval(repr(x)) == x``." msgid "" "Historically, the Python prompt and built-in :func:`repr` function would " "choose the one with 17 significant digits, ``0.10000000000000001``. " "Starting with Python 3.1, Python (on most systems) is now able to choose the " "shortest of these and simply display ``0.1``." msgstr "" "Історично склалося так, що підказка Python і вбудована функція :func:`repr` " "вибирали одну з 17 значущих цифр, ``0.10000000000000001``. Починаючи з " "Python 3.1, Python (у більшості систем) тепер може вибирати найкоротший із " "них і просто відображати ``0.1``." msgid "" "Note that this is in the very nature of binary floating point: this is not a " "bug in Python, and it is not a bug in your code either. You'll see the same " "kind of thing in all languages that support your hardware's floating-point " "arithmetic (although some languages may not *display* the difference by " "default, or in all output modes)." msgstr "" msgid "" "For more pleasant output, you may wish to use string formatting to produce a " "limited number of significant digits:" msgstr "" msgid "" ">>> format(math.pi, '.12g') # give 12 significant digits\n" "'3.14159265359'\n" "\n" ">>> format(math.pi, '.2f') # give 2 digits after the point\n" "'3.14'\n" "\n" ">>> repr(math.pi)\n" "'3.141592653589793'" msgstr "" msgid "" "It's important to realize that this is, in a real sense, an illusion: you're " "simply rounding the *display* of the true machine value." msgstr "" "Важливо усвідомлювати, що насправді це ілюзія: ви просто округлюєте " "*відображення* справжнього значення машини." msgid "" "One illusion may beget another. For example, since 0.1 is not exactly 1/10, " "summing three values of 0.1 may not yield exactly 0.3, either:" msgstr "" msgid "" ">>> 0.1 + 0.1 + 0.1 == 0.3\n" "False" msgstr "" msgid "" "Also, since the 0.1 cannot get any closer to the exact value of 1/10 and 0.3 " "cannot get any closer to the exact value of 3/10, then pre-rounding with :" "func:`round` function cannot help:" msgstr "" msgid "" ">>> round(0.1, 1) + round(0.1, 1) + round(0.1, 1) == round(0.3, 1)\n" "False" msgstr "" msgid "" "Though the numbers cannot be made closer to their intended exact values, " "the :func:`math.isclose` function can be useful for comparing inexact values:" msgstr "" msgid "" ">>> math.isclose(0.1 + 0.1 + 0.1, 0.3)\n" "True" msgstr "" msgid "" "Alternatively, the :func:`round` function can be used to compare rough " "approximations:" msgstr "" msgid "" ">>> round(math.pi, ndigits=2) == round(22 / 7, ndigits=2)\n" "True" msgstr "" msgid "" "Binary floating-point arithmetic holds many surprises like this. The " "problem with \"0.1\" is explained in precise detail below, in the " "\"Representation Error\" section. See `Examples of Floating Point Problems " "`_ for " "a pleasant summary of how binary floating point works and the kinds of " "problems commonly encountered in practice. Also see `The Perils of Floating " "Point `_ for a more complete " "account of other common surprises." msgstr "" msgid "" "As that says near the end, \"there are no easy answers.\" Still, don't be " "unduly wary of floating point! The errors in Python float operations are " "inherited from the floating-point hardware, and on most machines are on the " "order of no more than 1 part in 2\\*\\*53 per operation. That's more than " "adequate for most tasks, but you do need to keep in mind that it's not " "decimal arithmetic and that every float operation can suffer a new rounding " "error." msgstr "" msgid "" "While pathological cases do exist, for most casual use of floating-point " "arithmetic you'll see the result you expect in the end if you simply round " "the display of your final results to the number of decimal digits you " "expect. :func:`str` usually suffices, and for finer control see the :meth:" "`str.format` method's format specifiers in :ref:`formatstrings`." msgstr "" "Хоча патологічні випадки дійсно існують, для більшості випадкового " "використання арифметики з плаваючою комою ви побачите очікуваний результат, " "якщо просто округлите відображення кінцевих результатів до очікуваної " "кількості десяткових цифр. :func:`str` зазвичай достатньо, і для більш " "точного контролю дивіться специфікатори формату методу :meth:`str.format` у :" "ref:`formatstrings`." msgid "" "For use cases which require exact decimal representation, try using the :mod:" "`decimal` module which implements decimal arithmetic suitable for accounting " "applications and high-precision applications." msgstr "" "Для випадків використання, які вимагають точного десяткового представлення, " "спробуйте використовувати модуль :mod:`decimal`, який реалізує десяткову " "арифметику, придатну для програм бухгалтерського обліку та програм високої " "точності." msgid "" "Another form of exact arithmetic is supported by the :mod:`fractions` module " "which implements arithmetic based on rational numbers (so the numbers like " "1/3 can be represented exactly)." msgstr "" "Інша форма точної арифметики підтримується модулем :mod:`fractions`, який " "реалізує арифметику на основі раціональних чисел (таким чином числа, такі як " "1/3, можуть бути представлені точно)." msgid "" "If you are a heavy user of floating-point operations you should take a look " "at the NumPy package and many other packages for mathematical and " "statistical operations supplied by the SciPy project. See ." msgstr "" msgid "" "Python provides tools that may help on those rare occasions when you really " "*do* want to know the exact value of a float. The :meth:`float." "as_integer_ratio` method expresses the value of a float as a fraction:" msgstr "" msgid "" ">>> x = 3.14159\n" ">>> x.as_integer_ratio()\n" "(3537115888337719, 1125899906842624)" msgstr "" msgid "" "Since the ratio is exact, it can be used to losslessly recreate the original " "value:" msgstr "" msgid "" ">>> x == 3537115888337719 / 1125899906842624\n" "True" msgstr "" msgid "" "The :meth:`float.hex` method expresses a float in hexadecimal (base 16), " "again giving the exact value stored by your computer:" msgstr "" msgid "" ">>> x.hex()\n" "'0x1.921f9f01b866ep+1'" msgstr "" msgid "" "This precise hexadecimal representation can be used to reconstruct the float " "value exactly:" msgstr "" msgid "" ">>> x == float.fromhex('0x1.921f9f01b866ep+1')\n" "True" msgstr "" msgid "" "Since the representation is exact, it is useful for reliably porting values " "across different versions of Python (platform independence) and exchanging " "data with other languages that support the same format (such as Java and " "C99)." msgstr "" "Оскільки представлення є точним, воно корисне для надійного перенесення " "значень між різними версіями Python (незалежність від платформи) та обміну " "даними з іншими мовами, які підтримують той самий формат (такими як Java і " "C99)." msgid "" "Another helpful tool is the :func:`sum` function which helps mitigate loss-" "of-precision during summation. It uses extended precision for intermediate " "rounding steps as values are added onto a running total. That can make a " "difference in overall accuracy so that the errors do not accumulate to the " "point where they affect the final total:" msgstr "" msgid "" ">>> 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 == 1.0\n" "False\n" ">>> sum([0.1] * 10) == 1.0\n" "True" msgstr "" msgid "" "The :func:`math.fsum` goes further and tracks all of the \"lost digits\" as " "values are added onto a running total so that the result has only a single " "rounding. This is slower than :func:`sum` but will be more accurate in " "uncommon cases where large magnitude inputs mostly cancel each other out " "leaving a final sum near zero:" msgstr "" msgid "" ">>> arr = [-0.10430216751806065, -266310978.67179024, 143401161448607.16,\n" "... -143401161400469.7, 266262841.31058735, -0.003244936839808227]\n" ">>> float(sum(map(Fraction, arr))) # Exact summation with single rounding\n" "8.042173697819788e-13\n" ">>> math.fsum(arr) # Single rounding\n" "8.042173697819788e-13\n" ">>> sum(arr) # Multiple roundings in extended " "precision\n" "8.042178034628478e-13\n" ">>> total = 0.0\n" ">>> for x in arr:\n" "... total += x # Multiple roundings in standard " "precision\n" "...\n" ">>> total # Straight addition has no correct " "digits!\n" "-0.0051575902860057365" msgstr "" msgid "Representation Error" msgstr "Помилка представлення" msgid "" "This section explains the \"0.1\" example in detail, and shows how you can " "perform an exact analysis of cases like this yourself. Basic familiarity " "with binary floating-point representation is assumed." msgstr "" "У цьому розділі детально пояснюється приклад \"0.1\" і показано, як ви " "можете самостійно виконати точний аналіз подібних випадків. Передбачається " "базове знайомство з двійковим представленням із плаваючою комою." msgid "" ":dfn:`Representation error` refers to the fact that some (most, actually) " "decimal fractions cannot be represented exactly as binary (base 2) " "fractions. This is the chief reason why Python (or Perl, C, C++, Java, " "Fortran, and many others) often won't display the exact decimal number you " "expect." msgstr "" ":dfn:`Representation error` стосується того факту, що деякі (насправді " "більшість) десяткових дробів не можуть бути представлені точно як двійкові " "(за основою 2) дроби. Це головна причина, чому Python (або Perl, C, C++, " "Java, Fortran та багато інших) часто не відображає точне десяткове число, " "яке ви очікуєте." msgid "" "Why is that? 1/10 is not exactly representable as a binary fraction. Since " "at least 2000, almost all machines use IEEE 754 binary floating-point " "arithmetic, and almost all platforms map Python floats to IEEE 754 binary64 " "\"double precision\" values. IEEE 754 binary64 values contain 53 bits of " "precision, so on input the computer strives to convert 0.1 to the closest " "fraction it can of the form *J*/2**\\ *N* where *J* is an integer containing " "exactly 53 bits. Rewriting ::" msgstr "" msgid "1 / 10 ~= J / (2**N)" msgstr "" msgid "as ::" msgstr "як ::" msgid "J ~= 2**N / 10" msgstr "" msgid "" "and recalling that *J* has exactly 53 bits (is ``>= 2**52`` but ``< " "2**53``), the best value for *N* is 56:" msgstr "" msgid "" ">>> 2**52 <= 2**56 // 10 < 2**53\n" "True" msgstr "" msgid "" "That is, 56 is the only value for *N* that leaves *J* with exactly 53 bits. " "The best possible value for *J* is then that quotient rounded:" msgstr "" msgid "" ">>> q, r = divmod(2**56, 10)\n" ">>> r\n" "6" msgstr "" msgid "" "Since the remainder is more than half of 10, the best approximation is " "obtained by rounding up:" msgstr "" msgid "" ">>> q+1\n" "7205759403792794" msgstr "" msgid "" "Therefore the best possible approximation to 1/10 in IEEE 754 double " "precision is::" msgstr "" msgid "7205759403792794 / 2 ** 56" msgstr "" msgid "" "Dividing both the numerator and denominator by two reduces the fraction to::" msgstr "Ділення чисельника та знаменника на два скорочує дріб до:" msgid "3602879701896397 / 2 ** 55" msgstr "" msgid "" "Note that since we rounded up, this is actually a little bit larger than " "1/10; if we had not rounded up, the quotient would have been a little bit " "smaller than 1/10. But in no case can it be *exactly* 1/10!" msgstr "" "Зауважте, що оскільки ми округлили в більшу сторону, це насправді трохи " "більше, ніж 1/10; якби ми не округляли в більшу сторону, частка була б трохи " "меншою за 1/10. Але ні в якому разі не може бути *рівно* 1/10!" msgid "" "So the computer never \"sees\" 1/10: what it sees is the exact fraction " "given above, the best IEEE 754 double approximation it can get:" msgstr "" msgid "" ">>> 0.1 * 2 ** 55\n" "3602879701896397.0" msgstr "" msgid "" "If we multiply that fraction by 10\\*\\*55, we can see the value out to 55 " "decimal digits:" msgstr "" msgid "" ">>> 3602879701896397 * 10 ** 55 // 2 ** 55\n" "1000000000000000055511151231257827021181583404541015625" msgstr "" msgid "" "meaning that the exact number stored in the computer is equal to the decimal " "value 0.1000000000000000055511151231257827021181583404541015625. Instead of " "displaying the full decimal value, many languages (including older versions " "of Python), round the result to 17 significant digits:" msgstr "" msgid "" ">>> format(0.1, '.17f')\n" "'0.10000000000000001'" msgstr "" msgid "" "The :mod:`fractions` and :mod:`decimal` modules make these calculations easy:" msgstr "" msgid "" ">>> from decimal import Decimal\n" ">>> from fractions import Fraction\n" "\n" ">>> Fraction.from_float(0.1)\n" "Fraction(3602879701896397, 36028797018963968)\n" "\n" ">>> (0.1).as_integer_ratio()\n" "(3602879701896397, 36028797018963968)\n" "\n" ">>> Decimal.from_float(0.1)\n" "Decimal('0.1000000000000000055511151231257827021181583404541015625')\n" "\n" ">>> format(Decimal.from_float(0.1), '.17')\n" "'0.10000000000000001'" msgstr ""