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Translate library/math.po rst: 87-102, 239-374 #992

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reviewed~

library/math.po Outdated Show resolved Hide resolved
library/math.po Outdated
@@ -135,6 +134,8 @@ msgid ""
"``float`` format. This operation often provides better accuracy than the "
"direct expression ``(x * y) + z``."
msgstr ""
"融合乘加運算。回傳 ``(x * y) + z``,用近似於無限精度及範圍的方式計算,而後一"
"次轉換為浮點數格式。此操作通常能提供比運算式 ``(x * y) + z`` 更高的精度。"
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這邊的 round 是四捨五入的那個 round 嗎?

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然後這邊的 precision 跟 accuracy 的翻譯可能還是要分開,精確度跟準確度應該還是不太一樣
precision 精確度
accuracy 準確度

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根據wiki對Fused multiply-add的說明,我認為此處的round含義上確實類似四捨五入

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accuracy已修

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@@ -425,6 +433,10 @@ msgid ""
"*even* integer is used for ``n``. The remainder ``r = remainder(x, y)`` "
"thus always satisfies ``abs(r) <= 0.5 * abs(y)``."
msgstr ""
"回傳 *x* 對 *y* 根據 IEEE 754 定義的餘數。對有限數 *x* 及有限非零數 *y*,該餘"
"數值為 ``x - n*y``,``n`` 是最接近 ``x / y`` 精確值的整數。若 ``x / y`` 剛好"
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quotient 沒翻到,然後這句也沒有精確度歐

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我是將the exact value of the quotient x / y合翻為x / y精確值
quotient主要是擔心商可能會誤解為僅指運算結果的整數部分,這部分我會再思考一下

@@ -425,6 +433,10 @@ msgid ""
"*even* integer is used for ``n``. The remainder ``r = remainder(x, y)`` "
"thus always satisfies ``abs(r) <= 0.5 * abs(y)``."
msgstr ""
"回傳 *x* 對 *y* 根據 IEEE 754 定義的餘數。對有限數 *x* 及有限非零數 *y*,該餘"
"數值為 ``x - n*y``,``n`` 是最接近 ``x / y`` 精確值的整數。若 ``x / y`` 剛好"
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這邊的 difference 應該不是餘數?
看起來比較接近
"這是 x - n*y 的另一種寫法" 之類的

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因為上一句定義了x,y兩數,所以我將this翻為該餘數值與第一句的根據 IEEE 754 定義的餘數對應,此處的difference我想應該是指差(減法運算的結果),這部分我會再想一下

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library/math.po Outdated
@@ -525,6 +553,9 @@ msgid ""
"pair of values, rather than returning their second return value through an "
"'output parameter' (there is no such thing in Python)."
msgstr ""
"請注意 :func:`frexp` 及 :func:`modf` 的呼叫/回傳模式與 C 語言中相應函式不同:"
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very nit

Suggested change
"請注意 :func:`frexp` 及 :func:`modf` 的呼叫/回傳模式與 C 語言中相應函式不同"
"請注意 :func:`frexp` 及 :func:`modf` 的呼叫/回傳模式與 C 語言中的相應函式不同"

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這部分我想了一下,覺得與C語言中相應的函式不同比較順,已修

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@mattwang44
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上游原文有更新,再請幫忙修 conflict 🙏🏽

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