这个Fortran 77代码的意图是什么？ | 77代码的意图是什么

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问题描述

我正在尝试Pythonize FORTRAN77代码。有一段代码，我似乎无法把握它的意图。

I'm trying to Pythonize a FORTRAN77 code. There's a block of code that I just can't seem to grasp the intent of.

ZM只是0和1之间的一个标量.Z是一个数字的一维数组用NJ元素在0和1之间。 J，J1和J1M是INTEGER类型。 PDFZ是另一个带有NJ元素的一维数组。

ZM is just a scalar between 0 and 1. Z is a 1D array of numbers between 0 and 1 with NJ elements. J, J1, and J1M are type INTEGER. PDFZ is another 1D array with NJ elements. I'm having trouble mapping out the flow of execution.

     DO 18 J=2,NJ
     IF(ZM.GT.Z(J)) GOTO 18
     J1=J
     J1M=J-1
     GOTO 20
18   CONTINUE
20   CONTINUE

     DO 22 J=1,NJ
     PDFZ(J)=0.D0
22   CONTINUE

     PDFZ(J1)=(ZM-Z(J1M))/(Z(J1)-Z(J1M))
     PDFZ(J1M)=1.D0-PDFZ(J1)

我创建了我认为是Python2.7中的等价物。但我不再那么肯定，我的Python代码捕获了Fortran77代码的行为。

I created what I thought was the equivalent in Python2.7. But I'm not so sure anymore that my python code captures the behavior of the Fortran77 code.

loc = np.where(z < z_mean)[0][0]
pdf_z[loc] = (z_mean - z[loc-1])/(z[loc] - z[loc-1])
pdf_z[loc-1] = 1.0 - pdf_z[loc]

推荐答案

当1977年推出时，我已经编程了大约八年。幸运的是，这个代码是没有深奥或复杂的基础。不是说我可以辨别它的功能。

I had already been programming for about eight years when 1977 rolled in. Fortunately this code is bedrock with nothing abstruse or complicated. Not that I can discern what it does either.

然而，我可以翻译它。这里有一个你可以试验的形式。

However, I can translate it. Here it is in a form in which you can experiment with it.

Fortran数组是1个相对的;即一维数组的第一个元素是第一个元素。

正如你所知道的，Python浮点数是双精度浮点数。
Fortran DO-循环变量假设从第一个到最后一个值的每一个值，不像Python for-loop变量。
<$> GOTO 18 DO循环结束。循环将继续使用DO循环变量的下一个值， J 。

相反， GOTO 20 定位在循环之外的一行，因此，就像一个Python break 。

Fortran arrays were 1-relative; ie, the first element of a 1-D array was number one.
As you already know, Python floats are doubles.
Fortran DO-loop variables assume each and every value from the first to the last, unlike Python for-loop variables.
The GOTO 18 targets the end of the DO-loop. The loop will continue with the next value of the DO-loop variable, J.
In contrast, GOTO 20 targets a line outside of the loop and is, hence, like a Python break.

def sample(ZM):
    Z = [_/10 for _ in range(0,11)]

    NJ = len(Z)
    for J in range(1, NJ):
        if ZM > Z[J]:
            continue
        J1 = J
        J1M = J - 1
        break

    PDFZ = NJ * [0]

    PDFZ[J1] = (ZM - Z[J1M])/(Z[J1] - Z[J1M])
    PDFZ[J1M] = 1 - PDFZ[J1]

    print (ZM, PDFZ)

for ZM in [0, .1, .2, .3, ]:
    sample(ZM)

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