Question

所以我有以下代码。 merge_files（）函数基本上合并两个文件（字符串）并返回一个列表。此列表用于从对象提供参数。

file1

['2.02', '-3.88', '15.25']
['4.40', '-2.36', '14.97']

file2

['P1']
['P2']

问题是当直接将值提供给对象时，函数grade_compass（）可以正常工作。当对象从解析器中提取相同的值时，它似乎没有通过此函数。

整个代码：

    from ast import literal_eval


class Compass:

   #initialize the attributes
    def __init__(self, coordX, coordY, coordZ, Pos):
        self.coordx = coordX
        self.coordy = coordY
        self.coordz = coordZ
        self.coord_pos = Pos



def merge_files(infile1, infile2):
    with open(infile1) as f1, open(infile2) as f2:
        l3 = [] # this is a list for storing all the lines from both files
        f1_lists = (literal_eval(line) for line in f1)
        f2_lists = (literal_eval(line) for line in f2)
        for l1, l2 in zip(f1_lists, f2_lists):
            l3.append(l1 + l2)
        return l3 # Return all the lines from both text files


def check_step_xrange(coordx):
    if coordx > 0.00 and coordx < 7.99:
        return "short"

def check_step_zrange(coordz):
    if coordz > 4.40 and coordz < 20.00:
        return "short"

# Ignore coordy for now
def grade_compass(coordx, coordy, coordz):
    if check_step_xrange(coordx) == "short" and check_step_zrange(coordz) == "short":
        compass_degree = "compass degree is short"
        return compass_degree


def main():
    file1 = "coordinates1.txt"
    file2 = "coordinates2.txt"
    args = merge_files(file1, file2)

    # List of instances
    compasslist = [Compass(args[i][0], args[i][1], args[i][2], args[i][3]) for i in range(len(args))]

    # I just pull the first instance
    print "\nThis is the object from the object list"
    print compasslist[0].coordx + ' ' + compasslist[0].coordy + ' ' + compasslist[0].coordz + ' ' + compasslist[0].coord_pos
    print grade_compass(compasslist[0].coordx, compasslist[0].coordy, compasslist[0].coordz)

    print "\nThis is the object manually given"
    h = Compass(2.02, -3.88, 15.25, 'P1')
    print h.coordx, h.coordy, h.coordz, h.coord_pos
    print grade_compass(h.coordx, h.coordy, h.coordz)
if __name__ == '__main__':
    main()

第一个输出返回None。如果手动给定，它将起作用。

输出：

This is the object from the object list
2.02 -3.88 15.25 P1
None


This is the object manually given
2.02 -3.88 15.25 P1
compass degree is short

所需的输出：

This is the object from the object list
2.02 -3.88 15.25 P1
compass degree is short


This is the object manually given
2.02 -3.88 15.25 P1
compass degree is short

Answer 1

我猜这是Python 2.x，而不是Python 3.x。

在Python 2中，比较运算符始终成功-字符串始终大于整数（或浮点数）。在Python 3中，这是TypeError。

>>> 2 > "2"
False
>>> 2 < "2"
True
>>> 2 < "1"
True
>>> 2 < "-500"

因此，基本上，这段代码中的问题是您永远不会将从文件中读取的字符串转换为浮点数。

Answer 2

这是两种方法之间的区别：

Compass(args[i][0], args[i][1], args[i][2], args[i][3])

Compass(2.02, -3.88, 15.25, 'P1')

在前者中，所有参数都是字符串。因此，您需要强制转换（或解析，如果不能保证输入始终正确：

Compass(float(args[i][0]), float(args[i][1]), float(args[i][2]), args[i][3])

Python：操纵对象列表中的参数。

2 个答案: