如果列表中的整数为零,如何跳过字符串创建?

2024-09-29 23:29:55 发布

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在这个示例代码中,有时“差异”为0。如何删除差异为0的“语句”?(详情见下文)

classlist = ['CLASS1', 'CLASS2', 'CLASS3', 'CLASS4', 'CLASS5', 'CLASS6', 'CLASS7']
gradelist1 = ['69.8', '73', '89', '0', '93', '57']
gradelist2 = ['95.3', '79', '84', '0', '68', '63']

def gradeChangeShow():
    difference = []
    try:
        for i in range(len(gradelist1)):
            difference.append(int(float(gradelist2[i])) - float(gradelist1[i]))
        difference = ["{:.1f}".format(x) for x in difference]
    except ValueError:
        difference.append('0')
    difference = [x for x in difference if x != '0']
    statement = [f'\n{c.rstrip()}: {d}' for c, d in zip(classlist, difference)]
    comma_delete = ','.join(statement)
    return comma_delete.replace(',', '')

print(gradeChangeShow())

这是从上述代码中得出的结果:

CLASS1: 25.2 
CLASS2: 6.0  
CLASS3: -5.0 
CLASS4: 0.0  
CLASS5: -25.0
CLASS6: 6.0

我希望它是怎样出来的:

CLASS1: 25.2 
CLASS2: 6.0  
CLASS3: -5.0 
CLASS5: -25.0
CLASS6: 6.0

Tags: 代码infor差异class1differenceclass2class3
1条回答
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1楼 · 发布于 2024-09-29 23:29:55

当您尝试筛选结果时,使用了整数0值作为目标。因为您的值是带小数点的字符串,所以这不可能对您有效

相反,请确保与相同类型进行比较。您可以针对'0.0'进行筛选,但如果更改字符串中的精度,这将失败。我建议您将该值转换为数值,并与之进行比较

statement = [f'\n{c.rstrip()}: {d}'
               for c, d in zip(classlist, difference) if float(d) != 0]

输出:

CLASS1: 25.2
CLASS2: 6.0
CLASS3: -5.0
CLASS5: -25.0
CLASS6: 6.0

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