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
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,并与之进行比较输出:
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