根据其他值的结果计算值
根据其他值的结果计算值
提问于 2016-02-20 12:17:31
我正在写一个小程序，以帮助确定一个新的焊接是否涵盖了一个现有的焊接程序。逻辑是这样的：
焊接程序测试需要一个测试件。如果该测试件通过，这意味着公司可以焊接某些范围的材料。
如果钢板的thickness是<= 3:你可以焊接从0.7 * thickness到2* thickness
如果3< thickness < 30，则可以从0.5 * thickness焊接到2* thickness。
如果thickness > 30，则可以焊接任何大于5的焊缝。
throat_thickness值在results dict中取决于运行是's‘(单)还是'm’(多)。
我遇到的问题是，如果运行== 'm‘(用于多个)，那么throat_thickness的值应该是“无限制”。
# fillet weld
def fw_material_qual(throat_thickness, runs):
results = {}
if throat_thickness <= 3 and runs == 's':
results['mat_thickness_qual'] = (0.7 * throat_thickness,
1.5 * throat_thickness)
results['throat_thickness_qual'] = (0.75 * throat_thickness,
1.5 * throat_thickness)
elif 3 < throat_thickness < 30 and runs == 's':
results['mat_thickness_qual'] = (0.5 * throat_thickness,
2 * throat_thickness)
results['throat_thickness_qual'] = (0.75 * throat_thickness,
1.5 * throat_thickness)
elif throat_thickness >= 30:
results['mat_thickness_qual'] = '5 and up'
results['throat_thickness_qual'] = throat_thickness
return results所以我的问题是，我能用一种更干净的方式来做这件事吗？
elif throat_thickness <= 3 and runs == 'm': # only m has changed!
results['mat_thickness_qual'] = (0.7 * throat_thickness,
1.5 * throat_thickness)
results['throat_thickness_qual'] = 'no restriction' # only this value changed!等等。
回答 1
Stack Overflow用户
回答已采纳
发布于 2016-02-20 12:32:09
你可以这样做，
def fw_material_qual(throat_thickness, runs):
results = {}
if throat_thickness <= 3:
results['mat_thickness_qual'] = (0.7 * throat_thickness,
1.5 * throat_thickness)
if runs == 's':
results['throat_thickness_qual'] = (
0.75 * throat_thickness, 1.5 * throat_thickness)
else:
results['throat_thickness_qual'] = 'no restriction'页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/35523134
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