#!/usr/bin/python
#encoding:utf8
#ph值判别程序
print "Please input a ph value: "
def phchoose(ph):
'''ph value and choose acidic or basic'''
if ph < 7.0:
print "ph value %s is acidic." % ph
elif ph > 7.0:
print "ph value %s is basic." % ph
else:
print "ph value %s is neutral." % ph
ph = float(raw_input())
phchoose(ph)
#年龄体重指数和患心脏病的关系
#---------------年龄---
#------------< 45 >=45
#体重< 22.0 低 中
#指数>=22.0 中 高
#method_1
def method1(age,bmi):
if age < 45:
if bmi < 22.0:
return 'risk = low'
else:
return 'risk = medium'
else:
if bmi < 22.0:
return 'risk = medium'
else:
return 'risk = high'
print "Please input age and bmi: "
age = int(raw_input())
bmi = float(raw_input())
print "Age %d Bmi %s" % (age,bmi),
risk = method1(age,bmi)
print risk
#method_2
def method2(age,bmi):
young = age < 45
slim = bmi < 22.0
if young:
if slim:
return 'risk = low'
else:
return 'risk = medium'
else:
if slim:
return 'risk = medium'
else:
return 'risk = high'
risk2 = method2(age,bmi)
print 'method2 : ', risk2
#method_3
def method3(age,bmi):
young = age < 45
slim = bmi < 22.0
if young and slim:
return 'risk = low'
elif young and not slim:
return 'risk = medium'
elif not young and slim:
return 'risk = medium'
elif not young and not slim:
return 'risk = high'
risk3 = method3(age,bmi)
print 'method3 : ',risk3
#mehtod_4
young = age < 45
heavy = bmi >= 22.0
table = [['medium','high'],['low','medium']]
risk = table[young][heavy]
print 'method4 : risk =',risk
<<Python编程实践>>之Choose
原创
©著作权归作者所有:来自51CTO博客作者Digital2Slave的原创作品,请联系作者获取转载授权,否则将追究法律责任

提问和评论都可以,用心的回复会被更多人看到
评论
发布评论
相关文章
-
Python数据库编程全指南SQLite和MySQL实践
本文介绍了使用Python进行数据库连接与操作的多种方法和技术。
SQL mysql Python SQLite -
敏捷实践之结对编程Pair
本文是
Pair 解决方案 编写代码