Question

我希望第一个if ... else语句同时检查a和b。我想要一种有效的计数no_of_good和no_of_poor的方法，以最大程度地减少代码行数。可能吗？

如何在if语句中同时打印a和b？还有如何计算好坏计数？

a = 1  
b = 3  
good_count=0  
poor_count=0  
if a in range(0,6) and b in range(0,6):   
    //print and count
elif a in range(6,10) and b in range(6,10):  
    //pprint and count

这是完成工作的一种很长的方法：

a = 1  
b = 3  
good_count=0  
poor_count=0  
if a in range(0,6):  
    print("a = GOOD")
    good_count+=1`    
elif a in range(6,10):
    print("a = POOR")
    poor_count+=1
if b in range(0,6):    
    print("b = GOOD") 
    good_count+=1 
elif b in range(6,10):   
    print("b = POOR ")  
    poor_count+=1
print("no_of_good=",good_count," ","no_of_poor=",poor_count)

实际输出：

a = GOOD  
b = GOOD  
no_of_good= 2   no_of_poor= 0

预期的输出也应该相同，但是使用任何不同的方法

Answer 1

If you don't want to repeat your if statements then you could look into having a datatstructure to hold your values and iterate over all the values.

I also changed the condition to use integer comparisson instead of a range; it seames cleaner in my opinion.

Here is an example using a dict to hold the data:

data = {
    'a': 1,
    'b': 3,
    'c': 5,
    'd': 6,
}

good_count = 0
poor_count = 0
for k, v in data.items():
    if 0 <= v <= 5:
        good_count += 1
        result = 'GOOD'
    else:
        poor_count += 1
        result = 'POOR'
    print('{} = {}'.format(k, result))

print('no_of_good = {}'.format(good_count))
print('no_of_poor = {}'.format(poor_count))

And this is the output:

a = GOOD
b = GOOD
c = GOOD
d = POOR
no_of_good = 3
no_of_poor = 1

Does that help you?

Answer 2

如果您不关心变量a,b,c,d等，则可以使用列表，并使用lis t的索引来打印GOOD或POOR

li = [1,3,5,6]

good_count = 0
poor_count = 0

#Iterate through the list
for idx, item in enumerate(li):

    #Check the condition on the item and print accordingly
    if 0 <= item <= 5:
        good_count+=1
        print('index {} = GOOD'.format(idx))
    elif 6<= item <= 10:
        poor_count+=1
        print('index {} = POOR'.format(idx))

#Print final good and poor count
print('no_of_good = {}'.format(good_count))
print('no_of_poor = {}'.format(poor_count))

输出如下：

index 0 = GOOD
index 1 = GOOD
index 2 = GOOD
index 3 = POOR
no_of_good = 3
no_of_poor = 1

在一个条件语句中检查多个变量

2 个答案: