Question

我有一个完全不同的名称列表，例如：木工，工程总共15。在下面的代码中，我将其命名为name1，但是变量名称没有序列。除了将代码编写15次外，我怎样才能创建一个函数，每次更改名称时为每个函数执行相同的条件。

name1= int(raw_input("Please enter your value for name1: "))
if name1 <5:
status = "nill"
total = 0
if name1 >4 and name1 <8:
status = "valid"
total=1
if name1 >7 and name1 <12:
status = "superb"
total=5
if name1 >11:
status = "over qualified"
total=10
print "this is your current status ",status 
print "this equals ",total

Answer 1

我相信这可以满足您的需求。请注意，这可能不是您想要存储各个状态的方式（defaultdict可能更有意义），但希望从概念的角度来看这是有道理的：

def comparisons(value):
    """Only change here is using x <= y < z syntax."""
    if value < 5:
        status = 'nill'
        total = 0
    elif 5 <= value < 8:
        status = 'valid'
        total = 1
    elif 8 <= value < 12:
        status = 'superb'
        total = 5
    else:
        status = 'over-qualified'
        total = 10
    # Here we return the status and the total for each item
    # This may not be what you want, so this can be adjusted
    return status, total


# Create a list that will contain your 15 items
items = ['Engineering', 'Carpentry']

# Create a container to hold the results.
# Note that we don't use different variables names each time -
# instead, we create an entry in a dictionary corresponding to
# our 'items' values.
results = {}

# Now we iterate through each item in our items list, running
# our function and storing the results. Note that this is a guess
# as to how you want to handle 'status' - a more useful solution
# could be to use a nested dictionary where each item has a dictionary
# with two sub-fields: 'status' and 'total'.
for item in items:
    status, total = comparisons(int(raw_input('{0}: '.format(item))))
    results[item] = [status, total]

# Print out each item
print '\nIndividual items:'
for item, values in results.iteritems():
    print item, values

# Example of how you could sum the totals
print '\nTotal score:'
print sum((results[x][1] for x in results))

输出：

Engineering: 10
Carpentry: 5

Individual items:
Engineering ['superb', 5]
Carpentry ['valid', 1]

Total scores:
6

用不同的变量创建函数

1 个答案: