如何简化循环和条件语句

Question

有更好或更短的编写此代码的方式吗？

def business_role_code_to_name(x):
    y = []

    for position, code in enumerate(x):
        if position == 0 and code == 1.0: 
            y.append("Owner") 
        elif position == 1 and code == 1.0: 
            y.append("Manager")
        elif position == 2 and code == 1.0:
            y.append("Employee")
        elif position == 3 and code == 1.0:
            y.append("Other")
    return y

assert business_role_code_to_name([1.0, 1.0, 1.0, 0.0]) == ['Owner', 'Manager', 'Employee']
assert business_role_code_to_name([0.0, 1.0, 0.0, 1.0]) == ['Manager', 'Other']

我是编程新手，我认为有更好的方法来编写这段代码。谢谢！

- srisays

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由于在每个 if 中您都检查 code==1.0，因此您可以在 for 循环开始时仅检查一次，这样更有效率。 - Omri Attiya

输入列表是否保证有4个元素？ - user202729

很好的问题@user202729。我甚至没有考虑过这个。在我的数据中，是保证有4个元素的。 - srisays

这些数字是浮点数（1.0）还是整数（1）？ - Andrew Jaffe

Answer 1

尝试使用：

def business_role_code_to_name(x):
    y = []
    d = {(0, 1): 'Owner', (1, 1): 'Manager', (2, 1): 'Employee', (3, 1): 'Other'}
    for i in enumerate(x):
        if d.get(i):
            y.append(d[i])
    return y

- U13-Forward

Answer 2

使用 itertools.compress：

from itertools import compress

names = ['Owner', 'Manager', 'Employee', 'Other']
positions_0 = [1.0, 1.0, 1.0, 0.0]
positions_1 = [0.0, 1.0, 0.0, 1.0]

输出：

list(compress(names, positions_0))
# ['Owner', 'Manager', 'Employee']
list(compress(names, positions_1))
# ['Manager', 'Other']

- Chris

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如果代码中除了0或1之外还有其他内容，则必须检查是否等于1。 - user202729

Answer 3

该函数 business_role_code_to_name 的逻辑如下。

def business_role_code_to_name(x):
    z = ['Owner', 'Manager', 'Employee' ,'Other']
    y = [z[position]  for position, code in enumerate(x) if code==1.0]
    return y

- ComplicatedPhenomenon

Answer 4

您可以使用以下方式使用zip来完成该任务：

roles = ["Owner","Manager","Employee","Other"]
code1 = [1.0, 1.0, 1.0, 0.0]
code2 = [0.0, 1.0, 0.0, 1.0]
def decode_role(rolecode):
    return [role for code,role in zip(rolecode,roles) if code==1.0]
print(decode_role(code1)) # ['Owner', 'Manager', 'Employee']
print(decode_role(code2)) # ['Manager', 'Other']

- Daweo

Answer 5

你可以不用循环来完成这个操作，像这样：

roles = ('Owner', 'Manager', 'Employee', 'Other')
def business_role_code_to_name(x):
    positions = filter(lambda k: x[k] == 1.0, range(len(x)))
    return list(map(roles.__getitem__, positions))

- Victor Ruiz

Answer 6

我写了一个带有错误处理的示例：

功能：

def business_role_code_to_name(x):
    y = []
    positions = {0: "Owner", 1: "Manager", 2: "Employee", 3: "Other"}
    for position, code in enumerate(x):
        if code != 1.0:
            continue
        try:
            y.append(positions[position])
        except KeyError as key_err:
            print("{} is a wrong index.".format(position))
            raise key_err
    return y

测试：

print(business_role_code_to_name([1.0, 1.0, 1.0, 0.0]))
print(business_role_code_to_name([0.0, 1.0, 0.0, 1.0]))
print(business_role_code_to_name([0.0, 1.0, 0.0, 1.0, 1.0]))

输出：

>>> python3 test.py 
['Owner', 'Manager', 'Employee']
['Manager', 'Other']
4 is a wrong index.
Traceback (most recent call last):
  File "test.py", line 21, in <module>
    print(business_role_code_to_name([0.0, 1.0, 0.0, 1.0, 1.0]))
  File "test.py", line 11, in business_role_code_to_name
    raise key_err
  File "test.py", line 8, in business_role_code_to_name
    y.append(positions[position])
KeyError: 4

- milanbalazs