对简单脚本使用非常低级别的 Python 的微弱尝试

Question

我似乎无法理解为什么这段代码没有运行，我是 Python 和整个编程的新手。它返回的错误是语法错误之一，我们将不胜感激任何帮助。这是我的代码：heightDescription = ["short", "average", "tall", "very tall"]height = 0if int(height) <= 188:    print(heightDescription[-1])if int(height) in range(176, 187)    print(heightDescription[2])if int(height) in range(161, 175)    print(heightDescription[1])if int(height) in range(1, 174)    print(heightDescription[0])

Answer 1

您在语句后忘记了冒号if：

heightDescription = ["short", "average", "tall", "very tall"]

height = 0

if int(height) <= 188: 

    print(heightDescription[-1])

if int(height) in range(176, 187):  # <-- Added colon

    print(heightDescription[2])

if int(height) in range(161, 175):  # <-- Added colon

    print(heightDescription[1])

if int(height) in range(1, 174):  # <-- Added colon

    print(heightDescription[0])

Answer 2

确保所有条件的末尾if都有一个冒号 ( ):

if abcdef: 
   #

以下是一些非常接近您尝试的示例。

Answer 3

我还修复了你的程序。你的范围是重叠的，所以有时你会得到两个描述。

所以我修复了范围，这里是完整的代码：

heightDescription = ["short", "average", "tall", "very tall"]

height = 1000

if int(height) <= 188:

    print(heightDescription[0])

if int(height) in range(188, 198):

    print(heightDescription[1])

if int(height) in range(198, 208):

    print(heightDescription[2])

if int(height) in range(208, 228):

    print(heightDescription[3])

else:

    print(heightDescription[3])

Answer 4

所以你得到的语法错误是由于条件语句中逻辑语句后缺少冒号if。

此外，您的逻辑需要工作，因为您最终将获得某些值的多个打印输出。

这是您的代码的逻辑更合理的表述：

heightDescription = ["short", "average", "tall", "very tall"]

height = 0

if int(height) >= 187:

    print(heightDescription[-1])

elif int(height) in range(175, 187):

    print(heightDescription[2])

elif int(height) in range(161, 175):

    print(heightDescription[1])

elif int(height) in range(1, 161):

    print(heightDescription[0])

else: 

    print('no height')