Python 3.4 用户输入
Python 3.4 User Input
我想写一个小脚本来根据用户输入告诉我低音音量是否合适。
我只是在学习用户输入,这是我目前所掌握的:
def crisp():
bass = input("Enter bass level on a scale of 1 to 5>>")
print ("Bass level is at") + bass
if bass >=4:
print ("Bass is crisp")
elif bass < 4:
print ("Bass is not so crisp")
转换为整数:
bass = int(input("Enter bass level on a scale of 1 to 5>>"))
我真的没有看到这里有问题,只是一个简单的程序可以做到这一点,而且只有这样:
a=1
while a==1:
try:
bass = input('Enter Bass Level: ')
print('Bass level is at ' + str(bass))
if bass >=4:
print("Bass is crisp")
elif bass < 4:
print('Bass is not so crisp')
a=0
except ValueError:
print('Invalid Entry')
a=1
跟函数没太大区别:
def Bass():
a=1
while a==0:
try:
bass = input('Enter Bass Level: ')
print('Bass level is at ' + str(bass))
if int(bass) >=4:
print("Bass is crisp")
elif bass < 4:
print('Bass is not so crisp')
a=0
except ValueError:
print('Invalid Entry')
a=1
当您通过内置函数接收 input()
时,它会将输入作为字符串。
>>> x = input('Input: ')
Input: 1
>>> x
"1"
相反,将 int()
转换为您的 input()
:
>>> x = int(input('Input: '))
Input: 1
>>> x
1
否则,在您的代码中,您正在检查 if "4" == 4:
,这永远不会是真的。
因此,这是您编辑的代码:
def crisp():
bass = int(input("Enter bass level on a scale of 1 to 5>>"))
print ("Bass level is at") + bass
if bass >=4:
print ("Bass is crisp")
elif bass < 4:
print ("Bass is not so crisp")
我想写一个小脚本来根据用户输入告诉我低音音量是否合适。
我只是在学习用户输入,这是我目前所掌握的:
def crisp():
bass = input("Enter bass level on a scale of 1 to 5>>")
print ("Bass level is at") + bass
if bass >=4:
print ("Bass is crisp")
elif bass < 4:
print ("Bass is not so crisp")
转换为整数:
bass = int(input("Enter bass level on a scale of 1 to 5>>"))
我真的没有看到这里有问题,只是一个简单的程序可以做到这一点,而且只有这样:
a=1
while a==1:
try:
bass = input('Enter Bass Level: ')
print('Bass level is at ' + str(bass))
if bass >=4:
print("Bass is crisp")
elif bass < 4:
print('Bass is not so crisp')
a=0
except ValueError:
print('Invalid Entry')
a=1
跟函数没太大区别:
def Bass():
a=1
while a==0:
try:
bass = input('Enter Bass Level: ')
print('Bass level is at ' + str(bass))
if int(bass) >=4:
print("Bass is crisp")
elif bass < 4:
print('Bass is not so crisp')
a=0
except ValueError:
print('Invalid Entry')
a=1
当您通过内置函数接收 input()
时,它会将输入作为字符串。
>>> x = input('Input: ')
Input: 1
>>> x
"1"
相反,将 int()
转换为您的 input()
:
>>> x = int(input('Input: '))
Input: 1
>>> x
1
否则,在您的代码中,您正在检查 if "4" == 4:
,这永远不会是真的。
因此,这是您编辑的代码:
def crisp():
bass = int(input("Enter bass level on a scale of 1 to 5>>"))
print ("Bass level is at") + bass
if bass >=4:
print ("Bass is crisp")
elif bass < 4:
print ("Bass is not so crisp")