def convBin():
cont = []
rest = []
dev = []
decimal = []
print("Give me a number: ")
valor = input()
if isinstance(valor, int):
while valor > 0:
z = valor // 2
resto = x%2
valor = valor // 2
cont.append(z)
rest.append(resto)
cont.reverse()
rest.pop()
dev.append(cont[1])
for i in rest:
dev.append(rest[i])
print(" ")
print("Lista de devoluciones: ")
print(dev)
print("")
elif isinstance(valor, float):
a = valor // 1
b = valor % 1
while a > 0:
z = a // 2
resto = a%2
a = a // 2
cont.append(z)
rest.append(resto)
cont.reverse()
rest.pop()
dev.append(cont[1])
for i in rest:
dev.append(rest[i])
print("How many decimals do you want?")
num = input()
while num > 0:
dec = b * 1
dec2 = dec//1
dec %= 1
decimal.append(dec2)
print("Full part: ")
print(dev)
print("Decimal part:")
print(num)
else:
print("An error has appeared")
我正在独自学习Python,所以我知道我在代码中有很大的错误。欢迎任何建议。
此代码用于二进制转换器。
isinstance()
有问题。当我尝试代码时,在通过键盘读取的那一刻,它会忽略“ if”,而直接进入“ else”。
例如:
1. It asks you a number.
2. It goes to the first if and compare the x type with int(for some reason it is false).
3. It goes to the `elif` and does the same(check if its float).
4. Both are false so it goes to else and prints the error.
答案 0 :(得分:1)
您可以使用ast.literal_eval()
来将input()
函数返回的字符串解析为由字符串内容表示的对象,以便您可以使用isinstance()
来测试其类型如您所愿:
import ast
while True:
try:
valor = ast.literal_eval(input("Give me a number: "))
break
except SyntaxError, ValueError:
print("Please enter a valid number.")