Python列表读取整数

Question

我正在尝试创建一个简单的脚本，该脚本应该计算列表中有多少 integers 或 strings。 起初列表是空的，然后要求用户用数字或字符串填充它，这是脚本：

lis = []            # list name

num, lett = 0, 0    # init counters of numbers and letters
while True:
   x = input("digit an int or string, 'stop' returns values: ")
   if(x=='stop'):
      False
      break

   i = lis.append(x)
   if isinstance(i, int): # check whether is integer
       num += 1
   else:
       lett += 1    

print(lis)
print("there are " + str(num) + " numbers")
print("there are " + str(lett) + " strings")

程序可以运行，但问题最终出现，因为当我打印列表时，它只看到字符串，例如，偶数返回为 '10'。

我需要解释器自动识别整数。

Answer 1

lis = []            # list name

num, lett = 0, 0    # init counters of numbers and letters
while True:
  x = input("digit an int or string, 'stop' returns values: ")
  if(x=='stop'):
    break
  if x.isdigit():
    lis.append(int(x))
    num += 1
  elif x.isalpha(): 
    lis.append(x)
    lett += 1

print(lis)
print("there are " + str(num) + " numbers")
print("there are " + str(lett) + " strings")

结果

digit an int or string, 'stop' returns values:  1
digit an int or string, 'stop' returns values:  2
digit an int or string, 'stop' returns values:  3
digit an int or string, 'stop' returns values:  a
digit an int or string, 'stop' returns values:  b
digit an int or string, 'stop' returns values:  c
digit an int or string, 'stop' returns values:  stop
[1, 2, 3, 'a', 'b', 'c']
there are 3 numbers
there are 3 strings

Answer 2

您正在检查list.append 的返回值是否为整数，但不是。因此，它将它视为一个字符串。

lis = []            # list name

num, lett = 0, 0    # init counters of numbers and letters
while True:
   x = input("digit an int or string, 'stop' returns values: ")
   lis.append(x)
   if(x=='stop'):
      break

for items in lis:
    if items.isdigit(): # check whether is integer
        num += 1
    else:
        lett += 1   

print(lis)
print("there are " + str(num) + " numbers")
print("there are " + str(lett) + " strings")

Answer 3

d = l = 0
res = []
while True:
    s = input("input string or digit\n")
    if s == 'exit':
        break

    ## this is one way, might be faster to do it using isdigit suggested in the comment
    try:            
        temp = int(s)
        d += 1
    except ValueError:
        l += 1
    res.append(s)

print(d)
print(l)
print(res)

Answer 4

这是适用于我的 Python IDLE 版本 3.6.0 的解决方案（如果它不适合您，请回复）：

lis = []            # list name

num, lett = 0, 0    # init counters of numbers and letters

while True:
    try:
        x = input("digit an int or string, 'stop' returns values: ")
        i = lis.append(x)
        if(x=='stop'):
            False
            break   
    except ValueError:
        print("Not an integer!")
        continue
    else:
        num += 1

print(lis)
print("there are " + str(num) + " numbers")
print("there are " + str(lett) + " strings")

Answer 5

这是我的代码

while True:
    l,num,lett = [],0,0
    while True:
        x = input('digit an int or string, "stop" returns values: ').lower().strip()
        try:
            x = int(x)
        except:
            pass
        if x == ('stop'):
            break
        l.append(x)

    for element in l:
        if isinstance(element, int):
            num += 1
        else:
            lett += 1

    print (l)
    print ("there are " + str(num) + " numbers")
    print ("there are " + str(lett) + " strings")
    l,num,lett = [],0,0     #reset to go again

Answer 6

您可以使用isdigit 并将您的代码更改为：

lis = []            # list name

num, lett = 0, 0    # init counters of numbers and letters
while True:
    x = input("digit an int or string, 'stop' returns values: ")
    if(x=='stop'):
      False
      break
    if x.isdigit(): # check whether is integer
        lis.append(int(x))
        num += 1
    else:
        lis.append(x)
        lett += 1   

print(lis)
print("there are " + str(num) + " numbers")
print("there are " + str(lett) + " strings")

另外，你可以用这个来修复你的代码（它需要一个缩进并将它移动到你的主while）：

if isinstance(int(x), int): # check whether is integer
    num += 1
    lis.append(int(x))
else:
    lett += 1
    lis.append(x)