Python列表讀取整數

Question

我想創建一個簡單的腳本，應該計算列表中有多少個integers或strings。 起初列表爲空，則該用戶被要求用數字或字符串來填補它，這是腳本：

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") 

該項目工程，但問題就來了，最終，因爲當我打印列表，它只能看到字符串，偶數例如返回爲'10'。

我需要解釋器來自動識別整數。

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") 

結果

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

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 3

您正在檢查list.append的返回值是否爲整數，而不是。因此，它將其視爲一個字符串。

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 4

這裏是我的解決方案，它在我的Python IDLE版本3.6.0工程（請回復，如果它不爲你工作）：

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

這裏是我的代碼

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和你的代碼改成這樣：

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") 

而且，你可以用這個解決您的代碼（它需要一個縮進，並將其移動到主while）：

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