for 循环同时递减列表的大小

Question

so I have a list of 5 or fewer elements and the elements are just integers from 0-9 and the elements are randomly assigned and its possible for the list to have 5 zeros or 5 ones etc, and I have the following function to check if there is a zero in the list. this will return the index of the first zero it finds. just ignore the .getValue()

    def check0(self):
        '''
        check for the index of the first card with value 0 in hand
        :return:
        '''
        index = 0
        found = False
        while not found and index<len(self.hand):
            if self.hand[index].getValue() == 0:
                found = True
            index += 1
        if not found:
            index = -1
        return index

but the problem is that it always returns the first zero it finds in the list. in another class I am using this function to check if the hand has any zeros.

I need to write a for loop or some other loop that will traverse the list hand and tell me if all the elements in the hand are zeros.

so the only solution I can think of for this problem is to traverse the list once and when the first zero is found increment the counter and then traverse the list again this time excluding the zero that had already been found.

for example:

I have the list
[0,0,0,0,0]

in the first traversal, the check0() method will return the index 0 for the first zero but then I traverse the list again this time excluding the first zero and repeating that until I reach the last element.

I was thinking something like this:

def find_zeros():
counter = 0
     for I in some_list(0,len(some_list),-1):
          if I.check0() != -1:
              counter += 1
          if counter == len(some_list):
             return True
     return False

can anyone help me with this issue? let me know if anything is unclear also I'm not allowed to import anything and time complexity isn't an issue

Answer 1

"I need to write a for loop or some other loop that will traverse the list hand and tell me if all the elements in the hand are zeros." (OP)

Well, to check if all elements in your list are zero you could use count :

lst1 = [0,0,0,0,0]
print(len(lst1) == lst1.count(0))

---

Or maybe list comprehension:

lst1 = [0,0,0,0,0]
print(lst1 == [nr for nr in lst1 if nr == 0])

probably better written using all like:

lst1 = [0,0,0,0,0]
print(all(i==0 for i in lst1))

---

Or maybe create a second list the same size:

lst1 = [0,0,0,0,0]
print(lst1 == [0]*len(lst1))

Answer 2

You can use enumerate for this type of problem.

for index, ch in enumerate(list_name):
   print(i, ch)

This will give you the index of each and every character in the list. You can use an 'if' statement later to check if 'ch' is a zero. Hope it helped.

Answer 3

listt=[1,0,2,0,1]

for i in range(len(listt)):
    if listt[i]==0:
        print(i)
        break #if you want to find first occurence

To check all ekements are 0,

  if len(set(listt))==1 and listt[0]==0:   
           print("All index have 0 ")

Answer 4

You could define the function like this:

def check0(self):
    index = (self.hand+[0]).index(0)
    return -1 if not any(self.hand) else index