How to Check If Numbers Are in a List in Python

check if a number already exist in a list in python

You could do

if item not in mylist:
     mylist.append(item)

But you should really use a set, like this :

myset = set()
myset.add(item)

EDIT: If order is important but your list is very big, you should probably use both a list and a set, like so:

mylist = []
myset = set()
for item in ...:
    if item not in myset:
        mylist.append(item)
        myset.add(item)

This way, you get fast lookup for element existence, but you keep your ordering. If you use the naive solution, you will get O(n) performance for the lookup, and that can be bad if your list is big

Or, as @larsman pointed out, you can use OrderedDict to the same effect:

from collections import OrderedDict

mydict = OrderedDict()
for item in ...:
    mydict[item] = True

Python check if list items are integers?

Try this:

mynewlist = [s for s in mylist if s.isdigit()]

From the docs:

str.isdigit()
Return true if all characters in the string are digits and there is at least one character, false otherwise.
For 8-bit strings, this method is locale-dependent.

As noted in the comments, isdigit() returning True does not necessarily indicate that the string can be parsed as an int via the int() function, and it returning False does not necessarily indicate that it cannot be. Nevertheless, the approach above should work in your case.

how to check if a list of numbers is between 2 values in python?

you can use filter to get a list but first you should change it value to float and in addition don't use list as variable name. list is a built-in operation you need and shouldn't be overridden can cause some bad side effects:

list_ = ['3,25', '3,15', '1,78', '2,10', '1,06', '1,58', '1,88', '1,19', '4,00', '2,45', '2,25', '3,00', '2,95', '2,45', '2,30', '1,52', '1,96', '6,50', '4,20', '1,27']

list_number = [float(number.replace(",",".")) for number in list_]    
found = filter(lambda x: 1.50 <= x <= 2.50, list_number)

for value in found:
    print(value)

But of course if you don't want to cast it to float you can do the following as well without a problem:

found = filter(lambda x: "1,50" <= x <= "2,50", list_)

The output will be the same. (except in one it has "," and ending zero and the other result use "." without ending zeros)

Output:

1.78
2.1
1.58
1.88
2.45
2.25
2.45
2.3
1.52
1.96

How to check if list is a list of integers

Try:

all(isinstance(n, int) for n in a)  # False for your example

python: Check if all the elements in the list are only numbers [closed]

You can use the type function in conjunction with an all to gather the results.

l1 = [1, 2, 3, 4, 5, 6]
l2 = ["abc", "xyz", "pqr"]

print(all(type(e) in (int, float) for e in l1))
print(all(type(e) in (int, float) for e in l2))

Results in:

True
False

Python - Check if all n numbers are present in a list

There is no need to use zip with multiple zip function.You can use sorted :

if sorted(l)==list(range(max(l)+1))

example :

>>> sorted(l)==list(range(max(l)+1))
False
>>> l= [0,2,1,7,6,5,4,3] 
>>> sorted(l)==list(range(max(l)+1))
True

How to check if a number is in a list of lists?

You'll have to loop through each list in maze and check whether there is a 0 or not:

if not any(0 in i for i in maze):
    ...

The great thing about the any() function is that it stops looping through maze once it finds a 0.

Python: check if a number is already in a grid of list of lists

If the grid is a list of lists you can use a for/else block:

def possible(n, grid):
    for x in grid:
        if n in x:
            return False
    else:
        return True

grid = [[0, 2, 3],
        [4, 0, 5],
        [6, 7, 0]]

print(possible(1, grid))  # ---> True
print(possible(2, grid))  # ---> False