Check If Something Is (Not) in a List in Python

Check if something is (not) in a list in Python

The bug is probably somewhere else in your code, because it should work fine:

>>> 3 not in [2, 3, 4]
False
>>> 3 not in [4, 5, 6]
True

Or with tuples:

>>> (2, 3) not in [(2, 3), (5, 6), (9, 1)]
False
>>> (2, 3) not in [(2, 7), (7, 3), "hi"]
True

python - if not in list

How about this?

for item in mylist:
    if item in checklist:
        pass
    else:
       # do something
       print item

Checking if type == list in python

Your issue is that you have re-defined list as a variable previously in your code. This means that when you do type(tmpDict[key])==list if will return False because they aren't equal.

That being said, you should instead use isinstance(tmpDict[key], list) when testing the type of something, this won't avoid the problem of overwriting list but is a more Pythonic way of checking the type.

Python: how to check if an element is not in a list?

You can use a simple if-else pair for this:

l = ['a',12,'b',16,'c',20]
roll = int(input())
if roll in l:
    del (l[l.index(roll) - 1: l.index(roll) + 1])
    print("Succes! Your list is ",l)
else:
    print("Number not in list.")

If the number one enters is 12, the result is Succes! Your list is ['b', 16, 'c', 20].

As @Rawing pointed out in the comments, this method has too much linear time complexity, and thus I recommend this method:

l = ['a',12,'b',16,'c',20]
roll = int(input())
i = 0
try:
    i = l.index(roll)
except ValueError:
    print("The given number is not in the list.")
else:
    del(l[i-1:i+1])
    print("Succes, your list is: ",l)

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Quick note: You should not use list as a variable name, because it is the keyword for a list (array).

Python find elements in one list that are not in the other

TL;DR:

SOLUTION (1)

import numpy as np
main_list = np.setdiff1d(list_2,list_1)
# yields the elements in `list_2` that are NOT in `list_1`

SOLUTION (2) You want a sorted list

def setdiff_sorted(array1,array2,assume_unique=False):
    ans = np.setdiff1d(array1,array2,assume_unique).tolist()
    if assume_unique:
        return sorted(ans)
    return ans
main_list = setdiff_sorted(list_2,list_1)

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EXPLANATIONS:

(1) You can use NumPy's setdiff1d (array1,array2,assume_unique=False).

assume_unique asks the user IF the arrays ARE ALREADY UNIQUE.
If False, then the unique elements are determined first.

If True, the function will assume that the elements are already unique AND function will skip determining the unique elements.

This yields the unique values in array1 that are not in array2. assume_unique is False by default.

If you are concerned with the unique elements (based on the response of Chinny84), then simply use (where assume_unique=False => the default value):

import numpy as np
list_1 = ["a", "b", "c", "d", "e"]
list_2 = ["a", "f", "c", "m"] 
main_list = np.setdiff1d(list_2,list_1)
# yields the elements in `list_2` that are NOT in `list_1`

(2)
For those who want answers to be sorted, I've made a custom function:

import numpy as np
def setdiff_sorted(array1,array2,assume_unique=False):
    ans = np.setdiff1d(array1,array2,assume_unique).tolist()
    if assume_unique:
        return sorted(ans)
    return ans

To get the answer, run:

main_list = setdiff_sorted(list_2,list_1)

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SIDE NOTES:

(a) Solution 2 (custom function setdiff_sorted) returns a list (compared to an array in solution 1).

(b) If you aren't sure if the elements are unique, just use the default setting of NumPy's setdiff1d in both solutions A and B. What can be an example of a complication? See note (c).

(c) Things will be different if either of the two lists is not unique.

Say list_2 is not unique: list2 = ["a", "f", "c", "m", "m"]. Keep list1 as is: list_1 = ["a", "b", "c", "d", "e"]
Setting the default value of assume_unique yields ["f", "m"] (in both solutions). HOWEVER, if you set assume_unique=True, both solutions give ["f", "m", "m"]. Why? This is because the user ASSUMED that the elements are unique). Hence, IT IS BETTER TO KEEP assume_unique to its default value. Note that both answers are sorted.

pythonnumpy

Fastest way to check if a value exists in a list

7 in a

Clearest and fastest way to do it.

You can also consider using a set, but constructing that set from your list may take more time than faster membership testing will save. The only way to be certain is to benchmark well. (this also depends on what operations you require)

Check if any item in Python list is None (but include zero)

For list objects can simply use a membership test:

None in list_1

Like any(), the membership test on a list will scan all elements but short-circuit by returning as soon as a match is found.

any() returns True or False, never None, so your any(list_1) is None test is certainly not going anywhere. You'd have to pass in a generator expression for any() to iterate over, instead:

any(elem is None for elem in list_1)

Using any() and all() to check if a list contains one set of values or another

Generally speaking:

all and any are functions that take some iterable and return True, if

• in the case of all(), no values in the iterable are falsy;
• in the case of any(), at least one value is truthy.

A value x is falsy iff bool(x) == False.
A value x is truthy iff bool(x) == True.

Any non-booleans in the iterable will be fine — bool(x) will map (or coerce, if you prefer) any x according to these rules: 0, 0.0, None, [], (), [], set(), and other empty collections get mapped to False, anything else to True. The docstring for bool uses the terms 'true'/'false' for 'truthy'/'falsy', and True/False for the concrete boolean values.

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In your specific code samples:

You misunderstood a little bit how these functions work. Hence, the following does something completely not what you thought:

if any(foobars) == big_foobar:

...because any(foobars) would first be evaluated to either True or False, and then that boolean value would be compared to big_foobar, which generally always gives you False (unless big_foobar coincidentally happened to be the same boolean value).

Note: the iterable can be a list, but it can also be a generator/generator expression (≈ lazily evaluated/generated list) or any other iterator.

What you want instead is:

if any(x == big_foobar for x in foobars):

which basically first constructs an iterable that yields a sequence of booleans—for each item in foobars, it compares the item to big_foobar and emits the resulting boolean into the resulting sequence:

tmp = (x == big_foobar for x in foobars)

then any walks over all items in tmp and returns True as soon as it finds the first truthy element. It's as if you did the following:

In [1]: foobars = ['big', 'small', 'medium', 'nice', 'ugly']                                        

In [2]: big_foobar = 'big'                                                                          

In [3]: any(['big' == big_foobar, 'small' == big_foobar, 'medium' == big_foobar, 'nice' == big_foobar, 'ugly' == big_foobar])        
Out[3]: True

Note: As DSM pointed out, any(x == y for x in xs) is equivalent to y in xs but the latter is more readable, quicker to write and runs faster.

Some examples:

In [1]: any(x > 5 for x in range(4))
Out[1]: False

In [2]: all(isinstance(x, int) for x in range(10))
Out[2]: True

In [3]: any(x == 'Erik' for x in ['Erik', 'John', 'Jane', 'Jim'])
Out[3]: True

In [4]: all([True, True, True, False, True])
Out[4]: False

See also: http://docs.python.org/2/library/functions.html#all

How to check if an element from List A is not present in List B in Python?

Using list comprehension:

truthy answer

any([True for x in [1, 2, 3, 4] if x in [4, 5, 6, 7]])

list of elements not present in the second list

[x for x in [1, 2, 3, 4] if x not in [4, 5, 6, 7]]

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