Get Number of Items from List (Or Other Iterable) with Certain Condition

Count how many values in a list that satisfy certain condition

I take compactness, you mentioned in the question, as shorter code. So, I present

sum(float(num) >= 1.3 for num in mylist)

This takes advantage of the fact that, in python True values are taken as 1 and False as 0. So, whenever float(num) >= 1.3 evaluates to Truthy, it will be 1 and if it fails, result would be 0. So, we add all the values together to get the total number of items which are greater than or equal to 1.3.

You can check that like this

True == 1
# True
True + True
# 2
False * 10
# 0

Get the first item from an iterable that matches a condition

Python 2.6+ and Python 3:

If you want StopIteration to be raised if no matching element is found:

next(x for x in the_iterable if x > 3)

If you want default_value (e.g. None) to be returned instead:

next((x for x in the_iterable if x > 3), default_value)

Note that you need an extra pair of parentheses around the generator expression in this case − they are needed whenever the generator expression isn't the only argument.

I see most answers resolutely ignore the next built-in and so I assume that for some mysterious reason they're 100% focused on versions 2.5 and older -- without mentioning the Python-version issue (but then I don't see that mention in the answers that do mention the next built-in, which is why I thought it necessary to provide an answer myself -- at least the "correct version" issue gets on record this way;-).

Python <= 2.5

The .next() method of iterators immediately raises StopIteration if the iterator immediately finishes -- i.e., for your use case, if no item in the iterable satisfies the condition. If you don't care (i.e., you know there must be at least one satisfactory item) then just use .next() (best on a genexp, line for the next built-in in Python 2.6 and better).

If you do care, wrapping things in a function as you had first indicated in your Q seems best, and while the function implementation you proposed is just fine, you could alternatively use itertools, a for...: break loop, or a genexp, or a try/except StopIteration as the function's body, as various answers suggested. There's not much added value in any of these alternatives so I'd go for the starkly-simple version you first proposed.

How do I check the list elements satisfy a given condition?

Maybe something like this? Make function that will be called for list and return True if list satisfies condition and False if it doesn't.

def some_function(nums):
   for i in range(len(nums) - 1):
      y = nums[i] + nums[i + 1]
      z = y ** 0.5
      #till here found the square root of the sum of the adjacent numbers in list
      if z.is_integer() not True:
         # if there is some two numbers that don't meet condition, function will return False
         return False
   return True

You call it like this: meet_condition = some_function(x)
After that just check if it's True and if it is print list and appropriate text.

How do I get the number of elements in a list (length of a list) in Python?

The len() function can be used with several different types in Python - both built-in types and library types. For example:

>>> len([1, 2, 3])
3

Replacing list elements that satisfy some condition with items from another list in Python

you can do like this.

list1=[0,1,6,4,2,8]
list2=[10,20,30,40]
j=0
ans=[0]*len(list1)
for i in range(len(list1)):
    if list1[i] <= 4:
        ans[list1.index(list1[i])]+=list2[j]
        if(j<len(list2)):
            j+=1
    else:
        ans[i]+=list1[i]

print(ans)

that should work.

How to check if all elements of a list match a condition?

The best answer here is to use all(), which is the builtin for this situation. We combine this with a generator expression to produce the result you want cleanly and efficiently. For example:

>>> items = [[1, 2, 0], [1, 2, 0], [1, 2, 0]]
>>> all(flag == 0 for (_, _, flag) in items)
True
>>> items = [[1, 2, 0], [1, 2, 1], [1, 2, 0]]
>>> all(flag == 0 for (_, _, flag) in items)
False

Note that all(flag == 0 for (_, _, flag) in items) is directly equivalent to all(item[2] == 0 for item in items), it's just a little nicer to read in this case.

And, for the filter example, a list comprehension (of course, you could use a generator expression where appropriate):

>>> [x for x in items if x[2] == 0]
[[1, 2, 0], [1, 2, 0]]

If you want to check at least one element is 0, the better option is to use any() which is more readable:

>>> any(flag == 0 for (_, _, flag) in items)
True

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