Using list comprehension as condition for if else statement-CodePudding

I have the following code which works well

list = ["age", "test=53345", "anotherentry", "abc"]

val = [s for s in list if "test" in s]
if val != " ":
    print(val)

But what I'm trying to do is using the list comprehension as condition for an if else statement as I need to proof more than one word for occurence. Knowing it will not work, I'm searching for something like this:

PSEUDOCODE
if (is True = [s for s in list if "test" in s])
print(s)
elif (is True = [l for l in list if "anotherentry" in l])
print(l)
else:
print("None of the searched words found")

CodePudding user response：

First of all, avoid using reserved words like "list", to name variables. (Reserved words are always branded as blue).

If you need something like this:

mylist = ["age", "test=53345", "anotherentry", "abc"]
keywords = ["test", "anotherentry", "zzzz"]
    
    for el in mylist:
        for word in words:
            if (word in el):
                print(el)

Use this:

[el for word in keywords for el in mylist if (word in el)]

CodePudding user response：

any in python allows you to see if an element in a list satisfies a condition. If so it returns True else False.

if any("test" in s for s in list): # Returns True if "test" in a string inside list
    print([s for s in list if "test" in s])

CodePudding user response：

First, please don't use "list" as a variable. There is a built-in function called list()...

Could work like this:

list_ = ["age", "test=53345", "anotherentry", "abc"]
val = [s for s in list_ if "test" in s]                #filters your initial "list_" according to the condition set

if val:
    print(val)                                         #prints every entry with "test" in it
else:
    print("None of the searched words found")

CodePudding user response：

Since a non-empty list tests as True (e.g. if [1]: print('yes') will print 'yes') you can just see if your comprehension is empty or not:

>>> alist = ["age", "test=53345", "anotherentry", "abc"]
>>> find = 'test anotherentry'.split()
>>> for i in find:
...   if [s for s in alist if i in s]:i
...
'test'
'anotherentry'

But since it is sufficient to find a single occurance, it would be better to use any like this:

>>> for i in find:
...   if any(i in s for s in alist):i
...
'test'
'anotherentry'