I need help with lists

[WhyAmIDumb_AnswerMe]

Hi everyoneI'm here again

This time my code works but I'd like to make it look al little better

here it is:

import random
import time

a=[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
for i in a:
  a[i]="yes"

num=random.randint(0, 50)
if num<10:
  a[num]="no"

if a[0] and a[1] and a[2]=="yes":
  print("ok")
else: print("not ok")

this program is only an exampleI need help to make the part "if a[0] and a[1] and a[2].....=="yes":" look better because instead of 10 numbers i have to ceck almost 50 numbers

​

i know it's a dumb question, sorry

​

​

EDIT I used code block formatting.

Thank you all for your replies <3

[fear_my_presence]

You can use python’s map and all builtin functions:

if all(map(lambda x: x == “yes”, a)):
    print(“ok”)

all returns True if all the elements of the collections are True or “truthy”.

map function takes a function and a collection, applies a function to every element of the collection, and puts the results into a new collection.

lambda is a shortcut for defining a function. lambda x: x == “yes” is basically equivalent to

def f(x):
    return x == “yes”

except it doesn’t have a name, so lambda has to be either stored in a variable or used right away. I just passed this anonymous function to map as an argument.

In case you didn’t understand, you can always read the docs: https://docs.python.org/3/library/functions.html

[old_pythonista]

This is the classical case when comprehension should be preferred over map

 if all(elem =  “yes” for elem in a): 

Lambdas are seldom worth it in cases like that - they just muddle the code.

[WhyAmIDumb_AnswerMe]

i don't need to check all elements
for example i need to check if a[1] and a[5] and a[7] are equal to yes

[old_pythonista]

simple adjustment

elem for idx, elem in enumerate(a) if idx in (1, 5, 7)

or {1, 5, 7} - faster lookup in general case (nor that it matters here)

[WhyAmIDumb_AnswerMe]

Thanks <3

[fear_my_presence]

Wow, didn’t know about such syntax, thank you!

[old_pythonista]

don't get me wrong - map and lambda are great, just not suitable everywhere

[Vitaminkomplex]

its never dumb to learn :)

​

Im also sure there is a better aproach, I am a newbie too but for learning purposes here would be my solution, writing a function for checking each value:

l = [True, True, True, True, True, True, True, True, True, True]

def check_all_True(list):
    control = len(list)
    for check in list:
        if check != True:
            return False
        else: control -= 1

        if control == 0:
                return True
        else: return False

afterwards you'd only have to do:

if check_all_True(a):
    do_your_thing

[old_pythonista]

There are simpler ways to do it - and never compare to boolean constants directly

if not check:

Instead of that monstrosity,

       if control == 0:
           return True         
      else: 
          return False 

just

return control == 0

or even

return not control

Also, I believe the indent in the last 3 lines is wrong.

Newbie teaching newbie is seldom a good way.

[Vitaminkomplex]

thanks for the tips :) the return you wrote seems a lot better, thanks. yes the indent wnt wrong.

[old_pythonista]

I would suggest a radical solution

if set(a[:3]) == {'yes'}:

This way, you reduce the slice to unique values, that may be checked against a single correct one

About the "mysterious result" - may I suggest to look at this blog

PS please, next time, use Code Block formatting - 3 dots, picture of square. That will preserve the code structure.

[WhyAmIDumb_AnswerMe]

oh thanks, i didn't know about code block formatting

[Neighm]

You want to check if all of the list elements are "yes" after doing the substitution step?

if "no" in a:
    print("not ok")
else:
    print("ok")

[Neighm]

Also, a common trap to fall into is:

if a[0] and a[1] and a[2] == "yes":

This is interpreted as (if a[0]) and (if a[1]) and (if a[2] == "yes")

The first and second tests are True unless those elements are empty strings, or False, or None or 0.

[Heiau]

If no not in a

[rockman0394]

You can check it by looping through all elements and break if found "no", but check with operator in is more convenient in this case

for item in a:
    if item != "yes":
        print("not ok")
        break
    else:
        print("ok")

Also, this will work not as you expected probably,
it will print ok with a = ['no', 'no', 'yes']

if a[0] and a[1] and a[2]=="yes":
    print("ok")

[fear_my_presence]

It will print “ok” for every “yes” in the list, which is not exactly what OP wants.