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Help need for repeat task(solved)
import random A = [10,20,30,40,50,60] random.shuffle(A) X = (A[0] / 10) if X > 2: print("Ok" ,X) else: print("Not Ok", X) # How i get always > 2.0 in this code Dont want to print Not Ok #EDIT SOLVED #It will give always print Ok import random A = [10,20,30,40,50,60] random.shuffle(A) X = (A[0] / 10) if X >= 3: print("Ok" ,X) else: print("Not Ok and Retrying....") X = (A[1] / 10) if X >= 3: print("Ok" ,X) else: print("Not Ok and Retrying....") X = (A[2] / 10) if X >= 3: print("Ok" ,X)
5 Answers
+ 1
Why? May I what exactly you are trying here?
Edit: Ňá´É´ 3x
import random
A = [10,20,30,40,50,60]
while True:
random.shuffle(A)
X = (A[0] / 10)
if X > 2:
print("Ok" ,X)
break
#2nd way
import random
A = [10,20,30,40,50,60]
random.shuffle(A)
i=0
while True :
X = (A[i] / 10)
if X >= 3:
print("Ok" ,X)
break
else:
print("Not Ok and Retrying....")
i++
+ 1
You can do it more ways.. Ňá´É´ 3x
Put X=A[0] only.. Instead A[0]/10 //OK?
But still it is not clear...
Edit :
Is you want to print otherthan 10 and 20 from list....?
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Want to give me ans only 2 or up
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After random.suffle(A) please add print(A).
Your code is working but you have a probability of 1/6 to output Not Ok. Do it ten times
import random
A = [10,20,30,40,50,60]
for i in range(10):
random.shuffle(A)
print(A)
X = (A[0]/10)
if X > 2:
print("Ok" ,X)
else:
print("Not Ok", X)
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JayakrishnađŽđł
Thanks A Lot Bro đđđ
I like first one while loop đ