#21 Python Tutorial for Beginners | For Loop in Python

[@kartikiyer3067]

for i in range(1,500):
if i**2

[@BibekGhoshh1]

for i in range(1,500):
if i**2

[@shivaprasadbh6013]

bro can i get explanation

[@BibekGhoshh1]

@@shivaprasadbh6013 after getting all the posible output loop will break help us to come out from the loop..... otherwise it will check until the condition become false

[@samrathot]

@@BibekGhoshh1 the code does the same even without break in this case right?
Ooh yeah got it! You're right without break there's increase in cost in terms of time. Thanks ✌️

[@gregoryopondi3652]

beautiful and simple... one problem, I think range should be (1, 501)

[@ashitshekhar847]

Hey Navin, you have explained the for loop in a very simple and easy to understand manner for a beginner. Thanks a lot.

[@lovesick9907]

Yeah I saw code with Harry’s video on this topic,he’s really made a hash of it

[@randyelliott2494]

sorry to be off topic but does anyone know of a way to get back into an Instagram account?
I was dumb forgot the password. I would appreciate any assistance you can give me

[@shepardlucian4485]

@Randy Elliott instablaster =)

[@randyelliott2494]

@Shepard Lucian i really appreciate your reply. I got to the site thru google and I'm waiting for the hacking stuff now.
Looks like it's gonna take quite some time so I will reply here later with my results.

[@randyelliott2494]

@Shepard Lucian It did the trick and I actually got access to my account again. Im so happy:D
Thank you so much you really help me out :D

[@jessekyomuhendot4436]

Man I love your tutorials. You are using understandable words for beginners. A lot of people teaching coding out here are using some terminolgy that a beginner can't even understand.

[@aswasshsutar5367]

It's been two years but still I am learning from your page.
Thank you so much Naveen Sir🙏

[@blackheart450-u4v]

It makes it so Easy to learn when there is a playlist instead of 1 six hour video

[@KunjaBihariKrishna]

There's so many videos for the basic python concepts, but I finally found someone who talks fast. When people talk too slow, it makes me nervous and agitated because I feel too much anticipation for what they're about to say. I have too much time to insert my own thoughts and it becomes harder to listen.
When someone talks fast enough, the information is denser. Then I can relax and absorb.
Fast talkers are soothing

[@vivekojha4807]

i=1
while i

[@sayanth.p3625]

Anyone in 2024?🙂

[@karthikhasri9264]

Yehh

[@firexex]

its 6 years old but more updated than indian education syllabus 😂

[@slokshaw959]

Yeah

[@AdedokunMartins]

Yeah. Need free python complete course pls

[@vainayak30]

import math
for i in range(1,500):
a=math.sqrt(i)
if i%a==0:
print(i, end=' ')

[@DeshDeepak08]

your code is correct and same my code

[@adesinaolusegunabiola9615]

Thanks the code help me.... i have been on this for more than 3hrs but i got the hint fro your solution.

[@nmewar]

Can we do it without using math library...????

[@shreyaspatil24]

@@nmewar yep....
for i in range(1,500):
for x in range(1,25):
if i/x==x:
print(i)
i+=1
print()

[@BoomcoreIsLive]

@@nmewarmaybe

[@priyanshishukla6238]

for i in range(1,500):
from math import sqrt
root=sqrt(i)
if int(root+0.5)**2==i:
print(i)

[@krishwanthkishore8299]

Why did u print that fourth command ; if int( root + 0.5)**2==i .
Why is that 0.5 needed ?

[@jayakumarpurushothama6473]

# Print all perfect Square numbers between 1 to 500.
# Note : for efficiency of program we have to break the loop to avoid 500 iteration. Observe this by using debug. Post 22 iterations if we dont break the loop it does unwanted
477 iterations and increase CPU cycles.
for i in range(1,500):
if (i**2

[@harshavardhandsh5190]

Few reply are also supportive...

[@harishjagdale3893]

this one is the best solution....Thanks!!

[@bindassboss4842]

yes this is efficient program

[@himanshushekhardas1730]

we can take iteration upto 23, as 22^2 is less than 500

[@DevendraSingh-xt2wn]

it's a very useful trick, but in general, if you have to check if a given number (x) is a perfect square number or not then how you would proceed?

[@-Lauren-gs8mm]

I was having huge confusion on loops and I'm literally searching for good explanation on loops atlast I was able to find one and got a clear understanding on this concept. Even though this video is if 5 years old still is was very useful in 2024.Thanks a lot.

[@GurpreetSingh-gq4hx]

Solution to quiz question:
# To print perfect squares between 1 to 500
# Using for loop
from math import*
for i in range(1, 500):
if int(sqrt(i)) ** 2 == i:
print(i, end = ' ')
# Using while loop
from math import*
i = 1
while i

[@fitSarthak]

Name error aa raha h sqrt kya kru?

[@Akarsh-]

for i in range(50):
a = i * i
if a

[@Akarsh-]

@@rishikavlogss It's at the end of the video around last 10 seconds

[@bernardkimemia9471]

Alternative solution
from math import sqrt
for num in range(1, 501):
if sqrt(num).is_integer():
print(num)

[@koushikkankala1873]

@@bernardkimemia9471 for i in range(1,501):
if (i*i

[@SantoshSahu-nd1bw]

for i in range(1,501):
j=1
while(j*j

[@badrit5153]

Thank you so much navin. My small suggestion after completion of video ur giving some home work we are ready to do tat work my request is in ur next video plz tell me tat ur answer for that question we will also able to know tat answer & if someone did mistake also tey will rectify. Not someone i am also one to do mistakes i am also able to rectify

[@msrikavya9651]

Previous video homework is the example for next video

[@eddyr978]

@@msrikavya9651 where is that
please help i cant find it

[@gautam_anup]

Check out 100 days of code by Dr. Angela yu on udemy or app brewery. I'm learning from she teaches everything step by step. I'm sure you will get from 0 to a professional python developer after this course

[@siddharthgupta1058]

thats so mch btter teaching than my school teacher i have final exams coming this really helped me a lot. i had no idea of loops. i dont know why i had not seen you videos before

[@haha44261]

for i in range(1,50,1):
import math as m
if int(m.sqrt(i))==float(m.sqrt(i)):
print(i)

[@rishabhsharma3908]

could you explain it plz !

[@mrpubg6561]

@@rishabhsharma3908 it will print the square of all numbers from 1 to 50

[@tridevshrestha7124]

@@rishabhsharma3908 in if block, it will check square root of all number in both int and float..
lets suppose int(sqrt(4)) == float(sqrt(4.0)) but int(sqrt(5)) != float(sqrt(5.0)) because in int case , the result will be only 2 but in float case, result will be like this 2.23606797749979 which is not equal.

[@rohitn6910]

nice one Rishabh, as we can also have it without "import"

[@code6233]

@@rohitn6910 how?

[@hhcdghjjgsdrt235]

You are the best teacher cause a guy like me found your videos interesting as I'm a civil engineer.

[@MrObadebella]

for i in range(1,501):
if i ** 2

[@tanishkbhardwaj7043]

for i in range(1,501):
if i!=0:
print(i**2)
else:
print(i)
Hey Buddy Its gr8 learning from U U are SUPERB!!

[@poorna685]

i was following codewithharry python tutorials,and i didnt get the for loop concept explained by him,but thankyou navin for explaining it much more clearly and simply

[@faisal7980]

Same 👍

[@hackwithprogramming7849]

plz dont comparmise

[@amansura8754]

Same bro what a coincidence

[@kunalagrawal8084]

Same here

[@isrogaganyaan6923]

Same

[@k.manideep2543]

from math import sqrt
for i in range(1,501,1):
a = sqrt(i)
if(i%a == 0):
print(i , end="")

[@harshavardhandsh5190]

Can you please Help me here
# It should print found or not found...
arr=array('i',[5,7,9])
print (arr)
k =int(input('Enter the value to search'))
for i in arr:
if k==i:
print('value found')
break
else:
print('value not found')

[@priteshwadhwa8833]

using Math Function:
import math
for i in range(1,501):
sq=math.sqrt(i)
if i%sq==0:
print('The Prefect Square root of number',i,'is',sq)
-----------------------------OR-------------------------
without Using Math Function:
for i in range(1,501):
sq=i**(1/2)
if i%sq==0:
print('The Prefect Square root of number',i,'is',sq)

[@nikhilgupta2637]

thaks bro u also clear my doubt

[@TheOneMusketeer]

i = 1
while i**2

[@NoraAlfraidi]

Nice

[@sadatlucas4045]

very easy ...the same 4rm the lesson...thanks sparta...

[@1104944846]

Not working

[@janisilverlake0783]

you didnt use the for looooooooop!

[@vengamnaidu9273]

import math
for i in range(1,int(math.sqrt(500))):
print(i*i, end=" ")

[@niftypliers]

This is perfect

[@ritikagrawal892]

perfect code with minimal complexity :-)

[@riteshjat108]

while running it is not showing 484
please tell me why?

[@sandeepkumar3866]

@@riteshjat108. Here range means up to the value but not include that value that's why.

[@riteshjat108]

@@sandeepkumar3866 484 is very far from 500. What is the range you are talking about, I am not getting it.

[@ExploreMalaya]

for i in range(500):
r=pow(i,2)
if r

[@varunakavoor]

import math
for i in range(1,500):
perfect_square = pow(i,2)
if perfect_square

[@tommyshelbyobe9098]

Thanks for putting the solution in the comments 😅

[@jaheerkalanthar816]

This is my first answer sir,
num=1
for i in range(num,500):
sqrt=num*num
if sqrt

[@akankshakumari6327]

Sir your explanation is crystal clear which reflects your great knowledge about these things and moreover your interest pertaining to world of programmings. It's wonderful to learn from you.🎉🎉🎉🎉🎉🎉🎉

[@surabhirai3864]

Solution of the question at the end:
import math
for i in range(1,51):
if math.sqrt(i) - (math.sqrt(i)//1) == 0:
print(i)

[@kharithakharitha]

you are teaching in a way that listeners can really enjoy coding .Thank you so much Navin.

[@GMENavneetPiyushDhal]

u r great sir ur way of teaching is just incredible

[@shiili7699]

ans:
import math
for i in range(1, 501):
if math.sqrt(i) % 1 == 0:
print(i)

[@yogeshkumarshankariya642]

Superb

[@AnkitKumar-gg2zs]

Not working

[@seemarajput779]

after % their will be i maybe

[@shubhamsingh8817]

import math
for i in range(1,501)
X=int(math.pow(i,2))
Y=int(math.sqrt(x))
if y==i and x

[@aryaveergupta4269]

for i in range (1 ,23):
print ("i**2)
i think this is simpler

[@shubhamsingh8817]

@@aryaveergupta4269 Ofcourse 👍🏻👍🏻 , that was me one year ago 😅.

[@blood1612]

Took time but finally got it :)
Heres my code-
import math
for i in range(1,501):
a=math.sqrt(i)
if a/math.floor(a)==1:
print(i)

[@chandanmishra5934]

Can you please explain that a/

[@blood1612]

@@chandanmishra5934 The loop goes frm 1 to 500
A is square root of the number
If the number is divided by the floor of that number(floor means converting the decimal value to integer value
For ex: 2.6 is converted to 2)
Then it's a perfect square

[@abhaysaini9136]

Sirrr thankyou soooo much you are building humans future...
Your teaching style and every thing is superb.... keeep it in pro level sir....god bless you... #2020👏🙏

[@closethedoorman9158]

sir the way you are teaching is awesome

[@nalinmahajan1337]

from math import sqrt
for i in range(1,501):
if sqrt(i)==int(sqrt(i)) :
print(i)

[@mannythoughts5822]

for i in range(1,501):
print(i*i)
try this one your code is not giving the square of last (500)

[@sumanshu.nankana]

for i in range(1,501):
s = pow(i,2)
if s < 501:
print(s)

[@taehyungsmochisunshine6576]

how can people dislike this😲😭😔...you are sooo amazing sir...🥺❤️

[@closethedoorman9158]

yo ..army!!

[@deeshantgupta3629]

Thank u sir ji
Answer of Assignment
from math import sqrt
for i in range(51):
if int(sqrt(i)) - sqrt(i) == 0:
print(i)

[@Pinkberry381]

I really enjoy the way you teach; I understand the concept easily.

[@simranjitkaur5364]

for i in range(1,50):
b=i*I
if b

[@amandamartinez7179]

Thank you! I didn’t understand this at all and now it seems clear. Great teacher!

[@amarrokade1600]

the general solution :
import math
x= int(input("enter any positive no." ))
for i in range (1,x+1,1):
if math.floor(math.sqrt(i))==math.ceil(math.sqrt(i)):
print(i)

[@saulenisilaschisale5464]

of all the tutorials, yours are the best. i understand everything more than i do in class. thank you so much.

[@mr8tweaker]

asli dakait boltee

[@vardaanthakur5513]

yo, this video was really helpful for me , although I am watching this after 5 years but still got my answers , thank you !!!!!!

[@varunkrishnaKyathanpally]

The answer to this quiz question is in the next video.
for i in range(1, 500):
i = i ** 2
if i >500:
break
print(i)
Got this after trying it for couple of hrs :)

[@vaibhavbopte9970]

import math
for i in range (1,51):
square = i ** 2
print(square)

[@kohlbovee7592]

similarly:
import math
for i in range(1,501):
if math.sqrt(i) % 1 == 0:
print(i)

[@kandras]

@@vaibhavbopte9970 what is import math? Is it the same to c lang lib math.h?

[@Lifeofskr9]

@@kohlbovee7592 can u please explain me the functioning of if line clearly? I am a bit confused 😕

[@vedanthrocks2215]

for i in range(1,51,1):
if (i*i

[@sylvo1328]

for i in range(1,500,1):
a=i**2
if a

[@kumaraakarshan6264]

You're so helpful. I am learning at 11 and you teach so well.

[@babyjarchive5823]

Dont stop

[@monkeyturntable]

Holy shit dude youve just cleared up so many things in this video def subbing.

[@omsaichand9798]

from math import *
n = [ ]
for i in range(1, 501):
n.append(sqrt(i))
for j in n:
if j == int(j):
print("%d"%(j*j))
else:
continue

[@aishikg.5580]

Thanks a lot This was really helpful.
I was searching all other places to learn some of these, then I found this and right when I saw the video I immedietly Subscribes and liked the video I think everyone else should subscribe and like and share.
Keep Up The Great Work!!!

[@mariaxia2687]

from math import sqrt
for n in range(1, int(sqrt(500))+1):
print(n * n)

[@topgameplay4897]

I love your way of explaining,as you first do it wrong,then tell us the write method.
Thank You 😀😀

[@hungamamachao7808]

for i in range(1,501):
y=i*i
if y

[@akshatuniyal6033]

Very Well explained, I am laying my hands on programming and i find it very interesting and better than any classroom lecture though i am a math expert, i can easily understand.

[@suramyaverma7063]

import math
i=1
while i

[@mohankumarm1543]

My Code for Quiz Question :
x=range(23)
for i in x:
if(i*i0):
print(i*i)
THANK U NAVIN SIR

[@mrjh8467]

Another excellent explanation. Highly encourage people who are studying python at college to definitely add these videos to supplement their learning.

[@eventhorizon69]

you explained it better than my schoolteacher. THANKYOU

[@riserich]

for i in range(1,500):
i *= i
if i

[@sameer281187]

Hey Richie, tried your version and it worked just fine. However the 2nd line in your code changes the value of i to (i) square and then when we iterate back to the for statement for next iteration, i is getting reset to the number in range. would like to confirm if this is something that range() offers us that it sets the variable back to its value in the range from the value that it attained during the running of program.
Eg. When i=2, the 2nd line in the code sets the value of i to 4 but when we go back to the for statement after this iteration, i is reset to 3

[@riserich]

@@sameer281187 Yes,thats how it suppose to happen. Eg. When i in range is 2 : i *= i is 4 ; i in range is 3 : i *= i is 9 ; i in range is 22 : i *= i is 484 .
Since when the value of i >= 500 in i in range of 23, The if condition won't be calculated and no print is done anymore but still python will calculate for loop from 23 to 500.
Thats why while loop is better for this assignment
n = 1
i = 1
while i = 0:
i = n*n
n += 1
if i

[@hola_gracias165]

Best...The best... way to learn coding is by watching ur vdo.

[@selinjodhani2634]

from math import sqrt
for i in range(1,501):
if(i%(sqrt(i))==0):
print(i)

[@DevSharma-eg3sx]

thank you sir,
ans>>> for a in range(1,26):
if a**2>=500:
continue
print(a**2)

[@yashkhanna2602]

import math
for i in range(1,501):
if(i**2

[@varsinisivabalu9245]

Why we are printing i**2??? Pls explain it..

[@manojs6647]

** indicates power of , start from initial vedios

[@krishashrimankar8390]

We dont need 'import math' here

[@RohitSingh-zo8ci]

It is wrong see the question b/w 500 not 501

[@sreemanbadugu9781]

@@varsinisivabalu9245 it means ** power so we are squaring it

[@aleezaali-v2r]

for i in range(1,501):
if i*i

[@chagaletiramamohanreddy332]

for i in range(1,100):
j = i * i
if j

[@mohamedfaalil]

from math import pow,sqrt
for i in range(1,30,1):
i = pow(i,2)
if i >= 1 and i

[@arulvalananto]

answer is:
for i in range(1, 501, 1):
if i**2

[@muneraalkurbi4583]

I feel it's not right to eliminate that one from the 500 even if it gives you the same result :)

[@amitkabi]

for i in range(0,500,1):
if i*i

[@hoot2416]

I am taking a python class and your videos is helping me a lot. Thank you so much for this!

[@gregoryopondi3652]

same

[@papun__kumar]

import math
for i in range(1,501):
sqr=math.sqrt(i)
res=sqr-int(sqr)
if res==0:
print (i,"=",sqr)

[@nidhigoswami2526]

Thanks a lot for this wonderful series, sir. It's easy and interesting. I have recommended it to my colleagues as well. They also love it.

[@xavif7166]

This also works and you don't have to import math :)
for i in range(1,501,1):
j= i**0.5
if j==int(j):
print(i)

[@bilalrehman989]

code dosent work friend

[@deepkamalsingh636]

We are enjoying this series!!!!

[@harshavardhandsh5190]

Can you please Help me here
# It should print found or not found...
arr=array('i',[5,7,9])
print (arr)
k =int(input('Enter the value to search'))
for i in arr:
if k==i:
print('value found')
break
else:
print('value not found')

[@yannamsettiaishwarya7448]

import math
for i in range(1,51):
x=math.pow(i,2)
print(i)

[@GurpreetSingh-td4es]

for i in range(1,50):
if(i*i==i*i):
print(i*i)

[@harshabasavaraju4609]

for I in range (1,501):
s=I*I
If I

[@gauravthapa3596]

import math
for i in range(1,501,1):
if i*i

[@nabinkumarteemilsina]

import math as m
for i in range(1,50):
a=m.sqrt(i)
if i%a==0:
print(i)
output:
1
4
9
16
25
36
49

[@simranjitkaur5364]

After using this code the ouput coming is :- 1

[@atulanand4249]

Viewing at 0.75x

[@abhiviews1]

True😂😂

[@anjankumarkantepalli8623]

me at 2x...😅

[@aadarshm2008]

exactly 😂

[@AdamuJebuni]

Thanks for meeting you ❤

[@clinicalsas-zs3fx]

Me at 2.0x😂

[@bodalasagarika3539]

for i in range(1,501):
y=pow(i,2)
if y

[@ZinduZatism]

cannot wait for pattern teach us diamond and circle pls a d all shapes also pls teach us merge sort thank u

[@msrikavya9651]

No

[@pavitrataro2777]

@@msrikavya9651 What no?

[@suvarnakumari936]

Answer-1
for i in range(1,501):
if (i*i)

[@agirlfrom_mountains]

Import math
For I in range(1, 500):
Print(math.pow(i, 2))

[@adarshsahani1947]

u can not use capital letter & it will print 1,4,9 upto 249001 which is not askesd.

[@saivikasreddy7692]

from math import sqrt, ceil
for i in range(1,ceil(sqrt(501)),1):
print("square of : ", i, "is", i*i)

[@SuRaj-ho9dq]

Import math
for i in range(500):
If math.sqrt(i)%1==0:
Print(i)

[@Simsom]

I saw that most people were doing the for loop from 1 to 500 which seemed a bit redundant to me. I set the for loop to go up to what I aproximated (cause I'm lazy that way) to be the closest number which multiplied by itself would result in a number close to 500, and in order to not go overboard with the results I used the if condition to only display the results less than 500
for i in range(1,25,1):
if i**2 < 500:
print(i**2)

[@abhishekkakeri8994]

Anna..... you are awesome!! No one can explain things this clearly!!

[@ishanthmishra7658]

For i in range (1,23)
print (i**2)
Note: Since 23 square is greater than 500 so value of i is restricted to 22

[@VivekSingh-xf5pi]

find perfect square not square of number bro

[@anshitasingla2334]

Sir u r awsm i mean really , i print a pattern by my own first tym after learning two languages before this... Thank you sir for such wonderful tutorials.. 🤗🤗

[@milanbariya4914]

Quiz Solution
# Print all the perfect square number between 1 to 500.
from math import sqrt, floor
for i in range(1, 500):
x = sqrt(i)
if x - floor(x) == 0:
print(i)

[@rajeshrao7]

# printing all perfect square numbers between 1 and 500
i=1
for i in range (1,30,1):
if i*i

[@harshvora1922]

Perfect squares between 1 and 500:
x = range(1, 501)
for i in x:
print(i * i)

[@souravnandy1308]

for i in range(1,500):
if i*i>=500:
break
print ("Square num ",i*i)

[@kishankini3406]

for i in range(1, 501) :
if(i*i

[@khushiverma8454]

for i in range(1,500):
if(i*i

[@LightYagami-pw4wu]

Anyone here in 2024 july or thereafter ?

[@Sarcastic_sia]

Yosss