#23 Python Tutorial for Beginners | Printing Patterns in Python
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- เผยแพร่เมื่อ 26 ก.ย. 2024
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In this lecture we will learn:
- How to implement the logic for pattern printing?
- Pattern printing through python
- Print the pattern of a square
- Print the pattern of the right triangle and reverse right-triangle
- Different approaches to printing a pattern
#1
- Pattern printing improves logical thinking capability.
- Programming is all about solving the problem, we have to understand the problem statement and wrote code for it.
- A pattern can be printed in many ways.
- We can print multiple hashes(#) in a single statement as a string.
- If you are allowed to print only hash in a statement at a time, then you have to write the same print statement multiple times.
- On writing a statement multiple times, it will give an output in a column.
- To convert a column output into a row format, we can use the end=" "parameter in every statement.
- To will restrict the cursor to move on a new line and hashes(#) will be printed in a single row.
#2
- Instead of writing the same code multiple times, we can use a for a loop.
- For a range(n) in a for loop, it will start from 0 and end at (n-1).
- We can use two loops, for printing two rows separated by a new line.
- For printing multiple rows, we have to write the same for loop code repeatedly.
- So, we can write the above for loop block inside another outer loop.
- In nested loops, an outer loop contains a variable that represents the number of rows and an inner loop contains a variable that represents the number of columns.
#3
- We can apply the concept to different patterns in a similar way.
e.g., for printing a right -triangle, we can restrict the inner loop till the value of a number of rows.
- For the reverse of the right triangle, we can restrict the inner loop till (n-i), where i is the number of rows.
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In this video we will see:
- Printing patterns
- Using loop
- For loop
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I can bet that no courses can be better then this even that is paid. I have purchased almost 5-6 courses form udemy but i couldn't find anyone better then this. A huge thanks to Mr Naveen Reddy Sir for providing this awesome courses and sir we really appreciate you for providing us such valuable resources.
even i had some paid courses of udemy but this is the best ......
Riyaz mhd hats off*
@@sss2393 lol
Yes,sssssssssssss
sure he is out to create people's mind by removing litter of technology in our mind, thank sir
str1='ABCD'
str2='PQR'
for i in range(4):
print(str1[ : i+1 ] + str2[ i : ])
Too good and easy method just loved it
that's great :)
Dont understand it properly...
What does that statement means
Print(str1[:i+1])
:
@@sonikunj696 it means from zero index to i+1 index all element.namely as range(0,i+1)
@@sonikunj696 this sign defines interval i.e if u want to go from begining to 4th one element in the list,
then u have to use like [ :5]
that man needs much more recognition , seriously i learned from this guy in a couple of months what i learned from college in one academic year , not just in python , but his other topics and his logic
I've watched a 6 hour course and used sololearn and this was never properly explained. You explained perfectly. I commend you
You definitely saw Mosh classes
i know the sololearn pain
@@mysteri0us_998 they give you a pretty decent framework, but the application isnt taught well imo
@@brodycates8472 ok
i can understand i couldnt understand object oriented programming , generators and stuff from that course
For the reverse I used range and gotten the same result which is pretty cool. Code:
for i in range(5,0,-1):
for j in range(i):
print("# ", end="")
print()
I used this:
x=4
while x>0:
for i in range(x):
print('# ',end="")
print()
x-=1
@@sriabhirammuthadi4249 smort 🗿
Awesome...
Better than paid courses...
exactly
I don't know if there is any paid courses as I am new on Net
Yes
#Solution for 2nd problem
for i in range(4):
for j in range(4):
if(i
Nice thinking
good work
How to work this code plz understand me how to implement this logic plz help me🤔🤔
You are intelligent ❤
Very good.....
Dear Sir. This is a very valuable series. I got the same result with below code.
for i in range(4, 0, -1): # starting each row from the heighest size
for j in range(i): # Printing each column inside each row according to row size
print("#", end="")
print() # Printing a new line to start a new row
Please do more stuff like this. Respect from a SriLankan..............!!!!
1).
for i in range(4):
for j in range(i+1,5):
print(j, end="")
print()
2).
s1="ABCD"
s2="PQR"
for i in range(4):
print(s1[0:i+1]+s2[i:])
thanks dude , its nice logic . how did u come up with this idea
s1[(0:i+1)]+s2[i:] ye thoda samjha do pls
@@subhashis3675 It states the indexing for printing : s1[(indexing start:ending index number]. So first strings indexing starts from 0 and ends in i+1. The second start from i and end in the last index number. I hope you understand this :)
Answers:
Asgn1:
for i in range(1,5):
for j in range(i,5):
print(j,end=" ")
print()
Asgn2:
x='ABCD'
y='PQR'
for i in range(4):
print(x[:i+1]+y[i:])
can yo please explain this 2nd assingment logic use in print
@@alisabir2440 Debug the program
Or
write that print statement in idle
U will understand....
I actually don't understand it I did debug but that print statement I get confused
what will be the code for asgn 1 if the pattern is in reverse order
@@akshaypathak8280 for i in range(1,5):
for j in range(5-i,5):
print(j,end=" ")
print()
The reason for learning this explained in 45 seconds. This is hat I have found gold with Navin's videos, in addition the the solid walk through.
1st pattern:==>
for i in range(1,5):
for j in range(1,5):
k=i+j-1
if k>4:
print("",end="")
else:
print(k,end="")
print()
2nd pattern:==>
x='ABCD'
y='PQR'
for i in range(1,5):
print(x[:i]+y[i-1:])
1)
for i in range(1,6):
for j in range(i,5):
print(j,end="")
print()
2)
s1="ABCD"
s2="PQR"
l=len(s1)
for i in range(0,4):
for j in range(i+1):
print(s1[j],end="")
for k in range(i,3):
print(s2[k],end="")
print()
(1,5) *
@@sourabhvyas7097 at least try first
1)
for i in range(4):
for j in range(4-i):
print (i+j+1,end=' ')
print()
Dimag hi nahi chal raha hai aisa
Sb indentation ka khel hai
#1
For i in range (1,5):
For j in range (i,5):
Print(j,end="")
Print()
#2
a="ABCD"
b="PQR"
for i in range (4):
For j in range (3):
Print(a[0:i+1],end="")
Print(b[i:j+3],end="")
Break
Print()
@jssriharijaikumar7473 Yes, I also want to know the logic behind this.
@@SarbajitDey @jssriharijaikumar7473 range function always starts at 0 . That's why the j value remains 0 till the end of the loop
I think there is no need for j in the sixth line just i:3 is fine and this solution have no blank space in a and b list. I have a better one:
a = "APQR"
b = "ABCD"
for i in range(4):
for j in range(i+1):
print(b[j], end=" ")
for k in range(j, 3):
print(a[-(3-k)], end=" ")
print()
@@boyanghu1222bro that's so clever
Can u please tell me how to reach this level of reasoning
Does it come with practice or do I have to be gifted to crack it
Please respond if possible
@@imissedthepartwherethatsmy9305 Hey bro, thanks very much for the appriciation! But unfortunately I was just an amatuer, who studies Python for his research paper. However what I want to tell you is that, first of all you must know the fundamental knowlege very well and remember them deeply. And then do more practical exercises. This level is not that difficult to reach. By the way, I have learned Python on and off for about more than half a year.
Navin, you saved me. I owe you, sir. You don't know that you are my guide.
and, now he knows it 🙂🙂.
Trust me it's much better than any paid course's . Truely appreciable work. Thanks sir for giving such type of knowledge 🙂
I'm a beginner.I liked the way he teaches. Started watching his videos in the morning 11 a.m and now reached 21st video by this time on the very same day. wonderful job sir.Nobody can teach the way he teaches.
1.for i in range(1,5):
2. for j in range(i,5):
3. print( j , end=" ")
4. print()
i dont understand the second print, whats that for?
@@almirkazaferovic4041stands for new line
@@_harshasgowda_ tq brh 😜
Now I am getting this in my brain exactly just one day before my exams but now I am understanding every point and now I am gaining some confidence and if I will be practising this type of things ,my logic building will increase .
Third pattern can also be drawn in thus way also:
For i in range(4,0,-1):
For j in range(i):
print('*',end=" ")
print()
for i in range(4):
print("# " * 4)
i did the same
Answers for 1st and 2nd:
for i in range(5):
for a in range(4-i):
print(a+1+i,end='')
print()
😑2nd one:
a,b="ABCD","PQR"
for i in range(4):
print(a[:i+1]+b[i:])
Thank you SIR 😊
solution for second is
a="ABCD"
b="PQR"
for i in range (4):
print(a[:i+1]+ b[i:])
wonderful , genius
From Tamilnadu, Thanks a million to you to explain in such a great way.
Love you brother Navin reddy, Now a days addicted to watch about your python videos
You are a very good and passionate teacher. Indian people are the best in programming! Greetings and respect from Rome, Italy.
FOR 1st pattern
for i in range(5):
for j in range (4-i):
print(j+i+1,end =" ")
print()
By far the best technical teacher I have seen. Navin, keep your passion for technology and teaching going. You have a gift and you are a gift!
the answer to the third one can be simplified as:
for i in range(4):
print("#"*(4-i))
easy
in the exam they'll ask fr nested loop
I have started learning Python using this course and I can tell it is simply amazing! Thank you Navin! Short videos which are all easy to follow is the key of success of this course. I love it.
x = int(input("enter no or rows"))
for i in range(x):
for j in range(x-i):
print(j+i+1, end=" ")
print()
came late, but had fun.
Also I noticed for the same code in 5:45, My pattern is flipped upside down(which you reversed btw(which when i tried got you first pattern)). Do you think this is because of the updated pycharm versions.
I used
t=-1
for i in range(4):
t=t+1
s=t
for j in range(4-i):
S=S+1
print(s, end="")
Print()
Sir, Print triangle patterns it can also write 👇
For j in range (1,5):
For i in range (i):
Print ("#",end="")
Print()
for i in range(4):
print("# " * 4)
for i in range(1, 5):
for j in range(i, 5):
print(j, end=" ")
print()
2)
x,y='ABCD','PQR'
for i in range(4):
print(x[:i+1]+y[i:])
Perfect
How does last line worked
a=['A','B','C','D']
b=["P","Q","R"]
for i in range (4):
for j in range(4):
if j
for any number:
x=int(input('enter number'))
for i in range (x):
for j in range (i,x,1):
print(j+1,'',end='')
print()
# for assignment quest1
for i in range(1,6,1):
for j in range(1,6-i,1):
print(j,end="")
print()
ans for ass 1:
for i in range(0,5,1):
for j in range(1+i,5,1):
print(j,end="")
print()
for j in range(1,5):
for i in range(j, 5):
print(i, end=" ")
print()
sir my answer is this.
wrong answer
Correct answer
tq so much
Whats the use of print() in the end
for i in range(rows):
for k in range(rows):
if k >= i:
print('#', end = '')
print('
')
for i in range(4):
for j in range (4-i):
print(i+j+1,end="")
print()
thanks for such a wonderful course for free sir
2)Prob
str1 = ('APQR')
str2 = ('ABCD')
for i in range(1):
print(str1,end="")
print()
i=str1.replace("P", "B")
print(i,end="")
print()
j=i.replace("Q","C")
print(j,end="")
print()
print(str2)
Thanks
Correct..but 1 mistake..at 9th line str1 = I would come
solution for the first ques
for i in range (1,5):
for j in range(5-i):
print(i+j ,end= "")
print()
print(j+1) should be used instead
How colleges dont license this dude is beyond me, in depth explanation barney style, thank you for taking the time to do this.
Much more easy and worth-fame version of "Printing Patterns". 100% sure, it is an under-rated channel. Well-appreciated your work, sir!
Naveen Sir, You have a wonderful way of teaching..I will always remember you for the change you have made in my life. Learning any Programming language was always so boring and challenging task for me but now I always look forward to your python series as I find it very easy and interesting. God bless you, Sir!
Answer for 1st qst:
for j in range(0,5):
for i in range(j+1, 5):
print(i, end="")
print()
@ 04:24
we can use only one for loop for similar kind of values.
1) for i in range(4):
print("* " * 4)
o/p :
* * * *
* * * *
* * * *
* * * *
2) @ 5:42
row = 4
start_point = 1
for i in range(row):
print("* " * start_point)
start_point += 1
o/p :
*
* *
* * *
* * * *
# Problem1 solution
for i in range(1,5):
count=i
for j in range(5-i):
print(count, end='')
count += 1
print('')
No output 🙄
These sessions are really well articulated and explained to a newbie!
for i in range(4):
print("#"*4)
this also work
Great Teacher I got in TH-cam ,the best part you have mention "we don't teach we educate" it is actual true , I really enjoy your videos ,Thanks for teaching thank youuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu
answer of question no.2
for i in range (4):
for j in range (4):
if (i
Alternate way of printing second pattern:
a=('# ')
for i in range (1,5):
print(i*a,end='')
print()
thank you this was helpful😇
Bro you are just awesome! you made learning python so easy, your energy level and presentation is so good
If you understand in a batter way..
Can u help me out
I am getting an error msg in
print("#",end="")
print("#",end="")
@@letsviraltoday add a print() in last line if you use end=" "
@@arunrajak.b9170 thank u
Ultimate teacher, don't know how to pay gratitude... Sir, better than any paid course... 🥰
for j in range(1,5):
print(j, end="")
i = j+1
while i
(1)
string = '1234'
for i in range(len(string)):
print(string[i:])
(2)
string1 = 'APQR'
string2 = 'ABCD'
for i in range(len(string1)):
string1 = string1.replace(string1[i], string2[i])
print(string1)
Bro can u plz tell the meaning of statement
print(string[i:])
Actually i not getting the meaning of [i:]
@@ishaningale4360 i value will be incremented each time and range will also chnages. At first i=0 and [i:] means i value to end of list, When i is 0 it will print 1234 after that i will be increment to 1. so here [i:] means [1:] it will print values from first index to end so here it will print 234 and it goes on
Awesome Malik Sir..Great Work using only String makes a program very Simple.
2nd Problem:
a = ('A','B','C','D')
b = ('P','Q','R')
for i in range (4):
for j in range (4):
if i>=j:
print(a[j],end ="")
else:
print(b[j-1],end ="")
print()
wrong
Super bro
I can feel my brain get LARGER!
@lakshmi prasanna At the end of the vid..
@lakshmi prasanna hey where you find the assignment I am also struggling to find the assignment if you know please tell me
@@bluemeet8546 7:42 just pause vha dikh jayega top
Navin Reddy, I am watching you from Pakistan for a year. You improved my skill a lot. now I know that you are watching my comment, can I have a heart from u
1st Assignment:
for j in range(1,5):
for i in range(j,5):
print(i,end="")
print()
2nd Assignment:
i="ABCD"
j="PQR"
for l in range(1,5):
print(i[0:l]+j[l-1:len(j)])
Ur debugging part is the best visualisation i have ever seen. Thank you for putting such an effort in this course.
your teaching style is really different from others and you are awesome . keep teaching like this you are crushing it through your coarse.
Sir you are the best.
All courses are nothing before your videos.
Salute you sir.
Because of you now I'm the best python programmer in my class.
code for 1st assignment:
for i in range(1,5):
for j in range(i,5):
print(j,'',end='')
print()
solution1st:
for i in range(4):
for j in range(i,4):
print (j+1,end="")
print()
solution 2nd:
chr='ABCDPQR'
for i in range(4):
print(chr[0:i+1],end="")
print(chr[4+i:7]
In ur second solution exactly at 4th line why is that indexing is 7 ?
@@vishnupriyam2926 because there are 7 characters 'ABCDPQR' for position 0 to 6 . so when you give the end point at 7 it actually access the sixth character as the final character.
this guy is so good and the course is free oh my days. Learned so much from you thanks so much
for j in range(5):
for i in range(j+1,5):
print(i,end="")
print()
For 1st pattern
That's so cool.
for i in range(4,0,-1):
for j in range(i):
print('#',end='')
print()
for k in range(4):
print('# '*4)
--- > This will do for creating the square
This is way better than some paid courses. Thank you Navin Sir. :)
For Question 2:
This will be a generic solution where str1 and str2 can be replaced by any characters irrespective of the length of the strings.
str1="ABCD"
str2="PQR"
i=0
while i < len(str1) :
for j in range(0,i+1):
print(str1[j],end=" ")
for k in range(i,len(str2)):
print(str2[k],end=" ")
i+=1
print()
for j in range(1,5):
for i in range(5-j):
print(i+j, end="")
print()
for j in range(4):
for k in range(4-j):
print(j+k+1, end='')
print()
All the three patterns can be done using one loop:
1) for i in range(5):
print('#'*5)
2)for i in range(5):
print('#'*i)
3) for i in range(6,1,-1):
print('#'*i)
str="1234"
for i in range(4):
print(str[i:])
Bro explain this maan
Explain b krdo
it was cool bro
@@ektadhapola9698
i = 0 str[0:] 1234
i = 1 str[1:] 234
..
.
str[start_limit:end_Limit]
it's a Greattt solution..!! ;)
That's really fantastic master. Currently my platform is uipath but I have interest to learn more technologies. So finally I decided to learn python. So in that TH-cam is doing much favor the guys who want to learn anything. So in that you crated a good platform to learn a python such a easy way and very logically. Thank you very much Master. Finally my conclusion is in any schools, colleges and any where the persons who are being like you then every one will became a great learner and they will reach their goal.
Tomorrow I have an interview, Thank you sir for all these videos these lectures are better than any paid course or coaching.
Assignment1:
for i in range(1,5):
print(i,end="")
for j in range(1,5-i):
print(i+j,end="")
print()
cool answer bro, thanks
Another alternative :- for the 1st assignment:-
num1 = '1234'
for i in range(4):
print(num1[:4-i] and num1[i-4:])
print()
For perfect square pattern i.e 1
for i in range(4):
print('# ' *4)
It'll work definitely
Great explanation. I was able to simpify this even more using the following:
for i in range(4):
print("# " * 4)
for i in range(4):
print("# " * (i+1))
for i in range(4):
print("# " * (4-i))
for i in range(4):
j=1
while j
Solution for 1st assignment question.
# Print reverse triangle of range(4)
for i in range(4):
for j in range(4-i):
print(i+j+1, end='')
print()
bro i wanna ask why the print() is putted in the end of code? cuz i tried and it was so important but i dont get why
@@juscoding it's for new line ig
@@vipusingh8584 yeah that was the reason, thanks
Sure this guy will take my big, I mean big pocket soda when I reach him out, I pledge to touch a life like he has done to me. will mentor whole good individuals courtesy of Neveen and the team.
Amazing Series Navin Sir, thoroughly enjoyed it.Kudos!
Q1. print pattern
for i in range (1,5):
for j in range (i,5):
print(j, end="")
print()
1st PATTERN:
for i in range(1,5):
j=4
for j in range(i,j+1):
print(i,end="")
i+=1
print()
Print(j,end="")
Not i
😁
love the python series... must compliment you on creating such consistently amazing content... just a small observation, for the last example, would reversing the outside loop have created a more intuitive code?
Question 1:
str1='ABCD'
str2='PQR'
for i in range(4):
print(str1[ : i+1 ] + str2[ i : ])
Question 2:
for i in range(5):
while i
row=int(input("enter the no of row "))
j=row
while j>=1:
for i in range(j):
print("#",end=" ")
print()
j=j-1
for i in range(1, 5):
print(i * "#")
Thanks
I spent so much time on that much at last, i made it
for i in range(5):
for j in range(i+1,5):
print(j,end="")
print()
1234
234
34
4
I am watching this course 2 yrs after the date it was uploaded but it still seem to be up to date. Awesome teacher and very well explained. Can watch your videos whole day. Thx for this course and for helping us learn.👍👍👍👍
Q1)
a, b = 1, 5
for j in range(1, 5):
for i in range(a, b):
print(i, end=" ")
a += 1
print()
For second pattern
for i in range(1,5):
Print('#'*i,end=' ')
Print()
This is more efficient sir😁
For a in range (5)
print('*' * a)
Done
🙃👍
It can also be done in this way
for i in range(4):
print("#"*4)
and way better optimised
Cool
#Both solutions are written in a program below
#1
for i in range(5):
for j in range(4-i):
print(i+1,end='')
i+=1
print()
#2
li='ABCD'
lj='PQR'
for i in range(4):
chi=li[0:i+1:]
chj=lj[i::]
print(chi+chj)
Could you please explain the code for Q 2
for i in range(4):
for j in range (4):
if i
finally understood what is happening in the pattern code
for i in range (4):
print("#"*4)
it worked
THQ bro
for i in range(4):
print(("# "*4))