Develop Reference

"None" appears before matrix

I wrote a program that generates matrix 7x7 and then rotates it and transposing it. In output i see this "None"
import time
import numpy as np
matsize = np.array([7,7])
matrix = np.random.randint(1000,size=(matsize))
print('\nSwource matrix:\n', matrix, '\n')
rotmatr = np.rot90(matrix, k=-1)
print('Rotation to 90 degrees...\n')
print(time.sleep(2), rotmatr, '\n')
transpos = np.transpose(rotmatr)
print('Transposing...\n')
print(time.sleep(2), transpos)
example of how the code works
Swource matrix:
[[909 859 984 490 773 696 576]
[780 645 632 233 109 181 18]
[ 81 890 328 746 930 45 999]
[944 992 556 436 545 210 814]
[192 827 820 321 45 959 940]
[921 529 276 996 141 132 183]
[235 842 287 169 71 857 70]]
Rotation to 90 degrees...
None [[235 921 192 944 81 780 909]
[842 529 827 992 890 645 859]
[287 276 820 556 328 632 984]
[169 996 321 436 746 233 490]
[ 71 141 45 545 930 109 773]
[857 132 959 210 45 181 696]
[ 70 183 940 814 999 18 576]]
Transposing...
None [[235 842 287 169 71 857 70]
[921 529 276 996 141 132 183]
[192 827 820 321 45 959 940]
[944 992 556 436 545 210 814]
[ 81 890 328 746 930 45 999]
[780 645 632 233 109 181 18]
[909 859 984 490 773 696 576]]
Process finished with exit code 0
What this NONE is and how to delete it???

if you remove the time.sleep(2) from your print statements, that will remove the None

print(time.sleep(2), transpos)
Prints two items:
the return value of time.sleep(2) which is None
the matrix in transpos
To remove None, move the sleep to its own line:
time.sleep(2)
print(transpos)

the problem is the time.sleep(2) just remove it and put it above the print

Python list alignment

I have an assignment I am trying to complete.
I have 100 random int. From those 100 I have to create a 10x10 table. DONE..
within that table I have to align my values to the right side of the each column. That is the part I'm missing.
Below is the code for that:
print(num, end=(" " if counter < 10 else "\n"))

Late answer, but you can also use:
import random
rl = random.sample(range(100, 999), 100)
max_n = 10
for n, x in enumerate(rl, 1):
print(x, end=("\n" if n % max_n == 0 else " "))
440 688 758 837 279 736 510 706 392 631
588 511 610 792 535 526 335 842 247 124
552 329 245 689 832 407 919 302 592 385
542 890 406 898 189 116 495 764 664 471
851 728 292 314 839 503 691 355 350 213
661 489 800 649 521 958 123 205 983 219
321 633 120 388 632 187 158 576 294 835
673 470 699 908 456 270 220 878 376 884
816 525 147 104 602 637 249 763 494 127
981 524 262 915 267 873 886 397 922 932

You can just format the number before printing it.
print(f"{num:>5}", end=(" " if counter < 10 else "\n"))
Alternatively, if you wanna cast the numbers to string you can use the rjust method of string.

There is a simple way to do it. I hope I have made it clear.
import random
# Generate 100 random numbers in range 1 to 1000.
random_numbers = list(map(lambda x:random.randrange(1,1000), range(100)))
# Create an empty string to store the pretty string.
pretty_txt = ''
# Loop through random_numbers and use an enumerate to get iteration count.
for index, i in enumerate(random_numbers):
# Add the current number into the string with a space.
pretty_txt += str(i) + ' '
# Add a newline every ten numbers.
# If you don't add index != 0 it will put a space in first iteration
if index % 9 == 0 and index != 0:
pretty_txt += '\n'
print(pretty_txt)
The output is:
796 477 578 458 284 287 43 535 514 504
91 411 288 980 85 233 394 313 263
135 770 793 394 362 433 370 725 472
981 932 398 275 626 631 817 82 551
775 211 755 202 81 750 695 402 809
477 925 347 31 313 514 363 115 144
341 684 662 522 236 219 142 114 621
940 241 110 851 997 699 685 434 813
983 710 124 443 569 613 456 232 80
927 445 179 49 871 821 428 750 792
527 799 878 731 221 780 16 779 333

print each 10 numbers in line always printing the first value alone

I have this loop which print each 10 numbers in line then move to next line
for i in range(100, 201):
if i % 10 == 0:
print(i)
else:
print(i, end=" ", )
and this the result:
100
101 102 103 104 105 106 107 108 109 110
111 112 113 114 115 116 117 118 119 120
121 122 123 124 125 126 127 128 129 130
131 132 133 134 135 136 137 138 139 140
141 142 143 144 145 146 147 148 149 150
151 152 153 154 155 156 157 158 159 160
161 162 163 164 165 166 167 168 169 170
171 172 173 174 175 176 177 178 179 180
181 182 183 184 185 186 187 188 189 190
191 192 193 194 195 196 197 198 199 200
it printing first number in line alone, but the want the opposite, the last number alone, something like this
100 101 102 103 104 105 106 107 108 109
110 111 112 113 114 115 116 117 118 119
120 121 122 123 124 125 126 127 128 129
130 131 132 133 134 135 136 137 138 139
140 141 142 143 144 145 146 147 148 149
150 151 152 153 154 155 156 157 158 159
160 161 162 163 164 165 166 167 168 169
170 171 172 173 174 175 176 177 178 179
180 181 182 183 184 185 186 187 188 189
190 191 192 193 194 195 196 197 198 199
200

You have the correct code. The only thing is that you are breaking at mod 0. You should break at mod 9.
for i in range(100, 201):
if i % 10 == 9:
print(i)
else:
print(i, end=" ", )

Try this:
for i in range(100, 201):
if (i + 1) % 10 == 0:
print(i)
else:
print(i, end=" ")

Well, 100 % 10 is equals to zero.
Which means it's going to print 100 in a line, and 101 in another line.
If you want the new line to start from 100 then all you have to do is to make the statement i % 10 == 0 a false statement. So you will have to add one to the 100.
so just try this code instead
for i in range(100, 201):
if (i+1) % 10 == 0:
print(i)
else:
print(i, end=" ", )
or you could try changing i % 10 == 0 to i % 10 == 9

What I Can Say By Seeing Your Code Is That You Are Using 2 Print Statements. One Inside The If Statement And One Inside Else Statement.
So What Python Compiler Is Doing There Is Printing New Line When Each If Statement Is True.
So When Your Iterating Number%10 Will Be Zero It Will Print That Number And Add A New Line, Because After Each Print Statement Python Do So.
And As You Have end=" " in Else Statement So It Is Printing The Number In Same Line When IF Statements Are False.
So What You Can Do Is:
You Can Use if (i+1)%10: instead of i % 10 == 0:
When Your Iterating Number's(i's) One's Place Will Be 9 (i.e. 109,119,etc) The If Statement Will Be True For That One, The Number Will Be Printed On The Same Line And Will Add A New Line After That.

Don't understand the result I have with pytesseract

I'm trying to read the following image :
try:
import Image
except ImportError:
from PIL import Image
import pytesseract as tes
results = tes.image_to_string(Image.open('./test.png'),boxes=True)
print(results)
And here is the result I have :
_ 239 780 263 787 0
. 239 758 263 767 0
L 235 737 263 761 0
1 220 763 229 783 0
1 220 741 229 761 0
‘ 129 763 137 784 0
1 129 741 136 761 0
1 220 650 229 670 0
‘ 220 628 229 648 0
F 235 537 263 561 0
. 239 531 263 540 0
A 239 511 268 534 0
_ 199 554 223 561 0
I 260 401 268 421 0
r 235 424 263 448 0
. 239 418 263 427 0
_ 239 398 263 404 0
{ 220 424 229 444 0
I 220 401 229 421 0
“ 220 288 229 331 0
What does this mean ? How I can interpret this result ?
Thanks a lot!

As you set boxes=True in tes.image_to_string(), the output is in box file format which the first letter in the line is the character recognized and then the bounding box coordinates of an occurrence of that character in the image. If boxes=False, tesseract will only output the characters recognized.
The image you are trying to OCR is the 7-segment digits, you may need to have a trained (language) data for 7-segment digits in order to get a good result.

Adding consecutive x values to a list

Suppose I have a list with items [123, 124, 125, ... 9820] and from that list I want to append to a second list with a string of every 8 items separated by a space up until the end. For example the list would have:
["123 124 125 126 127 128 129 130", "131, 132, 133, 134, 135, 136, 137, 138",..] etc.
What is the best way to do this in python? I have tried a naive solution of looping from 123 to 9820 but this takes way too much runtime and times out some of my simple tests I have set up. Are there any functions that would be useful to me?

Collect the elements into chunks of length 8 and use join(). Here's an example using an adapted recipe from itertools:
from itertools import zip_longest
lst = [str(x) for x in range(123, 9821)]
def grouper(iterable, n, fillvalue=""):
"Collect data into fixed-length chunks or blocks"
args = [iter(iterable)] * n
return zip_longest(*args, fillvalue=fillvalue)
lst2 = [" ".join(x) for x in grouper(lst, 8)]

We have to jump by 8 index to get next item from items list.
Demo
Consider items list from 1 to 999 numbers, Length of items list is 999.
Then use for loop with range function to jump by 8 index in a items list.
Use append method of string to get final result.
code:
>>> items = range(1, 1000)
>>> len(items)
999
>>> output_str = ""
>>> for i in range(0, 999, 8):
... output_str += " " + str(items[i])
...
>>> output_str.strip()
'1 9 17 25 33 41 49 57 65 73 81 89 97 105 113 121 129 137 145 153 161 169 177 185 193 201 209 217 225 233 241 249 257 265 273 281 289 297 305 313 321 329 337 345 353 361 369 377 385 393 401 409 417 425 433 441 449 457 465 473 481 489 497 505 513 521 529 537 545 553 561 569 577 585 593 601 609 617 625 633 641 649 657 665 673 681 689 697 705 713 721 729 737 745 753 761 769 777 785 793 801 809 817 825 833 841 849 857 865 873 881 889 897 905 913 921 929 937 945 953 961 969 977 985 993'
>>>

I think this does the work you want:
The code:
list = [str(x) for x in range(123, 9821)]
results = []
for index in range(0, len(list), 8):
results.append(" ".join(list[index:index+8]))
print(results)
The output:
[
'123 124 125 126 127 128 129 130',
'131 132 133 134 135 136 137 138',
'139 140 141 142 143 144 145 146',
'147 148 149 150 151 152 153 154',
'155 156 157 158 159 160 161 162',
...
'9795 9796 9797 9798 9799 9800 9801 9802',
'9803 9804 9805 9806 9807 9808 9809 9810',
'9811 9812 9813 9814 9815 9816 9817 9818',
'9819 9820'
]