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math operation in list of list

I have created 1000 lists containing some values. I wanted to have a math operation on elements of each list and saving in another list that I want to plot. Each list has the shape like below and I want to subtract the ith element from the i-1th element(distance between two consecutive elements)
[[ 4 29 73 111 130 140 167 231 248 267 284 298 320 333
379 404 421 433 475 510 523 534 544 558 575 602 617 630
661 672 685 698 711 731 742 764 780 828 842 854 874 885
903 916 944 961 985 996 1013 1032 1054 1064 1077 1109 1122 1138
1205 1233 1249 1282 1299 1311 1326 1337 1372 1409 1426 1437 1511 1549
1578 1591 1604 1646]]
I have written the code below but it does not work and I got the error of index out of range.
import numpy as np
import os
import matplotlib.pyplot as plt
import pandas as pd
from scipy.signal import find_peaks
Cases = [f for f in sorted(os.listdir('.')) if f.startswith('config')]
plt.rcParams.update({'font.size': 14})
maxnum = np.max([int(os.path.splitext(f)[0].split('_')[1]) for f in CASES])
CASES = ['configuration_%d.out' % i for i in range(maxnum)]
gg = []
my_l_h = []
for i, d in enumerate(CASES):
a = np.loadtxt(d).T
x = a[3]
peaks, _ = find_peaks(x, distance=10)
gg = [peaks]
L_h = np.array(gg)
for numbers in gg:
jp = L_h[:,i]-L_h[:,i-1]
my_l_h.append(jp)
print(my_l_h)
t = np.arange(0,len(my_l_h)
plt.plot(t,my_l_h)
plt.show()

extract only integer from txt

I have a txt file which contains lost of information,I do not want its head and tail, I need only numbers in the middle. which is a 1x11200 matrix.
[txtpda]
LT=5.6
DATE=21.06.2018
TIME=14:11
CNT=11200
RES=0.00854518
N=5
VB=350
VT=0.5
LS=0
MEASTIME=201806211412
PICKUP=BFW-2
LC=0.8
[PROFILE]
255
256
258
264
269
273
267
258
251
255
259
262
260
256
255
260
264
266
265
263
261
263
267
275
280
280
280
280
283
284
283
277
279
280
283
285
283
282
280
280
286
288
298
299
299
299
304
303
300
297
295
296
299
301
303
301
299
296
298
299
302
303
304
307
308
312
313
314
312
311
311
310
312
310
309
305
303
299
297
294
288
280
270
266
250
242
222
213
199
180
173
...
-1062
-1063
[VALUES]
Ra;2;3;2;0.769;0;0;-1;0;-1;0
Rz;2;2;2;5.137;0;0;-1;0;-1;0
Pt;0;0;0;26.25;0;0;-1;0;-1;0
Wt;0;0;0;24.3;0;0;-1;0;-1;0
now I using the following method to extract numbers:
def OpenFile():
name=askopenfilename(parent=root)
f=open(name,'r')
originalyvec1=[]
yvec1=[]
if f==0:
print("fail to open the file")
else:
print("file successfully opened")
data=f.readlines()
for i in range(0,14):
del data[0]//delete its head（string)
del data[11204]//delete its tail（string)
del data[11203]//delete its tail（string)
del data[11202]//delete its tail（string)
del data[11201]//delete its tail（string)
del data[11200]//delete its tail（string)
for line in data:
for nbr in line.split(): //delete \n
yvec1.append(int(nbr))
if f.close()==0:
print("fail to close file")
else:
print("file closed")
I want to use numpy to manage it in a easy way. Is that possible?
like np.array or something like that.

You can use a alternative form of iter(), where you pass iter() a function and it will keep calling that function until it sees the value (2nd arg). You can use this to skip until you see [PROFILE]\n and then use that same form of iter() to read until [VALUES]\n. The function is just the one called by next(iterable), which is iterable.__next__, e.g.:
with open(name) as f:
for _ in iter(f.__next__, '[PROFILE]\n'): # Skip until PROFILE
pass
yvec1 = [int(d) for d in iter(f.__next__, '[VALUES]\n')]
yvec1 will now contain all values between [PROFILE] and [VALUES].
An alternative and potentially quicker way to consume the first iter() is to use collections.deque() instead of the for loop but this is likely over-kill for this problem, e.g.:
deque(iter(f.__next__, '[PROFILE]\n'), maxlen=0)
Note: using with will automatically close(f) at the end of the block.

You can simply replace everything from the line data=f.readlines() and below with:
data = [int(line) for line in map(str.strip, f.readlines()) if line.isdigit() or line.startswith('-') and line[1:].isdigit()]
And data will be the list of integers you're looking for.

Just to give you the idea this may help
The s3[0] will be all the numbers between PROFILE ans VALUES
#s=your data
s='sjlkf slflsafj[PROFILEl9723,2974982,2987492,886[VALUES]skjlfsajlsjal'
s2=s.split('[PROFILE]')
s3=s2[1].split('[VALUES]')

How to display a sequence of numbers in column-major order?

Program description:
Find all the prime numbers between 1 and 4,027 and print them in a table which
"reads down", using as few rows as possible, and using as few sheets of paper
as possible. (This is because I have to print them out on paper to turn it in.) All numbers should be right-justified in their column. The height
of the columns should all be the same, except for perhaps the last column,
which might have a few blank entries towards its bottom row.
The plan for my first function is to find all prime numbers between the range above and put them in a list. Then I want my second function to display the list in a table that reads up to down.
2 23 59
3 29 61
5 31 67
7 37 71
11 41 73
13 43 79
17 47 83
19 53 89
ect...
This all I've been able to come up with myself:
def findPrimes(n):
""" Adds calculated prime numbers to a list. """
prime_list = list()
for number in range(1, n + 1):
prime = True
for i in range(2, number):
if(number % i == 0):
prime = False
if prime:
prime_list.append(number)
return prime_list
def displayPrimes():
pass
print(findPrimes(4027))
I'm not sure how to make a row/column display in Python. I remember using Java in my previous class and we had to use a for loop inside a for loop I believe. Do I have to do something similar to that?

Although I frequently don't answer questions where the original poster hasn't even made an attempt to solve the problem themselves, I decided to make an exception of yours—mostly because I found it an interesting (and surprisingly challenging) problem that required solving a number of somewhat tricky sub-problems.
I also optimized your find_primes() function slightly by taking advantage of some reatively well-know computational shortcuts for calculating them.
For testing and demo purposes, I made the tables only 15 rows high to force more than one page to be generated as shown in the output at the end.
from itertools import zip_longest
import locale
import math
locale.setlocale(locale.LC_ALL, '') # enable locale-specific formatting
def zip_discard(*iterables, _NULL=object()):
""" Like zip_longest() but doesn't fill out all rows to equal length.
https://stackoverflow.com/questions/38054593/zip-longest-without-fillvalue
"""
return [[entry for entry in iterable if entry is not _NULL]
for iterable in zip_longest(*iterables, fillvalue=_NULL)]
def grouper(n, seq):
""" Group elements in sequence into groups of "n" items. """
for i in range(0, len(seq), n):
yield seq[i:i+n]
def tabularize(width, height, numbers):
""" Print list of numbers in column-major tabular form given the dimensions
of the table in characters (rows and columns). Will create multiple
tables of required to display all numbers.
"""
# Determine number of chars needed to hold longest formatted numeric value
gap = 2 # including space between numbers
col_width = len('{:n}'.format(max(numbers))) + gap
# Determine number of columns that will fit within the table's width.
num_cols = width // col_width
chunk_size = num_cols * height # maximum numbers in each table
for i, chunk in enumerate(grouper(chunk_size, numbers), start=1):
print('---- Page {} ----'.format(i))
num_rows = int(math.ceil(len(chunk) / num_cols)) # rounded up
table = zip_discard(*grouper(num_rows, chunk))
for row in table:
print(''.join(('{:{width}n}'.format(num, width=col_width)
for num in row)))
def find_primes(n):
""" Create list of prime numbers from 1 to n. """
prime_list = []
for number in range(1, n+1):
for i in range(2, int(math.sqrt(number)) + 1):
if not number % i: # Evenly divisible?
break # Not prime.
else:
prime_list.append(number)
return prime_list
primes = find_primes(4027)
tabularize(80, 15, primes)
Output:
---- Page 1 ----
1 47 113 197 281 379 463 571 659 761 863
2 53 127 199 283 383 467 577 661 769 877
3 59 131 211 293 389 479 587 673 773 881
5 61 137 223 307 397 487 593 677 787 883
7 67 139 227 311 401 491 599 683 797 887
11 71 149 229 313 409 499 601 691 809 907
13 73 151 233 317 419 503 607 701 811 911
17 79 157 239 331 421 509 613 709 821 919
19 83 163 241 337 431 521 617 719 823 929
23 89 167 251 347 433 523 619 727 827 937
29 97 173 257 349 439 541 631 733 829 941
31 101 179 263 353 443 547 641 739 839 947
37 103 181 269 359 449 557 643 743 853 953
41 107 191 271 367 457 563 647 751 857 967
43 109 193 277 373 461 569 653 757 859 971
---- Page 2 ----
977 1,069 1,187 1,291 1,427 1,511 1,613 1,733 1,867 1,987 2,087
983 1,087 1,193 1,297 1,429 1,523 1,619 1,741 1,871 1,993 2,089
991 1,091 1,201 1,301 1,433 1,531 1,621 1,747 1,873 1,997 2,099
997 1,093 1,213 1,303 1,439 1,543 1,627 1,753 1,877 1,999 2,111
1,009 1,097 1,217 1,307 1,447 1,549 1,637 1,759 1,879 2,003 2,113
1,013 1,103 1,223 1,319 1,451 1,553 1,657 1,777 1,889 2,011 2,129
1,019 1,109 1,229 1,321 1,453 1,559 1,663 1,783 1,901 2,017 2,131
1,021 1,117 1,231 1,327 1,459 1,567 1,667 1,787 1,907 2,027 2,137
1,031 1,123 1,237 1,361 1,471 1,571 1,669 1,789 1,913 2,029 2,141
1,033 1,129 1,249 1,367 1,481 1,579 1,693 1,801 1,931 2,039 2,143
1,039 1,151 1,259 1,373 1,483 1,583 1,697 1,811 1,933 2,053 2,153
1,049 1,153 1,277 1,381 1,487 1,597 1,699 1,823 1,949 2,063 2,161
1,051 1,163 1,279 1,399 1,489 1,601 1,709 1,831 1,951 2,069 2,179
1,061 1,171 1,283 1,409 1,493 1,607 1,721 1,847 1,973 2,081 2,203
1,063 1,181 1,289 1,423 1,499 1,609 1,723 1,861 1,979 2,083 2,207
---- Page 3 ----
2,213 2,333 2,423 2,557 2,687 2,789 2,903 3,037 3,181 3,307 3,413
2,221 2,339 2,437 2,579 2,689 2,791 2,909 3,041 3,187 3,313 3,433
2,237 2,341 2,441 2,591 2,693 2,797 2,917 3,049 3,191 3,319 3,449
2,239 2,347 2,447 2,593 2,699 2,801 2,927 3,061 3,203 3,323 3,457
2,243 2,351 2,459 2,609 2,707 2,803 2,939 3,067 3,209 3,329 3,461
2,251 2,357 2,467 2,617 2,711 2,819 2,953 3,079 3,217 3,331 3,463
2,267 2,371 2,473 2,621 2,713 2,833 2,957 3,083 3,221 3,343 3,467
2,269 2,377 2,477 2,633 2,719 2,837 2,963 3,089 3,229 3,347 3,469
2,273 2,381 2,503 2,647 2,729 2,843 2,969 3,109 3,251 3,359 3,491
2,281 2,383 2,521 2,657 2,731 2,851 2,971 3,119 3,253 3,361 3,499
2,287 2,389 2,531 2,659 2,741 2,857 2,999 3,121 3,257 3,371 3,511
2,293 2,393 2,539 2,663 2,749 2,861 3,001 3,137 3,259 3,373 3,517
2,297 2,399 2,543 2,671 2,753 2,879 3,011 3,163 3,271 3,389 3,527
2,309 2,411 2,549 2,677 2,767 2,887 3,019 3,167 3,299 3,391 3,529
2,311 2,417 2,551 2,683 2,777 2,897 3,023 3,169 3,301 3,407 3,533
---- Page 4 ----
3,539 3,581 3,623 3,673 3,719 3,769 3,823 3,877 3,919 3,967 4,019
3,541 3,583 3,631 3,677 3,727 3,779 3,833 3,881 3,923 3,989 4,021
3,547 3,593 3,637 3,691 3,733 3,793 3,847 3,889 3,929 4,001 4,027
3,557 3,607 3,643 3,697 3,739 3,797 3,851 3,907 3,931 4,003
3,559 3,613 3,659 3,701 3,761 3,803 3,853 3,911 3,943 4,007
3,571 3,617 3,671 3,709 3,767 3,821 3,863 3,917 3,947 4,013

How can I count the number of numbers in this line-separated string literal in python?

I want to find the number of numbers in the following string literal 'a'. What am I doing wrong in this code? Is there any way I can find 'count' without manually counting through the string?
I thought of adding commas after each number to make it an array but I am sure there has to be a better way to scrape individual numbers when text is given in such a way.
a = """
1004
1003
1003
1002
1001
1000
996
994
992
989
987
984
977
970
963
958
954
951
948
943
939
935
929
917
911
905
903
897
885
878
877
872
857
838
815
796
779
757
725
684
632
578
528
460
258
66
49
42
41
39
39
38
38
38
38
41
53
"""
count = 0
while a:
if a == '\n':
count+=1
print count

This gives you the number of lines excluding empty lines:
print(len([line for line in a.splitlines() if line.strip()]))
A solution without a list comprehension:
counter = 0
for line in a.splitlines():
if line.strip():
counter += 1

why not use the build-in function count
print a.count('\n') - 1
I wrote the code,but I don't know whether I can help U.

You can do it in following way.
l = a.strip().split('\n')
my_list = list(map(lambda x: l.count(x), set(l)))
print(my_list)

Removing a recurrant regular expression in a string - Python

I have the following collection of items. I would like to add a comma followed by a space at the end of each item so I can create a list out of them. I am assuming the best way to do this is to form a string out of the items and then replace 3 spaces between each item with a comma, using regular expressions?
I would like to do this with python, which I am new to.
179 181 191 193 197 199 211 223 227 229
233 239 241 251 257 263 269 271 277 281
283 293 307 311 313 317 331 337 347 349
353 359 367 373 379 383 389 397 401 409
419 421 431 433 439 443 449 457 461 463

Instead of a regular expression, how about this (assuming you have it in a file somewhere):
items = open('your_file.txt').read().split()
If it's just in a string variable:
items = your_input.split()
To combine them again with a comma in between:
print ', '.join(items)

data = """179 181 191 193 197 199 211 223 227 229
233 239 241 251 257 263 269 271 277 281 """
To get the list out of it:
lst = re.findall("(\d+)", data)
print lst
To add comma after each item, replace multiple spaces with , and space.
data = re.sub("[ ]+", ", ", data)
print data