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spreading range of IPs to a list

I have the following code, what I am trying to do is create a small function so a when an IP range (contains : in it) is inputted all range is appended to the list.
collected_ips = []
while True:
query = input("IP:\t")
if not query:
break
elif len(query.split('.'))==4:
temp = query.split('.')
#result1 = all([i.isdigit() for i in t]) #must be True
if query.find(":")==-1:
try:
result2 = all([0<=int(i)<=255 for i in temp])
if result2:
collected_ips.append(query)
except ValueError:
print("Please Fix Input")
elif len(x.split(":"))==2:
#function here
#append to collected_ips
else:
print("Please Fix Input")
else:
print("Please Fix Input")
example of input:
123.123.30.20:50
output:
['123.123.30.20,'123.123.30.21'...'123.123.30.50']
example of input:
123.123.20:50.30
output:
['123.123.20.30','123.123.21.30',...'123.123.50.30']

This is one approach using range to generate numbers between two ranges:
def spread_ip_range(ip):
splits = ip.split('.')
indx = next((i for i, x in enumerate(splits) if ':' in x), -1)
lst = []
if indx != -1:
_from, to = splits[indx].split(':')
ranges = range(max(0, int(_from)), min(255, int(to)) + 1))
for r in ranges:
s = '.'.join(splits[:indx]) + '.' + str(r)
if splits[indx+1:]:
s += '.' + '.'.join(splits[indx+1:])
lst.append(s)
return lst
Usage:
>>> spread_ip_range('123.123.20:50.30')
['123.123.20.30', '123.123.21.30', '123.123.22.30', ......, '123.123.49.30', '123.123.50.30']
-
>>> spread_ip_range('123.123.30.20:50')
['123.123.30.20', '123.123.30.21', '123.123.30.22', ......, '123.123.30.49', '123.123.30.50']

You could also do this more concisely using reduce (from functools):
from functools import reduce
def expandIp(ip):
nodes = [list(map(int,n.split(":"))) for n in ip.split(".")]
ranges = [range(max(n[0],0),min(n[-1]+1,256)) for n in nodes]
ips = reduce(lambda a,rng: [ip+[n] for ip in a for n in rng],ranges,[[]])
return [".".join(str(n) for n in ip) for ip in ips]
nodes converts the ip pattern into a list of range values [start] or [start,end]
ranges converts the nodes into actual ranges using the start number as the end when there is not a ':' specifier for the node (also caps the node to range 0...255)
ips combines each node range with all values of preceding nodes
The result is the concatenation of nodes in each combination with a "." as separator
note: this solution will work for multiple range specifiers in the ip parameter. e.g. expandIp("10.1.1:2.100:101") will produce: 10.1.1.100, 10.1.1.101, 10.1.2.100, 10.1.2.101. So if you intend to use it for subnets, you will be able to do expandIp("10.0.1:3.0:255")
By the way, you could validate the ip parameter with a single condition at the beginning of the function or before calling it (then you wouldn't need to use min/max to assign the ranges variable):
n0255 = { str(n) for n in range(256) }
if not all( i<4 and j<2 and r in n0255 for i,n in enumerate(ip.split(".")) for j,r in enumerate(n.split(":"))):
print("Invalid Input")
The final function would look like this:
from functools import reduce
n0255 = { str(n) for n in range(256) }
def expandIp(ip):
if not all( i<4 and j<2 and r in n0255 for i,n in enumerate(ip.split(".")) for j,r in enumerate(n.split(":"))):
return None
nodes = [list(map(int,n.split(":"))) for n in ip.split(".")]
ranges = [range(n[0],n[-1]+1) for n in nodes]
ips = reduce(lambda a,rng: [ip+[n] for ip in a for n in rng],ranges,[[]])
return [".".join(str(n) for n in ip) for ip in ips]
which would simplify your calling code down to :
collected_ips = []
while True:
query = input("IP:\t")
if not query:
break
ips = expandIp(query)
if not ips:
print("Invalid Input")
else:
collected_ips += ips

Addition of two binaries numbers in Python

Hey guys i have a trouble when i want to add two binaries numbers in Python, i mean i can enter a chain of character in a form of a string but i don't know how to select a specific value in the chain. Here is my code:
chaina = input('Enter your first binary number')
chainb = input('Enter your second binary number')
liste = str()
r = 0
for i in range [-1,chaina]:
t = 0
t = chaina() + chainb() + r
if t == 2 :
r = 1
liste = str(t) + liste
elif t == 0 or t == 1:
r = 0
liste = str(t) + liste

To add two binary numbers chaina and chainb:
bin(eval('0b{} + 0b{}'.format(chaina, chainb)))
Or, if you want the binary number without the leading '0b':
format(eval('0b{} + 0b{}'.format(chaina, chainb)), 'b')
Explanation
Assume for illustration that chaina = '1010' and chainb = '1111'. Then:
>>> '0b{} + 0b{}'.format(chaina, chainb)
'0b1010 + 0b1111'
By applying eval() on this string, we get the same result as if we typed the expression 0b1010 + 0b1111 directly into Python console.
>>> 0b1010 + 0b1111
25
>>> eval('0b1010 + 0b1111')
25
Finally, bin() produces a binary representation of the number passed to it as an argument:
>>> bin(25)
'0b11001'
The same thing is accomplished by calling format() with a 'b' argument:
>>> format(25, 'b')
'11001'
All put together, we are getting the expressions shown above.

Why don't you simply convert them into decimal and add them as you would do with decimals:
y = '0b101010'
z = '0b101010'
print(int(y,2) + int(z,2))
print(bin((int(y,2) + int(z,2))))

Assuming that you want to do a binary sum by hand, you must:
process both numbers starting from the end (reversed will help here)
consistently add bits processing carry until the lengther of both numbers is exhausted
reorder the result bits (here again reversed)
Code could be (assuming that you can be sure that chaina and chainb only consist in 0 and 1 characters, no test for it here):
def binsum(chaina, chainb):
def next0(it):
"""Retrieve next digit from a binary representation, 0 when exhausted"""
try:
return int(next(it))
except StopIteration:
return 0
a = reversed(chaina) # reverse chains to start with lowest order bit
b = reversed(chainb)
r = 0
result = [] # future result
for i in range(n):
t = next0(a) + next0(b) + r # add with carry
if t > 1:
t -= 2
r = 1
else:
r = 0
result.append('1' if t else '0')
if r != 0: # do not forget last carry
result.append('1')
return ''.join(result)

A couple of suggestions
normalize the lengths of the bit strings
l0, l1 = map(len, (str0, str1))
if l0 < l1:
str0 = "0"*(l1-l0) + str0
elif l1 < l0:
str1 = "0"*(l0-l1) + str1
do a loop on the reversed strings elements and construct the binary string bottom up
remainder = 0
result = ""
for bit_0, bit1 in zip(reversed(str0), reversed(str1)):
bit_0, bit_1 = map(int, (bit_0, bit_1))
new_bit, remainder = f(bit_0, bit_1, remainder)
result = str(new_bit) + result
if remainder != 0
...
writing f(bit_0, bit_1, remainder) and treating what to do if remainder is not null at the end of the loop is left as an exercise.

Python: replace string, matched from a list

Trying to match and mark character based n-grams. The string
txt = "how does this work"
is to be matched with n-grams from the list
ngrams = ["ow ", "his", "s w"]
and marked with <> – however, only if there is no preceding opened quote. The output i am seeking for this string is h<ow >does t<his w>ork (notice the double match in the 2-nd part, but within just 1 pair of expected quotes).
The for loop i’ve tried for this doesn’t, however, produce the wanted output at all:
switch = False
for i in txt:
if i in "".join(ngrams) and switch == False:
txt = txt.replace(i, "<" + i)
switch = True
if i not in "".join(ngrams) and switch == True:
txt = txt.replace(i, ">" + i)
switch = False
print(txt)
Any help would be greatly appreciated.

This solution uses the str.find method to find all copies of an ngram within the txt string, saving the indices of each copy to the indices set so we can easily handle overlapping matches.
We then copy txt, char by char to the result list, inserting angle brackets where required. This strategy is more efficient than inserting the angle brackets using multiple .replace call because each .replace call needs to rebuild the whole string.
I've extended your data slightly to illustrate that my code handles multiple copies of an ngram.
txt = "how does this work now chisolm"
ngrams = ["ow ", "his", "s w"]
print(txt)
print(ngrams)
# Search for all copies of each ngram in txt
# saving the indices where the ngrams occur
indices = set()
for s in ngrams:
slen = len(s)
lo = 0
while True:
i = txt.find(s, lo)
if i == -1:
break
lo = i + slen
print(s, i)
indices.update(range(i, lo-1))
print(indices)
# Copy the txt to result, inserting angle brackets
# to show matches
switch = True
result = []
for i, u in enumerate(txt):
if switch:
if i in indices:
result.append('<')
switch = False
result.append(u)
else:
result.append(u)
if i not in indices:
result.append('>')
switch = True
print(''.join(result))
output
how does this work now chisolm
['ow ', 'his', 's w']
ow 1
ow 20
his 10
his 24
s w 12
{1, 2, 10, 11, 12, 13, 20, 21, 24, 25}
h<ow >does t<his w>ork n<ow >c<his>olm
If you want adjacent groups to be merged, we can easily do that using the str.replace method. But to make that work properly we need to pre-process the original data, converting all runs of whitespace to single spaces. A simple way to do that is to split the data and re-join it.
txt = "how does this\nwork now chisolm hisow"
ngrams = ["ow", "his", "work"]
#Convert all whitespace to single spaces
txt = ' '.join(txt.split())
print(txt)
print(ngrams)
# Search for all copies of each ngram in txt
# saving the indices where the ngrams occur
indices = set()
for s in ngrams:
slen = len(s)
lo = 0
while True:
i = txt.find(s, lo)
if i == -1:
break
lo = i + slen
print(s, i)
indices.update(range(i, lo-1))
print(indices)
# Copy the txt to result, inserting angle brackets
# to show matches
switch = True
result = []
for i, u in enumerate(txt):
if switch:
if i in indices:
result.append('<')
switch = False
result.append(u)
else:
result.append(u)
if i not in indices:
result.append('>')
switch = True
# Convert the list to a single string
output = ''.join(result)
# Merge adjacent groups
output = output.replace('> <', ' ').replace('><', '')
print(output)
output
how does this work now chisolm hisow
['ow', 'his', 'work']
ow 1
ow 20
ow 34
his 10
his 24
his 31
work 14
{32, 1, 34, 10, 11, 14, 15, 16, 20, 24, 25, 31}
h<ow> does t<his work> n<ow> c<his>olm <hisow>

This should work:
txt = "how does this work"
ngrams = ["ow ", "his", "s w"]
# first find where letters match ngrams
L = len(txt)
match = [False]*L
for ng in ngrams:
l = len(ng)
for i in range(L-l):
if txt[i:i+l] == ng:
for j in range(l):
match[i+j] = True
# then sandwich matches with quotes
out = []
switch = False
for i in range(L):
if not switch and match[i]:
out.append('<')
switch = True
if switch and not match[i]:
out.append('>')
switch = False
out.append(txt[i])
print "".join(out)

Here's a method with only one for loop. I timed it and it's about as fast as the other answers to this question. I think it's a bit more clear, although that might be because I wrote it.
I iterate over the index of the first character in the n-gram, then if it matches, I use a bunch of if-else clauses to see whether I should add a < or > in this situation. I add to the end of the string output from the original txt, so I'm not really inserting in the middle of a string.
txt = "how does this work"
ngrams = set(["ow ", "his", "s w"])
n = 3
prev = -n
output = ''
shift = 0
open = False
for i in xrange(len(txt) - n + 1):
ngram = txt[i:i + n]
if ngram in ngrams:
if i - prev > n:
if open:
output += txt[prev:prev + n] + '>' + txt[prev + n:i] + '<'
elif not open:
if prev > 0:
output += txt[prev + n:i] + '<'
else:
output += txt[:i] + '<'
open = True
else:
output += txt[prev:i]
prev = i
if open:
output += txt[prev:prev + n] + '>' + txt[prev + n:]
print output

I'm getting index out of list Error [duplicate]

This question already has answers here:
Does "IndexError: list index out of range" when trying to access the N'th item mean that my list has less than N items?
(7 answers)
Closed 5 years ago.
def calcDistance(x1, y1, x2, y2):
distance = sqrt((x1-x2)**2 + (y1-y2)**2)
return distance
def make_dict():
return defaultdict(make_dict)
# Capture 1 input from the command line.
# NOTE: sys.argv[0] is the name of the python file
# Try "print sys.argv" (without the quotes) to see the sys.argv list
# 1 input --> the sys.argv list should have 2 elements.
if (len(sys.argv) == 2):
print "\tOK. 1 command line argument was passed."
# Now, we'll store the command line inputs to variables
myFile = str(sys.argv[1])
else:
print 'ERROR: You passed', len(sys.argv)-1, 'input parameters.'
quit()
# Create an empty list:
cities = []
# Create an empty dictionary to hold our (x,y) coordinate info:
myCoordinates = {}
# Open our file:
myFile = '%s.csv' % (myFile)
with open(myFile, 'rb') as csvfile:
spamreader = csv.reader(csvfile, delimiter=',', quotechar='|')
for row in spamreader:
# Only read rows that do NOT start with the "%" character.
if (row[0][0] != '%'):
# print row
id = int(row[0])
isHome = int(row[1])
x = float(row[2])
y = float(row[3])
myCoordinates[id] = {'x': x, 'y': y}
# print myCoordinates[id]['x']
# print myCoordinates[id]['y']
if (isHome == 1):
# Store this id as the home city
homeCity = id
cities.append(id)
print homeCity
print cities
# Create a TSP tour.
# VERSION 1 -- Using range() and for() loops:
myTour = []
for i in range(homeCity, len(cities)+1):
myTour.append(i)
for i in range(1, homeCity+1):
myTour.append(i)
print myTour
# VERSION 2 -- Using only range()
'''
firstPart = range(homeCity, len(cities)+1)
secondPart = range(1, homeCity+1)
myTour = firstPart + secondPart
print myTour
'''
tau = defaultdict(make_dict)
for i in cities:
# print "distance[%d][%d] = 0" % (i, i)
tau[i][i] = 0
for j in range(i+1, len(cities)+1):
# print "distance[%d][%d] > 0" % (i, j)
tau[i][j] = calcDistance(myCoordinates[i]['x'], myCoordinates[i]['y'], myCoordinates[j]['x'], myCoordinates[j]['y'])
# print "distance[%d][%d] = distance[%d][%d]" % (j, i, i, j)
tau[j][i] = tau[i][j]
# FIXME -- Edit the code below...
# Calculate the total distance of our TSP solution:
i = myTour[i]
for myIndex in range(1, len(myTour)+1):
j = myTour[myIndex]
print j
Function to calculate cost based on distance. Need to be modified.
def cost(rate,j):
cost = rate * j
cost = cost(1000,j)
print cost
Also I need to calculate cost based on distance traveled. with myIndex i am getting an error of list index out of range. I am not knowing what exactly is going there. The j is like total distance calculated.

List in python have 0 based index. If you add n elements to a list the indexes are from 0 to n-1. But you are running the loop from 1 to n. So, it getting list index out of range error.
You should do this-
for myIndex in range(0, len(myTour)):
j = myTour[myIndex]
print(j)
If you are getting list index out of range error then change the loop where you are getting the error and accessing a list using 1-based indexing, from range(1,len(some_list)+1) to range(0,len(some_list)). Or you can simply write range(len(some_list)). When there is no start value passed in the range function it starts from 0 by default.
To calculate cost try this -
for myIndex in range(0, len(myTour)):
j = myTour[myIndex]
cost = rate * j
print(cost)
Set the value of rate before starting the loop.

get nth line of string in python

How can you get the nth line of a string in Python 3?
For example
getline("line1\nline2\nline3",3)
Is there any way to do this using stdlib/builtin functions?
I prefer a solution in Python 3, but Python 2 is also fine.

Try the following:
s = "line1\nline2\nline3"
print s.splitlines()[2]

a functional approach
>>> import StringIO
>>> from itertools import islice
>>> s = "line1\nline2\nline3"
>>> gen = StringIO.StringIO(s)
>>> print next(islice(gen, 2, 3))
line3

`my_string.strip().split("\n")[-1]`

Use a string buffer:
import io
def getLine(data, line_no):
buffer = io.StringIO(data)
for i in range(line_no - 1):
try:
next(buffer)
except StopIteration:
return '' #Reached EOF
try:
return next(buffer)
except StopIteration:
return '' #Reached EOF

A more efficient solution than splitting the string would be to iterate over its characters, finding the positions of the Nth and the (N - 1)th occurence of '\n' (taking into account the edge case at the start of the string). The Nth line is the substring between those positions.
Here's a messy piece of code to demonstrate it (line number is 1 indexed):
def getLine(data, line_no):
n = 0
lastPos = -1
for i in range(0, len(data) - 1):
if data[i] == "\n":
n = n + 1
if n == line_no:
return data[lastPos + 1:i]
else:
lastPos = i;
if(n == line_no - 1):
return data[lastPos + 1:]
return "" # end of string
This is also more efficient than the solution which builds up the string one character at a time.

From the comments it seems as if this string is very large.
If there is too much data to comfortably fit into memory one approach is to process the data from the file line-by-line with this:
N = ...
with open('data.txt') as inf:
for count, line in enumerate(inf, 1):
if count == N: #search for the N'th line
print line
Using enumerate() gives you the index and the value of object you are iterating over and you can specify a starting value, so I used 1 (instead of the default value of 0)
The advantage of using with is that it automatically closes the file for you when you are done or if you encounter an exception.

Since you brought up the point of memory efficiency, is this any better:
s = "line1\nline2\nline3"
# number of the line you want
line_number = 2
i = 0
line = ''
for c in s:
if i > line_number:
break
else:
if i == line_number-1 and c != '\n':
line += c
elif c == '\n':
i += 1

Wrote into two functions for readability
string = "foo\nbar\nbaz\nfubar\nsnafu\n"
def iterlines(string):
word = ""
for letter in string:
if letter == '\n':
yield word
word = ""
continue
word += letter
def getline(string, line_number):
for index, word in enumerate(iterlines(string),1):
if index == line_number:
#print(word)
return word
print(getline(string, 4))

My solution (effecient and compact):
def getLine(data, line_no):
index = -1
for _ in range(line_no):index = data.index('\n',index+1)
return data[index+1:data.index('\n',index+1)]