I'm new to Python and I have this problem
I have a list of numbers like this:
n = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47]
I want to print from 11 to 37, that means the output = 11, 13,.... 37.
I tried to print(n[11:37]) but of course it will print [37, 41, 43, 47]
because that is range index.
Any ideas or does Python have any built-in method for this ?
This should do the job...
n = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47]
n.sort()
mylist = [x for x in n if x in range(11, 38)]
print(mylist)
Want to print that as comma separated string:
print(mylist.strip('[]'))
This will work. (Assuming list is sorted)
print n[n.index(11): n.index(37)+1]
Output:
[11, 13, 17, 19, 23, 29, 31, 37]
Considering your list is ordered and it has no duplicates:
n = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47]
print(",".join(map(str,n[n.index(11): n.index(37)+1])))
Here you have a live example
Using numpy:
import numpy as np
narr = np.array(n)
m = (narr >= 11) & (narr <= 37)
for v in narr[m]:
print(v)
# or, to get rid of the loop:
print('\n'.join(map(str, narr[m])))
it pretty simple, since your list is already sorted you can write
my_list = [x for x in n if x in range(11, 38)]
print(*my_list)
what the '*' does is that it unpacks the array into individual elements, a term known as unpacking.This will produce the actual result you wanted and not an array
If your data is sorted, you can use a generator expression with either a range object or chained comparisons:
n = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47]
print(*(i for i in n if i in range(11, 38)), sep=', ')
print(*(i for i in n if 11 <= i <= 37), sep=', ')
If your data is unsorted and you can use indices of the first occurrences of each value, you can slice your list:
print(*n[n.index(11): n.index(37)+1], sep=', ')
Result with the data you have provided:
11, 13, 17, 19, 23, 29, 31, 37
Related
I have a text file formatted like so
1-8
10-12
14-45
48-50
How do I go about getting the range of each line?
1, 2, 3, 4, 5, 6, 7, 8
10, 11, 12
I've tried splitting the file to get each number into a list. 1, 8, 10, 12, 14, 45, 48, 50, but I'm not sure how to translate that into the start and end numbers for range.
data = []
with open('file.txt','r') as myfile:
for line in myfile:
data.extend(map(int, line.split('-')))
print (data)
You can use a short list comprehension to extract the numbers from a given line. Since the ranges you listed seem to be inclusive on both ends, we need to add one to end interval (j) below.
for line in myfile:
i, j = [int(n) for n in line.split("-")]
for x in range(i, j + 1):
# do things
If you simply want to stick the two numbers in a range, here's an alternative:
range(*[int(n) for n in line.split("-")])
If you want to list out the numbers in the range, you need to wrap the expression in a list().
You just need to extract the start and end index from each line and use it in range to create your list
data = []
with open('file.txt','r') as myfile:
for line in myfile:
start,end = [int(item) for item in line.split('-')]
li = list(range(start ,end+1))
data.append(li)
print(data)
So if the input is:
1-8
10-12
14-45
48-50
The output will be:
[
[1, 2, 3, 4, 5, 6, 7, 8],
[10, 11, 12],
[14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45],
[48, 49, 50]
]
If you just want to print the ranges, you can use this:
with open('file.txt', 'r') as myfile:
for line in myfile:
nums = [int(i) for i in line.split('-')]
nums[1] += 1 # more on this in a second
my_range = range(*nums) # the * unpacks the two numbers into two arguments
print([i for i in my_range])
# [1, 2, 3, 4, 5, 6, 7, 8]
Walking through it, the first line in the for loop you pretty much have already. This just grabs the two numbers and interprets them as integers. After this line, nums = [1, 8] for your first line '1-8'. Next we add one to the the last element of the list, so nums = [1, 9] now. We do this because the range(a, b) builtin generates numbers from a to b-1.
Next we create a range that will generate the numbers that you want with range(*nums). The * in that statement unpacks the list into the two arguments that range() is expecting.
Finally we print all of the items in the range. Since my_range is now a generator, we need to unpack it to print, so we use a list comprehension to iterate over it and get all of the numbers.
You're very close, you just need to append the list range():
data = []
with open("file.txt") as f:
for line in f:
start, end = map(int, line.split("-"))
data.append(list(range(start, end + 1)))
print(data)
# [[1, 2, 3, 4, 5, 6, 7, 8], [10, 11, 12], [14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45], [48, 49, 50]]
And also making sure the end is incremented by one because it is not inclusive in range().
How about something like this?
Of course, it has no error checking, so you'd want to be mindful of the quality of your input data.
data = []
with open('file.txt','r') as myfile:
for line in myfile:
# Split the line and force conversion to int
start_int, end_int = map(int, line.split('-'))
# Get a Python 3 range - note this is not a list but a "range" type in Python 3, so we'll have to convert it before appending to our global list
int_range = range(start_int, end_int+1)
data.append(list(int_range))
I believe that you can match the pattern using capturing groups using Regex and add it into your list.
Regex: https://regex101.com/r/f1Ghff/1
(\d+)\-(\d+)
Explanation:
1st Capturing Group (\d+)
\d+ matches a digit (equal to [0-9])
+ Quantifier — Matches between one and unlimited times, as many times as possible, giving back as needed (greedy)
\- matches the character - literally (case sensitive)
2nd Capturing Group (\d+)
\d+ matches a digit (equal to [0-9])
+ Quantifier — Matches between one and unlimited times, as many times as possible, giving back as needed (greedy)
The code:
import re
str = """
1-8
10-12
14-45
48-50
"""
pattern = r"(\d+)\-(\d+)"
matches = re.findall(pattern, str)
ranges = []
# Convert Group1 and Group2 into integers
for tuple in matches:
low = int(tuple[0])
high = int(tuple[1])
ranges.append(list(range(low, high)))
print (ranges)
Outputting:
[[1, 2, 3, 4, 5, 6, 7], [10, 11], [14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44], [48, 49]]
I have a list of numbers as
N = [13, 14, 15, 25, 27, 31, 35, 36, 43]
After some calculations, for each element in N, I get the following list as the answers.
ndlist = [4, 30, 0, 42, 48, 4, 3, 42, 3]
That is, for the first index in N (which is 13), my answer is 4 in ndlist.
For some indices in N, I get the same answer in ndlist. For example, when N= 13 and 31, the answer is 4 in ndlist.
I need to find the numbers in N (13 and 31 in my example) such that they have the same answer in ndlist.
Can someone help me to that?
You can use a defaultdict and put those into a list keyed by the answer like:
Code:
N = [13, 14, 15, 25, 27, 31, 35, 36, 43]
ndlist = [4, 30, 0, 42, 48, 4, 3, 42, 3]
from collections import defaultdict
answers = defaultdict(list)
for n, answer in zip(N, ndlist):
answers[answer].append(n)
print(answers)
print([v for v in answers.values() if len(v) > 1])
Results:
defaultdict(<class 'list'>, {4: [13, 31], 30: [14],
0: [15], 42: [25, 36], 48: [27], 3: [35, 43]})
[[13, 31], [25, 36], [35, 43]]
Here is a way using only a nested list comprehension:
[N[idx] for idx, nd in enumerate(ndlist) if nd in [i for i in ndlist if ndlist.count(i)>1]]
#[13, 25, 31, 35, 36, 43]
To explain: the inner list comprehension ([i for i in ndlist if ndlist.count(i)>1]) gets all duplicate values in ndlist, and the rest of the list comprehension extracts the corresponding values in N where those values are found in ndlist
Suppose i have a a = range(1,51). How can i slice a to create a new list that look like this:
[1,2,3,11,12,13,21,22,23,31,32,33,41,42,43]
Is there a pythonic way that can help me do this without writing function?
I know that [start:stop:step] for periodically slicing one element but i'm not sure if i'm missing something obvious.
EDIT: The suggested duplicate question/answer is not the same as mine question. I simply asked to slice/extract periodically elements from a larger list/array. The suggested duplicate modifies elements of existing array.
Another option you can go with logical vector subsetting, something like:
a[(a - 1) % 10 < 3]
# array([ 1, 2, 3, 11, 12, 13, 21, 22, 23, 31, 32, 33, 41, 42, 43])
(a - 1) % 10 finds the remainder of array by 10 (period); and (a - 1) % 10 < 3 gives a logical vector which gives true for the first three elements of every ten elements.
What you want is more complicated than a simple slice, so you're going to need some kind of (likely fairly simple) function to do it. I'd look at using zip to combine multiple slices, something like:
reduce(lambda a,b:a+b, map(list, zip(a[1::10], a[2::10], a[3::10])))
Given:
>>> li=range(1,52)
You can do:
>>> [l for sl in [li[i:i+3] for i in range(0,len(li),10)] for l in sl]
[1, 2, 3, 11, 12, 13, 21, 22, 23, 31, 32, 33, 41, 42, 43, 51]
Or, if you want only full sublists:
>>> [l for sl in [li[i:i+3] for i in range(0,len(li),10)] for l in sl if len(sl)==3]
[1, 2, 3, 11, 12, 13, 21, 22, 23, 31, 32, 33, 41, 42, 43]
Or, given:
>>> li=range(1,51)
Then you do not need to test sublists:
>>> [l for sl in [li[i:i+3] for i in range(0,len(li),10)] for l in sl]
[1, 2, 3, 11, 12, 13, 21, 22, 23, 31, 32, 33, 41, 42, 43]
Psidom's answer's index math can be adapted to a list comprehension too
a = range(1,51)
[n for n in a if (n - 1) % 10 < 3]
Out[23]: [1, 2, 3, 11, 12, 13, 21, 22, 23, 31, 32, 33, 41, 42, 43]
I've a huge number of entries, every one is a float number. These data x are accesible with an iterator. I need to classify all the entries using selection like 10<y<=20, 20<y<=50, .... where y are data from an other iterables. The number of entries is much more than the number of selections. At the end I want a dictionary like:
{ 0: [all events with 10<x<=20],
1: [all events with 20<x<=50], ... }
or something similar. For example I'm doing:
for x, y in itertools.izip(variable_values, binning_values):
thebin = binner_function(y)
self.data[tuple(thebin)].append(x)
in general y is multidimensional.
This is very slow, is there a faster solution, for example with numpy? I think the problem cames from the list.append method I'm using and not from the binner_function
A fast way to get the assignments in numpy is using np.digitize:
http://docs.scipy.org/doc/numpy/reference/generated/numpy.digitize.html
You'd still have to split the resulting assignments up into groups. If x or y is multidimensional, you will have to flatten the arrays first. You could then get the unique bin assignments, and then iterate over those in conjunction with np.where to split the the assigments up into groups. This will probably be faster if the number of bins is much smaller than the number of elements that need to be binned.
As a somewhat trivial example that you will need to tweak/elaborate on for your particular problem (but is hopefully enough to get you started with with a numpy solution):
In [1]: import numpy as np
In [2]: x = np.random.normal(size=(50,))
In [3]: b = np.linspace(-20,20,50)
In [4]: assign = np.digitize(x,b)
In [5]: assign
Out[5]:
array([23, 25, 25, 25, 24, 26, 24, 26, 23, 24, 25, 23, 26, 25, 27, 25, 25,
25, 25, 26, 26, 25, 25, 26, 24, 23, 25, 26, 26, 24, 24, 26, 27, 24,
25, 24, 23, 23, 26, 25, 24, 25, 25, 27, 26, 25, 27, 26, 26, 24])
In [6]: uid = np.unique(assign)
In [7]: adict = {}
In [8]: for ii in uid:
...: adict[ii] = np.where(assign == ii)[0]
...:
In [9]: adict
Out[9]:
{23: array([ 0, 8, 11, 25, 36, 37]),
24: array([ 4, 6, 9, 24, 29, 30, 33, 35, 40, 49]),
25: array([ 1, 2, 3, 10, 13, 15, 16, 17, 18, 21, 22, 26, 34, 39, 41, 42, 45]),
26: array([ 5, 7, 12, 19, 20, 23, 27, 28, 31, 38, 44, 47, 48]),
27: array([14, 32, 43, 46])}
For dealing with flattening and then unflattening numpy arrays, see:
http://docs.scipy.org/doc/numpy/reference/generated/numpy.unravel_index.html
http://docs.scipy.org/doc/numpy/reference/generated/numpy.ravel_multi_index.html
np.searchsorted is your friend. As I read somewhere here in another answer to the same topic, it's currently a good bit faster than digitize, and does the same job.
http://docs.scipy.org/doc/numpy/reference/generated/numpy.searchsorted.html
I want to initialize an array with 10 values starting at X and incrementing by Y. I cannot directly use range() as it requires to give the maximum value, not the number of values.
I can do this in a loop, as follows:
a = []
v = X
for i in range(10):
a.append(v)
v = v + Y
But I'm certain there's a cute python one liner to do this ...
>>> x = 2
>>> y = 3
>>> [i*y + x for i in range(10)]
[2, 5, 8, 11, 14, 17, 20, 23, 26, 29]
You can use this:
>>> x = 3
>>> y = 4
>>> range(x, x+10*y, y)
[3, 7, 11, 15, 19, 23, 27, 31, 35, 39]
Just another way of doing it
Y=6
X=10
N=10
[y for x,y in zip(range(0,N),itertools.count(X,Y))]
[10, 16, 22, 28, 34, 40, 46, 52, 58, 64]
And yet another way
map(lambda (x,y):y,zip(range(0,N),itertools.count(10,Y)))
[10, 16, 22, 28, 34, 40, 46, 52, 58, 64]
And yet another way
import numpy
numpy.array(range(0,N))*Y+X
array([10, 16, 22, 28, 34, 40, 46, 52, 58, 64])
And even this
C=itertools.count(10,Y)
[C.next() for i in xrange(10)]
[10, 16, 22, 28, 34, 40, 46, 52, 58, 64]
[x+i*y for i in xrange(1,10)]
will do the job
If I understood your question correctly:
Y = 6
a = [x + Y for x in range(10)]
Edit: Oh, I see I misunderstood the question. Carry on.