I've created a list (which is sorted):
indexlist = [0, 7, 8, 12, 19, 25, 26, 27, 29, 30, 31, 33]
I want to extract the numbers from this list that are at least five away from each other and input them into another list. This is kind of confusing. This is an example of how I want the output:
outlist = [0, 7, 19, 25, 31]
As you can see, none of the numbers are within 5 of each other.
I've tried this method:
for index2 in range(0, len(indexlist) - 1):
if indexlist[index2 + 1] > indexlist[index2] + 5:
outlist.append(indexlist[index2])
However, this gives me this output:
outlist = [0, 12, 19]
Sure, the numbers are at least 5 away, however, I'm missing some needed values.
Any ideas about how I can accomplish this task?
You need to keep track of the last item you added to the list, not just compare to the following value:
In [1]: indexlist = [0, 7, 8, 12, 19, 25, 26, 27, 29, 30, 31, 33]
In [2]: last = -1000 # starting value hopefully low enough :)
In [3]: resultlist = []
In [4]: for item in indexlist:
...: if item > last+5:
...: resultlist.append(item)
...: last = item
...:
In [5]: resultlist
Out[5]: [0, 7, 19, 25, 31]
This should do the trick. Here, as I said in comment, the outlist is initialised with the first value of indexlistand iterated indexlist elements are compared to it. It is a rough solution. But works.
indexlist = [0, 7, 8, 12, 19, 25, 26, 27, 29, 30, 31, 33]
outlist = [indexlist[0]]
for index2 in range(1, len(indexlist) - 1):
if indexlist[index2] > (outlist[-1] + 5):
outlist.append(indexlist[index2])
output:
>>outlist
[0, 7, 19, 25, 31]
Tim Pietzcker's answer is right but this can also be done without storing the last added item in a separate variable. Instead you can read the last value in outlist:
>>> indexlist = [0, 7, 8, 12, 19, 25, 26, 27, 29, 30, 31, 33]
>>> outlist = []
>>> for n in indexlist:
... if not outlist or n > outlist[-1] + 5:
... outlist.append(n)
...
>>> outlist
[0, 7, 19, 25, 31]
I suppose your index_list is sorted. Then this will give you only indexes MIN_INDEX_OFFSET apart.
MIN_INDEX_OFFSET = 5;
index_list = [0, 7, 8, 12, 19, 25, 26, 27, 29, 30, 31, 33];
last_accepted = index_list[0];
out_list = [last_accepted];
for index in index_list:
if index-last_accepted > MIN_INDEX_OFFSET:
out_list.append(index);
last_accepted = index;
print(out_list)
Related
Given a list of integers, how can I group together consecutive numbers that share the same parity?
Here is an example input:
[5, 9, 11, 20, 24, 30, 31, 33, 39, 41]
And here is an example output:
[[5, 9, 11], [20, 24, 30], [31, 33, 39, 41]]
a simple loop does the job:
test = [5,7,9,11,13,20,22,24,31,33,39,41,43,44,46,50,52]
result = []
res = []
prevalue = None
for v in test:
if prevalue == None or v == prevalue + 2:
prevalue = v
res.append(v)
else:
prevalue = v
result+= [res]
res = [v]
result+= [res]
print(result)
result:
[[5, 7, 9, 11, 13], [20, 22, 24], [31, 33], [39, 41, 43], [44, 46], [50, 52]]
You can use itertools.groupby() to produce the groupings of odd / even numbers:
[list(group) for _, group in groupby(data, lambda x: x % 2 == 0)]
This outputs:
[[5, 9, 11], [20, 24, 30], [31, 33, 39, 41]]
Basically, what you want to do is track the current list you're looking at and modify it as you go and, when your condition no longer holds true, you save the list and create a new one:
results = []
current = [data[0]]
flag = data[0] % 2
for number in data[1:]:
if number % 2 == flag:
current.append(number)
else:
flag = 1 - flag
results.append(current)
current = [number]
results.append(current)
Suppose I have a list like this, where numbers increase in different steps:
[ 0, 4, 6, 8, 12, 15, 19, 21, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32]
I want to return the index for the first element in the list where the increase is incremental (+1 step only). In this case, 23 is the first location from which point the increase becomes incremental, and its index would be 8, which is what I want as an output.
What would be an elegant simple way to achieve this? This is what I have tried:
>>> for (a,b) in zip(l, l[1:]):
... if b-a == 1:
... print(l.index(a))
... break
UPDATE: In this particular setup, once the increase becomes incremental it will continue to stay that way. It is possible that the increase will never become incremental.
Solution 1: operator
from operator import sub, indexOf
L = [ 0, 4, 6, 8, 12, 15, 19, 21, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32]
print(indexOf(map(sub, L[1:], L), 1))
# prints 8
Raises ValueError: sequence.index(x): x not in sequence if difference 1 never occurs, so might want to use try/except for that.
Solution 2: bisect
This one only takes O(log n) time, using the monotonicity of incrementalness (as you commented "once the increase becomes incremental it will continue to stay that way").
from bisect import bisect
L = [ 0, 4, 6, 8, 12, 15, 19, 21, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32]
class IsIncremental:
def __getitem__(_, i):
return L[i+1] - L[i] == 1
print(bisect(IsIncremental(), False, 0, len(L) - 1))
# prints 8
Prints len(L) - 1 if difference 1 never occurs.
Btw... readability
As PEP 8 says:
Never use the characters 'l' (lowercase letter el), [...] as single character variable names. In some fonts, these characters are indistinguishable from the numerals one and zero. When tempted to use 'l', use 'L' instead.
Steps:
Iterate over the array until the second last element.
Check if next element value differs from current element value by exactly 1.
Print the index and break the loop.
Code:
my_list = [0, 4, 6, 8, 12, 15, 19, 21, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32]
for i in range(len(my_list)-1):
if my_list[i+1] - my_list[i] == 1:
print(i)
break
Result:
8
Here is an iterative approach. We can loop over the list and take the following action at each index:
If the current value is one plus the previous, then don't move the incremental index
Otherwise, reset the incremental index to the current position
If we reach the end of the list and we have an incremental index which is earlier than the last position, then we have a potential match.
lst = [0, 4, 6, 8, 12, 15, 19, 21, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32]
idx = 0
for i in range(1, len(lst)):
if lst[i] != lst[i-1] + 1:
idx = i
if idx < len(lst) - 1:
print("Found index: " + str(idx) + ", value: " + str(lst[idx]))
else:
print("No incremental index found")
This prints:
Found index: 8, value: 23
Do a for each loop and check the previous value with the current. Once you reach a point where your current value is only 1 greater than the previous value, return the index of the previous value in your array:
myList = [ 0, 4, 6, 8, 12, 15, 19, 21, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32]
lastVal = -1000
for i in myList:
if i - lastVal == 1:
print(myList.index(lastVal)) #will print your desired value's index. If this is in a function, replace print with return
break
lastVal = i
if myList.index(lastVal) == len(myList) - 1:
print("There is no incremental increase in your array")
(edited, replaced return with lastVal, fixed to print the index)
Output:
8
Here is a way to do this using list comprehension
lst = [ 0, 4, 6, 8, 12, 15, 19, 21, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32]
list2 = [i-1 for i,j in enumerate(lst) if j-lst[i-1]==1]
if len(list2)>0:
print(list2[0])
else:
print('No one up number exists')
Similar to the previous answer.
myList = [0, 4, 6, 8, 12, 15, 19, 21, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32]
l0 = 0 #suppose that the initial value is 0
for l1 in myList:
increment = l1 - l0
if increment == 1:
print(myList.index(l0)) #if you want to get the second element, use l1 instead.
break #if you want to get all the first elements that has 1 increment, remove break
l0 = l1 #memorize l1
This works fine for my purposes, except that when the value (in this case 20) is the same, it will only return the index from the front. Code is as follows, I'm not sure what the workaround is, but I need it to return the index of the value from reversed. I have some floats that would differ, but seem more difficult to work with.
lmt_lst = [-1, 5, 9, 20, 44, 37, 22, 20, 17, 12, 6, 1, -6]
lft_lmt = next(x for x in lmt_lst if x >=20)
rgt_lmt = next(x for x in reversed(lmt_lst) if x >= 20)
lft_idx = lmt_lst.index(lft_lmt)
rgt_lmt = lmt_lst.index(rgt_lmt)
print('Left Limit:', lft_lmt, 'Left Index:', lft_idx)
print('Right Limit:', rgt_lmt, 'Right Index:', rgt_idx)
If you change either of the values of '20' to 21, it works just fine
It does not return any errors, just returns the first index of the value, regardless of reversed
Your lft_lmt and rgt_lmt lists only contain values greater than or equal to 20, so your lists are [20, 44, 37, 22, 20] and [20, 22, 37, 44, 20] respectively.
>>> lmt_lst = [-1, 5, 9, 20, 44, 37, 22, 20, 17, 12, 6, 1, -6]
>>> [x for x in lmt_lst if x >=20]
[20, 44, 37, 22, 20]
>>> [x for x in reversed(lmt_lst) if x >= 20]
[20, 22, 37, 44, 20]
The first item of both of these lists is 20, and so when you search for them (using .index which searches from the beginning of the list) in the initial list you'll get 3 both times (because 20 is found at position 3 no matter how many times you search the list from beginning to end).
To find your right index you want to search the reversed list and account for the result being as if searching the list backwards:
>>> lft_idx = lmt_lst.index(lft_lmt)
>>> lft_idx
3
>>> rgt_idx = len(lmt_lst) - 1 - lmt_lst[::-1].index(rgt_lmt)
>>> rgt_idx
7
I want to know how to print numbers which are in sorted list. The interval will be given. For example:
list = [5, 10, 14, 18, 20, 30, 55]
and our interval input is between 11 and 29. So the program must print 14, 18,20.
You can simmply do as follows:
a_list = [5, 10, 14, 18, 20, 30, 55]
print([v for v in a_list if 11 <= v <= 29])
# Prints [14, 18, 20]
number_list = [5, 10, 14, 18, 20, 30, 55]
interval_list = [11,29]
result_list = []
for number in number_list:
if number in range(interval_list[0], interval_list[1]):
result_list.append(number)
print result_list
I Have the below data format.It is a dictionary with names as keys, and a list of week numbers as values.
{'Mali': [17, 16, 23, 18, 17, 16, 17, 18, 16],
'Gooki': [7, 8, 8, 15, 7, 7, 8],
'Piata': [85],
'Goerge': [82],
'Samoo': [106, 55],
'Marria: [101,39]}
I would like to count the number of weeks between the week numbers, and change the value of dictionary with the number of weeks instead of week numbers.This means that for instance in name 'Samoo' I have week 55 and week 106. I would like my code to count these two weeks plus the weeks between them(which is equal to 52 weeks ) and set it as the value for dictionary.
I have the below code, but I am not sure to make this above count.
datedict = defaultdict(set)
with open('d:/info.csv', 'r') as csvfile:
filereader = csv.reader(csvfile, 'excel')
#passing the header
read_header = False
start_date=date(year=2009,month=1,day=1)
#print((seen_date - start_date).days)
tdict = {}
for row in filereader:
if not read_header:
read_header = True
continue
# reading the rest rows
name,firstseen = row[0],row[3]
try:
seen_date = datetime.datetime.strptime(firstseen, '%d/%m/%Y').date()
deltadays = (seen_date-start_date).days
deltaweeks = deltadays/7 + 1
key = name
currentvalue = tdict.get(key, set())
currentvalue.add(deltaweeks)
tdict[key] = currentvalue
except ValueError:
print('Date value error')
pass
pprint.pprint(tdict)
Can anyone help me with this?
>>> d = {'Mali': [17, 16, 23, 18, 17, 16, 17, 18, 16],
'Gooki': [7, 8, 8, 15, 7, 7, 8],
'Piata': [85],
'Samoo': [47, 63, 48, 58, 49, 48],
'Goerge': [82],
'Samoo': [106, 55],
'Marria': [101,39]}
>>> dict((name, max(weeks) - min(weeks) + 1) for name, weeks in d.iteritems())
{'Samoo': 52, 'Gooki': 9, 'Mali': 8, 'Goerge': 1, 'Piata': 1, 'Marria': 63}