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I have the following array:
array = [
'javi#indaloymedia.com',
'caroline#grupoplatinum.com'
]
Then I have the following file:
javi#indaloymedia.com
asdsd#indaloymedia.com
jasdasd#indaloymedia.com
caroline#grupoplatinum.com
asdasde#grupoplatinum.com
wata#man.com
How can I do to eliminate all the elements that are in the array with domain 'indaloymedia.com' and grupoplatinum that is to say so that the file is as follows:
wata#man.com
The easiest way would be:
blacklist = [
'javi#indaloymedia.com',
'caroline#grupoplatinum.com'
]
domains = [e.split('#')[-1] for e in blacklist]
filtered_emails = []
with open("emails.txt") as f:
for line in f:
line = line.strip()
domain = line.split('#')[-1]
if domain not in domains:
filtered_emails.append(line.strip())
print(filtered_emails)
Note that this solution won't cover every corner case, but should be enough to get you started.
https://repl.it/repls/RealDramaticField
Also, if your blacklist is huge, domains should be a set instead of a list for fast lookups.
Only using basic for loops and an if statement:
def remove_domains(addresses, blacklist):
for a in addresses:
for b in blacklist:
if b in a:
addresses.remove(a)
return a
Related
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Given two lists, I need to count the frequency of the items in one list as they are found in the other list; and place the relative frequencies of each item inside frequencyList (where the
frequency of searchFor[0] is stored in frequencyList[0])
I am unable to import anything
textList=['a','b','a','c',...]
searchFor=['a','b']
frequencyList=[2,1]
Try:
[textList.count(i) for i in searchFor]
Or?
list(map(textList.count, searchFor))
The other answer is quite compact and very pythonic but this is an alternate solution that is slightly more efficient as it only requires one pass over the input list.
textList=['a','b','a','c']
output_dict = {}
for i in textList:
try:
output_dict[i] = d[i] + 1
except:
output_dict[i] = 1
print(output_dict['a'])
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I have a data like this, aka input data:
data = ['a-aa-aab', 'a-aa-aaa', 'b-ba', 'a-aa-aab-aaba', 'a-aa-aab-aabb']
And I want to transform that to the taxonomy string like this, aka output data:
root a b
a a-aa
a-aa a-aa-aab a-aa-aaa
a-aa-aab a-aa-aab-aaba a-aa-aab-aabb
b b-ba
I think there is a recursive solution in this sample, but I don't know how to achive this target. If you happen to know the answer, please tell me, god bless you!
from collections import defaultdict
data = ['a-aa-aab', 'a-aa-aaa', 'b-ba', 'a-aa-aab-aaba', 'a-aa-aab-aabb']
result = defaultdict(set)
for string in data:
parts = string.split('-')
for i in range(len(parts)):
key = '-'.join(parts[:i])
val = '-'.join(parts[:i+1])
result[key].add(val)
print(result)
for prefix, children in result.items():
print(prefix or 'root', *children)
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How would you turn a string that looks like this
7.11,8,9:00,9:15,14:30,15:00
into this dictionary entry
{7.11 : [8, (9:00, 9:15), (14:30, 15:00)]}?
Suppose that the number of time pairs (such as 9:00,9:15 and 14:30,15:00 is unknown and you want to have them all as tuple pairs.
First split the string at the commas, then zip cluster starting from the 3rd element and put it into a dictionary:
s = "7.11,8,9:00,9:15,14:30,15:00"
ss = s.split(',')
d = {ss[0]: [ss[1]] + list(zip(*[iter(ss[2:])]*2))}
Output:
{'7.11': ['8', ('9:00', '9:15'), ('14:30', '15:00')]}
If you need to convert it from string to appropiate data types (you'll have to adapt it according to your needs), then after getting the ss list:
time_list = [datetime.datetime.strptime(t,'%H:%M').time() for t in ss[2:]]
d = {float(ss[0]): [int(ss[1])] + list(zip(*[iter(time_list)]*2))}
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I have the following list which was created from a text file that I already have.
In the end I have a list that contains all the values that I need from the text, and now I am trying to have a more than one small lists from the list array where every small list start with switch and end where the value is empty.
with open("read.txt") as f:
for line in f:
if line.startswith('switch '):
array.append(line)
for line in f: # Continue iterating f for additional lines to keep
if not line.rstrip():
break # We hit an empty line, return to looking for switch
array.append(line)
i = 0
while i+4 <= len(my_list):
print(my_list[i+1:i+4])
i+=4
#['v1', 'v2', 'v3']
#['m1', 'm2', 'm3']
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I'm trying to find all possible groupings for a list so that the resulting list has a specified length, here's an example:
group([2,3,5,6,8], length=3)
would give
[[2,[3,5],[6,8]], [[2,3],5,[6,8]], [2,3,[5,6,8]], ...
what would be the best way to approach this?
Try this recursive solution, warning: if your list grows larger, this will grow exponentially! Also, this will have some duplicate in that the orders are different.
from itertools import permutations
master = []
def group(l, num, res=[]):
if num == 1:
res.append(l)
master.append(res)
return
for i in range(len(l)-num + 1):
firstgroup = list({e[:i+1] for e in permutations(l)})
for each in firstgroup:
myres = res[:]
myres.append(list(each))
lcp = l[:]
for j in each:
lcp.remove(j)
group(lcp, num-1, myres)