fh=open('asd.txt')
data=fh.read()
fh.close()
name=data.split('\n')[0][1:]
seq=''.join(data.split('\n')[1:])
print name
print seq
In this code, the 3rd line means "take only first line with first character removed" while the 4th line means "leave the first line and join the next remaining lines".
I cannot get the logic of these two lines.
Can anyone explain me how these two slice operators ([0][1:]) are used together?
Thanx
Edited: renamed file variable (which is a keyword, too) to data.
Think of it like this: file.split('\n') gives you a list of strings. So the first indexing operation, [0], gives you the first string in the list. Now, that string itself is a "list" of characters, so you can then do [1:] to get every character after the first. It's just like starting with a two-dimensional list (a list of lists) and indexing it twice.
When confused by a complex expression, do it it steps.
>>> data.split('\n')[0][1:]
>>> data
>>> data.split('\n')
>>> data.split('\n')[0]
>>> data.split('\n')[0][1:]
That should help.
lets do it by steps, (I think I know what name and seq is):
>>> file = ">Protein kinase\nADVTDADTSCVIN\nASHRGDTYERPLK" <- that's what you get reading your (fasta) file
>>> lines = file.split('\n') <- make a list of lines
>>> line_0 = lines[0] <- take first line (line numbers start at 0)
>>> name = line_0[1:] <- give me line items [x:y] (from x to y)
>>> name
'Protein kinase'
>>>
>>> file = ">Protein kinase\nADVTDADTSCVIN\nASHRGDTYERPLK"
>>> lines = file.split('\n')
>>> seqs = lines[1:] <- gime lines [x:y] (from x to y)
>>> seq = ''.join(seqs)
>>> seq
'ADVTDADTSCVINASHRGDTYERPLK'
>>>
in slice [x:y], x is included, y is not included. When you want to arrive to the end of the list just do not indicate y -> [x:] (from item of index x to the end)
Each set of [] just operates on the list that split returns, and the resulting
list or string then used without assigning it to another variable first.
Break down the third line like this:
lines = file.split('\n')
first_line = lines[0]
name = first_line[1:]
Break down the fourth line like this:
lines = file.split('\n')
all_but_first_line = lines[1:]
seq = ''.join(all_but_first_line)
take this as an example
myl = [["hello","world","of","python"],["python","is","good"]]
so here myl is a list of list. So, myl[0] means first element of list which is equal to ['hello', 'world', 'of', 'python'] but when you use myl[0][1:] it means selecting first element from list which is represented by myl[0] and than from the resulting list(myl[0]) select every element except first one(myl[0][1:]). So output = ['world', 'of', 'python']
Related
So I have a long list of column headers. All are strings, some are several words long. I've yet to find a way to write a function that extracts the first word from each value in the list and returns a list of just those singular words.
For example, this is what my list looks like:
['Customer ID', 'Email','Topwater -https:', 'Plastics - some uml']
And I want it to look like:
['Customer', 'Email', 'Topwater', 'Plastics']
I currently have this:
def first_word(cur_list):
my_list = []
for word in cur_list:
my_list.append(word.split(' ')[:1])
and it returns None when I run it on a list.
You can use list comprehension to return a list of the first index after splitting the strings by spaces.
my_list = [x.split()[0] for x in your_list]
To address "and it returns None when I run it on a list."
You didn't return my_list. Because it created a new list, didn't change the original list cur_list, the my_list is not returned.
To extract the first word from every value in a list
From #dfundako, you can simplify it to
my_list = [x.split()[0] for x in cur_list]
The final code would be
def first_word(cur_list):
my_list = [x.split()[0] for x in cur_list]
return my_list
Here is a demo. Please note that some punctuation may be left behind especially if it is right after the last letter of the name:
names = ["OMG FOO BAR", "A B C", "Python Strings", "Plastics: some uml"]
first_word(names) would be ['OMG', 'A', 'Python', 'Plastics:']
>>> l = ['Customer ID', 'Email','Topwater -https://karls.azureedge.net/media/catalog/product/cache/1/image/627x470/9df78eab33525d08d6e5fb8d27136e95/f/g/fgh55t502_web.jpg', 'Plastics - https://www.bass.co.za/1473-thickbox_default/berkley-powerbait-10-power-worm-black-blue-fleck.jpg']
>>> list(next(zip(*map(str.split, l))))
['Customer', 'Email', 'Topwater', 'Plastics']
[column.split(' ')[0] for column in my_list] should do the trick.
and if you want it in a function:
def first_word(my_list):
return [column.split(' ')[0] for column in my_list]
(?<=\d\d\d)\d* try using this in a loop to extract the words using regex
I would like to separate my string every both commas but I can not, can you help me.
This is what I want: ['nb1,nb2','nb3,nb4','nb5,nb6']
Here is what I did :
a= 'nb1,nb2,nb3,nb4,nb5,nb6'
compteur=0
for i in a:
if i==',' :
compteur+=1
if compteur%2==0:
print compteur
test = a.split(',', compteur%2==0 )
print a
print test
The result:
2
4
nb1,nb2,nb3,nb4,nb5,nb6
['nb1', 'nb2,nb3,nb4,nb5,nb6']
Thanks you by advances for you answers
You can use regex
In [12]: re.findall(r'([\w]+,[\w]+)', 'nb1,nb2,nb3,nb4,nb5,nb6')
Out[12]: ['nb1,nb2', 'nb3,nb4', 'nb5,nb6']
A quick fix could be to simply first separate the elements by commas and then join the elements by two together again. Like:
sub_result = a.split(',')
result = [','.join(sub_result[i:i+2]) for i in range(0,len(sub_result),2)]
This gives:
>>> result
['nb1,nb2', 'nb3,nb4', 'nb5,nb6']
This will also work if the number of elements is odd. For example:
>>> a = 'nb1,nb2,nb3,nb4,nb5,nb6,nb7'
>>> sub_result = a.split(',')
>>> result = [','.join(sub_result[i:i+2]) for i in range(0,len(sub_result),2)]
>>> result
['nb1,nb2', 'nb3,nb4', 'nb5,nb6', 'nb7']
You use a zip operation of the list with itself to create pairs:
a = 'nb1,nb2,nb3,nb4,nb5,nb6'
parts = a.split(',')
# parts = ['nb1', 'nb2', 'nb3', 'nb4', 'nb5', 'nb6']
pairs = list(zip(parts, parts[1:]))
# pairs = [('nb1', 'nb2'), ('nb2', 'nb3'), ('nb3', 'nb4'), ('nb4', 'nb5'), ('nb5', 'nb6')]
Now you can simply join every other pair again for your output:
list(map(','.join, pairs[::2]))
# ['nb1,nb2', 'nb3,nb4', 'nb5,nb6']
Split the string by comma first, then apply the common idiom to partition an interable into sub-sequences of length n (where n is 2 in your case) with zip.
>>> s = 'nb1,nb2,nb3,nb4,nb5,nb6'
>>> [','.join(x) for x in zip(*[iter(s.split(','))]*2)]
['nb1,nb2', 'nb3,nb4', 'nb5,nb6']
I want to use Python to create a file that looks like
# empty in the first line
this is the second line
this is the third line
I tried to write this script
myParagraph = []
myParagraph[0] = ''
myParagraph[1] = 'this is the second line'
myParagraph[2] = 'this is the third line'
An error is thrown: IndexError: list index out of range. There are many answers on similar questions that recommend using myParagraph.append('something'), which I know works. But I want to better understand the initialization of Python lists. How to manipulate a specific elements in a list that's not populated yet?
Since you want to associate an index (whether it exists or not) with an element of data, just use a dict with integer indexes:
>>> myParagraph={}
>>> myParagraph[0] = ''
>>> myParagraph[1] = 'this is the second line'
>>> myParagraph[2] = 'this is the third line'
>>> myParagraph[99] = 'this is the 100th line'
>>> myParagraph
{0: '', 1: 'this is the second line', 2: 'this is the third line', 99: 'this is the 100th line'}
Just know that you will need to sort the dict to reassemble in integer order.
You can reassemble into a string (and skip missing lines) like so:
>>> '\n'.join(myParagraph.get(i, '') for i in range(max(myParagraph)+1))
You can do a limited form of this by assigning to a range of indexes starting at the end of the list, instead of a single index beyond the end of the list:
myParagraph = []
myParagraph[0:] = ['']
myParagraph[1:] = ['this is the second line']
myParagraph[2:] = ['this is the third line']
Note: In Matlab, you can assign to arbitrary positions beyond the end of the array, and Matlab will fill in values up to that point. In Python, any assignment beyond the end of the array (using this syntax or list.insert()) will just append the value(s) into the first position beyond the end of the array, which may not be the same as the index you assigned.
A list doesn't have an unkown size - len(myParagraph) will give you its length
You can define a function that will do this for you:
def set_at(xs, idx, x, default=None):
if len(xs) <= idx:
xs.extend([default] * (idx - len(xs) + 1))
xs[idx] = x
Then to use it:
myParagraph = []
set_at(myParagraph, 1, 'this is the second line', default='')
set_at(myParagraph, 2, 'this is the third line')
set_at(myParagraph, 20, 'this is the twenty-first line', default='')
myParagraph = []
myParagraph.append('')
myParagraph.append('this is the second line')
myParagraph.append('this is the third line')
for i,item in enumerate(myParagraph):
print "i:"+str(i)+": item:"+item
result:
i:0: item:
i:1: item:this is the second line
i:2: item:this is the third line
append is the easiest way to get around this, but if it makes you more comfortable having those indices then you should consider using insert:
Insert an item at a given position. The first argument is the index of
the element before which to insert, so a.insert(0, x) inserts at the
front of the list, and a.insert(len(a), x) is equivalent to
a.append(x).
myParagraph = []
myParagraph.insert(0, '\n')
myParagraph.insert(1, 'this is the second line\n')
myParagraph.insert(2, 'this is the third line\n')
And don't forget the new line character '\n' when writing to a file.
I would like to sort out the first row of a given list.
I've been already tried to use python "replace" to remove the second row.
But the problem is that the replace function seems not work at all.
Here is the regular expression I used: replace(r'^ //.*$','')
Here is the list:
//SA/... //short_message/Saint/...
//SS-SA/... //long_message/wonder-girl/...
here is the output I am expecting:
//SA/...
//SS-SA/...
l = ["1 12","3 12","2 12"] # space separated
n = [x.split()[0] for x in l]
print sorted(n)
If i have a list strings:
first = []
last = []
my_list = [' abc 1..23',' bcd 34..405','cda 407..4032']
how would i append the numbers flanking the .. to their corresponding lists ? to get:
first = [1,34,407]
last = [23,405,4032]
i wouldn't mind strings either because i can convert to int later
first = ['1','34','407']
last = ['23','405','4032']
Use re.search to match the numbers between .. and store them in two different groups:
import re
first = []
last = []
for s in my_list:
match = re.search(r'(\d+)\.\.(\d+)', s)
first.append(match.group(1))
last.append(match.group(2))
DEMO.
I'd use a regular expression:
import re
num_range = re.compile(r'(\d+)\.\.(\d+)')
first = []
last = []
my_list = [' abc 1..23',' bcd 34..405','cda 407..4032']
for entry in my_list:
match = num_range.search(entry)
if match is not None:
f, l = match.groups()
first.append(int(f))
last.append(int(l))
This outputs integers:
>>> first
[1, 34, 407]
>>> last
[23, 405, 4032]
One more solution.
for string in my_list:
numbers = string.split(" ")[-1]
first_num, last_num = numbers.split("..")
first.append(first_num)
last.append(last_num)
It will throw a ValueError if there is a string with no spaces in my_list or there is no ".." after the last space in some of the strings (or there is more than one ".." after the last space of the string).
In fact, this is a good thing if you want to be sure that values were really obtained from all the strings, and all of them were placed after the last space. You can even add a try…catch block to do something in case the string it tries to process is in an unexpected format.
first=[(i.split()[1]).split("..")[0] for i in my_list]
second=[(i.split()[1]).split("..")[1] for i in my_list]