['column1:abc,def', 'column2:hij,klm', 'column3:xyz,pqr']
I want to get the values after the :. Currently if I split it takes into account column1, column2, column3 as well, which I dont want. I want only the values.
This is similar to key-values pair in dictionary. The only dis-similarity is that it is list of strings.
How will I split it?
EDITED
user_widgets = Widgets.objects.filter(user_id = user_id)
if user_widgets:
for widgets in user_widgets:
widgets_list = widgets.gadgets_list //[u'column1:', u'column2:', u'column3:widget_basicLine']
print [item.split(":")[1].split(',') for item in widgets_list] //yields list index out of range
But when the widgets_list value is copied from the terminal and passed it runs correctly.
user_widgets = Widgets.objects.filter(user_id = user_id)
if user_widgets:
for widgets in user_widgets:
widgets_list = [u'column1:', u'column2:', u'column3:widget_basicLine']
print [item.split(":")[1].split(',') for item in widgets_list] //prints correctly.
Where I'm going wrong?
You can split items by ":", then split the item with index 1 by ",":
>>> l = ['column1:abc,def', 'column2:hij,klm', 'column3:xyz,pqr']
>>> [item.split(":")[1].split(',') for item in l]
[['abc', 'def'], ['hij', 'klm'], ['xyz', 'pqr']]
Nothing wrong with a 'for' loop and testing if your RH has actual data:
li=[u'column1:', u'column2:', u'column3:widget_basicLine', u'column4']
out=[]
for us in li:
us1,sep,rest=us.partition(':')
if rest.strip():
out.append(rest)
print out # [u'widget_basicLine']
Which can be reduced to a list comprehension if you wish:
>>> li=[u'column1:', u'column2:', u'column3:widget_basicLine', u'column4']
>>> [e.partition(':')[2] for e in li if e.partition(':')[2].strip()]
[u'widget_basicLine']
And you can further split by the comma if you have data:
>>> li=[u'column1:', u'column2:a,b', u'column3:c,d', u'column4']
>>> [e.partition(':')[2].split(',') for e in li if e.partition(':')[2].strip()]
[[u'a', u'b'], [u'c', u'd']]
Related
I have a list of strings with a few unclean entries and I want to replace the unclean entries with clean entries
list = ['created_DATE', 'column1(case', 'timestamp', 'location(case']
I want to get a list that is like this
cleanList = ['created_DATE', 'column1', 'timestamp', 'location']
I tired the following:
str_match = [s for s in list if "(case" in s] *#find the intersecting elements*
print (str_match)
new=[]
for k in str_match:
a=k.replace("(case" , "")
new.append(a) *#make an list of the words without the substring*
print(new)
I am not sure how do I now replace the entries from the new list into the original list. Can someone please help.
Thank you
If you want to remove all occurrences of "case(" from your list's elements, then you could write it like this:
list = ['created_DATE', 'column1(case', 'timestamp', 'location(case']
clean = []
for n in list:
clean.append(n.replace("(case", ""))
print(clean)
You can either create a new list clean as told by #alani:
import re
myList = ['created_DATE', 'column1(case', 'timestamp', 'location(case']
clean = [re.sub("\(.*", "", s) for s in myList]
print(clean)
or iterate over elements of myList and update in place
for i in range(len(myList)):
if "(case" in myList[i]:
myList[i] = myList[i].replace("(case" , "")
print(myList)
I have a single list that could be any amount of elements.
['jeff','ham','boat','','my','name','hello']
How do I split this one list into two lists or any amount of lists depending on blank string elements?
All these lists can then be put into one list of lists.
If you are certain that there is only one blank string in the list, you can use str.index to find the index of the blank string, and then slice the list accordingly:
index = lst.index('')
[lst[:index], lst[index + 1:]]
If there could be more than one blank string in the list, you can use itertools.groupby like this:
lst = ['jeff','ham','boat','','my','name','hello','','hello','world']
from itertools import groupby
print([list(g) for k, g in groupby(lst, key=bool) if k])
This outputs:
[['jeff', 'ham', 'boat'], ['my', 'name', 'hello'], ['hello', 'world']]
Using itertools.groupby, you can do:
from itertools import groupby
lst = ['jeff','ham','boat','','my','name','hello']
[list(g) for k, g in groupby(lst, key=bool) if k]
# [['jeff', 'ham', 'boat'], ['my', 'name', 'hello']]
Using bool as grouping key function makes use of the fact that the empty string is the only non-truthy string.
This is one approach using a simple iteration.
Ex:
myList = ['jeff','ham','boat','','my','name','hello']
result = [[]]
for i in myList:
if not i:
result.append([])
else:
result[-1].append(i)
print(result)
Output:
[['jeff', 'ham', 'boat'], ['my', 'name', 'hello']]
Let list_string be your list. This should do the trick :
list_of_list=[[]]
for i in list_string:
if len(i)>0:
list_of_list[-1].append(i)
else:
list_of_list.append([])
Basically, you create a list of list, and you go through your original list of string, each time you encounter a word, you put it in the last list of your list of list, and each time you encounter '' , you create a new list in your list of list. The output for your example would be :
[['jeff','ham','boat'],['my','name','hello']]
i'm not sure that this is what you're trying to do, but try :
my_list = ['jeff','ham','boat','','my','name','','hello']
list_tmp = list(my_list)
final_list = []
while '' in list_tmp:
idx = list_tmp.index('')
final_list.append(list_tmp[:idx])
list_tmp = list_tmp[idx + 1:]
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 have a list of elements whose text is like the following:
aSampleElementText = "Vraj Shroff\nIndia" I want to have two lists now where the first list's element would have "Vraj Shroff" and the second list's element would have "India".
I looked at other posts about split and splitlines. However, my code below is not giving me expected results.
Output:
"V",
"r"
Desired output:
"Vraj Shroff",
"India"
My code:
personalName = "Something" #first list
personalTitle = "Something" #second list
for i in range(len(names)-1)
#names is a list of elements (example above)
#it is len - 1 becuase I don't want to do this to the first element of the list
i += 1
temp = names[i].text
temp.splitlines()
personName.append(temp[0])
personTitle.append(temp[1])
names is a string. names[I] is the character corresponding to that index in the string. Hence you are getting this kind of output.
Do something like,
x = names.splitlines()
x will be the list with the elements.
names = []
locations = []
a = ["Vraj Shroff\nIndia", "Vraj\nIndia", "Shroff\nxyz", "abd cvd\nUS"]
for i in a:
b = i.splitlines()
names.append(b[0])
locations.append(b[1])
print(names)
print(locations)
output:
['Vraj Shroff', 'Vraj', 'Shroff', 'abd cvd']
['India', 'India', 'xyz', 'US']
Is this what you were looking for?
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']