CODEWITHSUNDEEP
pythonlistpython-3.xreplacenested-lists
user1888390
user1888390

Reputation: 91

Replace a string in list of lists

I have a list of lists of strings like:

example = [["string 1", "a\r\ntest string:"],["string 1", "test 2: another\r\ntest string"]]

I'd like to replace the "\r\n" with a space (and strip off the ":" at the end for all the strings).

For a normal list I would use list comprehension to strip or replace an item like

example = [x.replace('\r\n','') for x in example]

or even a lambda function

map(lambda x: str.replace(x, '\r\n', ''),example)

but I can't get it to work for a nested list. Any suggestions?

Upvotes: 9

Views: 17855

Answers (7)

Kayvan Shah
Kayvan Shah

Reputation: 508

Here's am implementation with recusion and generators

def replace_nested(lst, old_str, new_str):
    for item in lst:
        if isinstance(item, list):
            yield list(replace_nested(item, old_str, new_str))
        else:
            yield item.replace(old_str, new_str).rstrip(':')

example = [
    ["string 1", "a\r\ntest string:"],
    ["string 1", "test 2: another\r\ntest string"]
]
new_example = list(replace_nested(example, '\r\n', ' '))
print(new_example)

Output

[['string 1', 'a test string'], ['string 1', 'test 2: another test string']]

Upvotes: 0

Stuart
Stuart

Reputation: 9858

In case your lists get more complicated than the one you gave as an example, for instance if they had three layers of nesting, the following would go through the list and all its sub-lists replacing the \r\n with space in any string it comes across.

def replace_chars(s):
    return s.replace('\r\n', ' ')

def recursive_map(function, arg):
    return [(recursive_map(function, item) if type(item) is list else function(item))
            for item in arg]

example = [[["dsfasdf", "another\r\ntest extra embedded"], 
         "ans a \r\n string here"],
        ['another \r\nlist'], "and \r\n another string"]
print recursive_map(replace_chars, example)

Upvotes: 3

Daniel Haycraft
Daniel Haycraft

Reputation: 1

For beginners looking!

Change hello to 'goodbye'.

step 1: index first

list3 = [1,2,[3,4,'hello']]
   index=0 1      2

then you are on to the second list in which it starts back at zero. the list is index 2 as a whole so when referring to index 2 it means you are at [3,4,'hello']

step 2 figure out the hello index

list3 = [1,2,[3,4,'hello']]
       index= 0 1    2

since you start at 0 again, you go from 0 until you get the number or string you want to change. you can also do -1 because hello is the last index of the list

list3[2][2] = 'goodbye'

result = [1,2,[3,4,'goodbye']]

i hope this clears things up!

Upvotes: -1

JAL IPO_
JAL IPO_

Reputation: 106

you can easily do it by the code below

example = [["string 1", "a\r\ntest string:"],["string 1", "test 2: another\r\ntest string"]]
example_replaced = []
def replace(n):
    new = n.replace("\r\n" , " ")
    if new[:-1] == ":" : new = new[:-1]
    return new

for item in example:
    Replaced = [replace(sub_item) for sub_item in item]
    example_replaced.append(Replaced)
print(example_replaced)

the output will be:

[['string 1', 'a test string:'], ['string 1', 'test 2: another test string']]

enjoy :)

Upvotes: 0

Estatistics
Estatistics

Reputation: 940

The following example, iterate between lists of lists (sublists), in order to replace a string, a word. 

myoldlist=[['aa bbbbb'],['dd myword'],['aa myword']]
mynewlist=[]
for i in xrange(0,3,1):
    mynewlist.append([x.replace('myword', 'new_word') for x in myoldlist[i]])

print mynewlist
# ['aa bbbbb'],['dd new_word'],['aa new_word']

Upvotes: 1

Io_
Io_

Reputation: 183

Well, think about what your original code is doing:

example = [x.replace('\r\n','') for x in example]

You're using the .replace() method on each element of the list as though it were a string. But each element of this list is another list! You don't want to call .replace() on the child list, you want to call it on each of its contents.

For a nested list, use nested list comprehensions!

example = [["string 1", "a\r\ntest string:"],["string 1", "test 2: another\r\ntest string"]]
example = [[x.replace('\r\n','') for x in l] for l in example]
print example


[['string 1', 'atest string:'], ['string 1', 'test 2: anothertest string']]

Upvotes: 18

sions
sions

Reputation: 107

example = [[x.replace('\r\n','') for x in i] for i in example]

Upvotes: 4

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