Can anybody explain me whats wrong i am doing here -
multiArray = [
['one', 'two', 'three', 'four', 'five'],
['one', 'two', 'three', 'four', 'five'],
['one', 'two', 'three', 'four', 'five']
]
search ='four'
p1 = list(filter(lambda outerEle: search == outerEle, multiArray[0]))
p = list(filter(lambda multiArrayEle: list(filter(lambda innerArrayEle: search == innerArrayEle, multiArrayEle)), multiArray))
print (p1)
print (p)
The result i am getting here is
['four']
[['one', 'two', 'three', 'four', 'five'], ['one', 'two', 'three', 'four', 'five'], ['one', 'two', 'three', 'four', 'five']]
while i am expecting
[['four'],['four'],['four']]
In your second filter, you are using a list as a predicate (as opposed to simply a bool as you do in the first filter); now, this implicitly applies the built-in method bool to each element list, and for a list l, bool(l) is true exactly when l is non-empty:
In [4]: bool([])
Out[4]: False
In [5]: bool(['a'])
Out[5]: True
This allows you to pick out, for example, all the non-empty lists in a list of lists:
In [6]: ls = [['a'], [], ['b']]
In [7]: list(filter(lambda l: l, ls))
Out[7]: [['a'], ['b']]
Thus, in your case, at the end of the day, your filter ends up giving you all lists for which 'four' appears, which is all of them.
From your given example, it's not immediately obvious what you are trying to achieve as all the inputs are identical, but my guess is that it's something like the following:
In [19]: multiArray = [
...: ['one', 'two', 'three', 'four', 'five', 'four'],
...: ['one', 'two', 'three', 'for', 'five'],
...: ['one', 'two', 'three', 'four', 'five']
...: ]
In [20]: [list(filter(lambda x: x == search, l)) for l in multiArray]
Out[20]: [['four', 'four'], [], ['four']]
While #fuglede's answer is really the answer to your question, you can archive the result you want by changing your outer filter to map:
p = list(map(lambda multiArrayEle: list(filter(lambda innerArrayEle: search == innerArrayEle, multiArrayEle)), multiArray))
Related
I have two list items and I want to generate a list based on non-matching items. Here is what I am trying to do:
from __future__ import print_function
import os
mainline_list = ['one', 'two', 'three', 'four']
non_main_list = ['two', 'seven', 'six', 'four', 'four-3.1', 'new_three']
itemized_list = [item for item in non_main_list if item not in mainline_list]
print(itemized_list)
What is returns is
['seven', 'six', 'four-3.1', 'new_three']
while what I want is:
['seven', 'six']
Regex is not necessary, you can use all() builtin function:
mainline_list = ['one', 'two', 'three', 'four']
non_main_list = ['two', 'seven', 'six', 'four', 'four-3.1', 'new_three']
print([item for item in non_main_list if all(i not in item for i in mainline_list)])
Prints:
['seven', 'six']
I have a list like this:
lst = [['one two', 'three'], ['four five', 'six']]
I need to make:
lst = [['one', 'two', 'three'], ['four', 'five', 'six']]
Tried
([i[0].split() for i in lst])
but it gives only
[['one', 'two'], ['four', 'five']]
Any ideas how to manage it?
Thanks in advance!
Maybe I'm oversimplifying this, but you can just join and re-split?
>>> [' '.join(x).split() for x in lst]
[['one', 'two', 'three'], ['four', 'five', 'six']]
Or, using the equivalent map method:
>>> list(map(str.split, map(' '.join, lst)))
[['one', 'two', 'three'], ['four', 'five', 'six']]
I have a string type list from bash which looks like this:
inp = "["one","two","three","four","five"]"
The input is coming from bash script.
In my python script I would like to convert this to normal python list in this format:
["one","two","three","four","five"]
where all elements would be string, but the whole thin is represented as list.
I tried: list(inp)
it does not work. Any suggestions?
Try this code,
import ast
inp = '["one","two","three","four","five"]'
ast.literal_eval(inp) # will prints ['one', 'two', 'three', 'four', 'five']
Have a look at ast.literal_eval:
>>> import ast
>>> inp = '["one","two","three","four","five"]'
>>> converted_inp = ast.literal_eval(inp)
>>> type(converted_inp)
<class 'list'>
>>> print(converted_inp)
['one', 'two', 'three', 'four', 'five']
Notice that your original input string is not a valid python string, since it ends after "[".
>>> inp = "["one","two","three","four","five"]"
SyntaxError: invalid syntax
The solution using re.sub() and str.split() functions:
import re
inp = '["one","two","three","four","five"]'
l = re.sub(r'["\]\[]', '', inp).split(',')
print(l)
The output:
['one', 'two', 'three', 'four', 'five']
you can use replace and split as the following:
>>> inp
"['one','two','three','four','five']"
>>> inp.replace('[','').replace(']','').replace('\'','').split(',')
['one', 'two', 'three', 'four', 'five']
So I have this list that looks like this:
['One', 'Two', 'Three', 'Four']
['Five', 'Six', 'Seven']
so, a list with 2 elements
lst = [['One', 'Two', 'Three', 'Four'], ['Five', 'Six', 'Seven']]
And then I have also a dictionary that I declared like this:
numberDict = dict()
numberDict["One"] = "First"
numberDict["Two"] = "Second"
numberDict["Three"] = "Third"
numberDict["Four"] = "Fourth"
numberDict["Five"] = "Fifth"
numberDict["Six"] = "Sixth"
numberDict["Seven"] = "Seventh"
My question: How can I get me list to look like this? To replace its values with the dictionary ones?
lst = [['First', 'Second', 'Third', 'Fourth'], ['Fifth', 'Sixth', 'Seventh']]
Use a list comprehension:
>>> list_of_list = [['One', 'Two', 'Three', 'Four'], ['Five', 'Six', 'Seven']]
>>> [[numberDict.get(value, "") for value in lst] for lst in list_of_list]
[['First', 'Second', 'Third', 'Fourth'], ['Fifth', 'Sixth', 'Seventh']]
As an aside, note that you can also initialize the numbersDict in a single go
>>> numbers_dict = {"One": "First",
... "Two": "Second",
... "Three": "Third",
... "Four": "Fourth",
... "Five": "Fifth",
... "Six": "Sixth",
... "Seven": "Seventh"}
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I have a list of lists. I want to get all unique lists based on just the first three elements. If there are duplicates, then it should just return the last item. So for instance based on this
[['one', 'two', 'three', 'teennn'], ['five', 'five', 'five', 'five'],
['seven', 'nine', 'ten', 'eleven'], ['one', 'two', 'three', 'four']]
I want to return this
[['five', 'five', 'five', 'five'],
['seven', 'nine', 'ten', 'eleven'], ['one', 'two', 'three', 'four']]
lst = [['one', 'two', 'three', 'teennn'],
['five', 'five', 'five', 'five'],
['seven', 'nine', 'ten', 'eleven'],
['one', 'two', 'three', 'four']]
output = []
seen = set()
lst.reverse()
for item in lst:
if not item[:3] in seen:
output.append(item)
seen.add(item[:3])
output.reverse()
This ensures that the first three items are always unique. Starting from the end of your list lst, using reverse, ensures that the last appearance of each starting set is included.
If order isn't important, then you can use a dict:
data = [['one', 'two', 'three', 'teennn'], ['five', 'five', 'five', 'five'], ['seven', 'nine', 'ten', 'eleven'], ['one', 'two', 'three', 'four']]
new = {tuple(el[:3]): el for el in data}.values()
# [['one', 'two', 'three', 'four'], ['seven', 'nine', 'ten', 'eleven'], ['five', 'five', 'five', 'five']]
Or, if you really wanted to maintain order, then something like:
new = [data[idx] for idx in sorted({tuple(el[:3]): idx for idx, el in enumerate(data)}.values())]