Python list with filter function (excel-way)

Question

I have a Python list:

listx = [["a", 127, "Blue", 0],
         ["b", 127, "Red", 1],
         ["b", 127, "Green", 0],
         ["b", 99, "Green", 1],
         ["c", 99, "Yellow", 0]]

This is table view for understanding the situation better way:

This is table view

I want to do some filter function. For example; I want to get a list with index 0 = "b" and index 1 = 127. So the results must be:

listxnew = [["b", 127, "Red", 1],
            ["b", 127, "Green", 0]]

Table view for listxnew to understand the situation better way:

enter image description here

So how can I do this with simple Python code? Thanks.

sshashank124 · Accepted Answer · 2014-06-28 13:51:01Z

4

You can do it using list comprehension as follows:

listxnew = [i for i in listx if i[0:2]==['b', 127]]

>>> print listxnew
[['b', 127, 'Red', 1]
 ['b', 127, 'Green', 0]]

answered Jun 28, 2014 at 13:51

sshashank124

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5 Comments

kadir_cakir Over a year ago

This is not a way that I want. Any def, function, or something like that? I will use this for a GUI. So, things can be filtered may be more than two, three or just one.

sshashank124 Over a year ago

@user3051668, I'm not sure exactly what you mean but you can easily make a def out of the code above. Sorry

arvindch Over a year ago

@user3051668 - Check if my answer works? I'm curious whether that's what you're looking for.

kadir_cakir Over a year ago

It works, but I don't want to write different codes for different filterings. If I want to filter with index 0 and index 2, it'll be different list comprehension.

Dan Oberlam Over a year ago

@user3051668 You could readily put the list comprehension in a function and have it based on certain inputs (such as the indices you want to check and their desired value)

arvindch · Accepted Answer · 2014-06-28 14:20:05Z

1

Here's an easy solution, which is readily extensible:

def filterls(ls, opts):
    """
    ls - list
    opts - dict - {id: match_info}
    """
    results = []
    for l in ls:
        for (i, t) in opts.items():
            if l[i] != t:
                break
        else:
            results.append(l)
    return results

For your example:

listx = [["a", 127, "Blue", 0],
         ["b", 127, "Red", 1],
         ["b", 127, "Green", 0],
         ["b", 99, "Green", 1],
         ["c", 99, "Yellow", 0]]

print filterls(listx, {0: 'b', 1: 127})
# [['b', 127, 'Red', 1], ['b', 127, 'Green', 0]]

edited Jun 28, 2014 at 14:20

answered Jun 28, 2014 at 13:55

arvindch

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2 Comments

kadir_cakir Over a year ago

I think this is only for Python 2.7. I got this with Python 3.4: AttributeError: 'dict' object has no attribute 'viewitems'

arvindch Over a year ago

@user3051668 - Yes, in python 3.4, viewitems is just items. Edited.

sundar nataraj · Accepted Answer · 2014-06-28 13:57:51Z

1

listx = [["a", 127, "Blue", 0],
         ["b", 127, "Red", 1],
         ["b", 127, "Green", 0],
         ["b", 99, "Green", 1],
         ["c", 99, "Yellow", 0]]

listnew = filter(lambda x: x[0]=='b' and x[1]==127,listx)

print listnew

try using Filter

answered Jun 28, 2014 at 13:57

sundar nataraj

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4 Comments

kadir_cakir Over a year ago

print listnew gives me <filter object at 0x0345DCF0>, am i missing something?

sundar nataraj Over a year ago

@user3051668 no it prints filtered list

thefourtheye Over a year ago

@sundarnatarajСундар OP is using Python 3.x, I think.

dawg Over a year ago

@user3051668: just put list(listnew) to see what filter is producing under Python 3.x. +1 -- this is the best way IMHO

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