CODEWITHSUNDEEP
pythonstringpython-2.7listdictionary
lane
lane

Reputation: 679

How to loop through a dictionary to remove values that match a string?

I want to remove values from a dictionary if they contain a particular string and consequentially remove any keys that have an empty list as value.

mydict = {
    'Getting links from: https://www.foo.com/': 
    [
        '+---OK--- http://www.this.com/',
        '+---OK--- http://www.is.com/',
        '+-BROKEN- http://www.broken.com/',
        '+---OK--- http://www.set.com/',
        '+---OK--- http://www.one.com/'
    ],
    'Getting links from: https://www.bar.com/': 
    [
        '+---OK--- http://www.this.com/',
        '+---OK--- http://www.is.com/',
        '+-BROKEN- http://www.broken.com/'
    ],
    'Getting links from: https://www.boo.com/':
    [
        '+---OK--- http://www.this.com/',
        '+---OK--- http://www.is.com/'
    ]
}

val = "is"

for k, v in mydict.iteritems():
   if v.contains(val):
     del mydict[v]

The result I want is:

{
    'Getting links from: https://www.foo.com/':
    [
        '+-BROKEN- http://www.broken.com/',
        '+---OK--- http://www.set.com/',
        '+---OK--- http://www.one.com/'
    ], 
    'Getting links from: https://www.bar.com/': 
    [
        '+-BROKEN- http://www.broken.com/'
    ]
}

How can I remove all dictionary values that contain a string, and then any keys that have no values as a result?

Upvotes: 4

Views: 2383

Answers (5)

j-i-l
j-i-l

Reputation: 10957

This is a one-liner:

{k: [e for e in v if val not in e] for k, v in mydict.items() if any([val not in e for e in v])}

The expected output:

Out[1]: {
    'Getting links from: https://www.bar.com/': 
    [
        '+-BROKEN- http://www.broken.com/'
    ],
    'Getting links from: https://www.foo.com/': 
    [
        '+-BROKEN- http://www.broken.com/',
        '+---OK--- http://www.set.com/',
        '+---OK--- http://www.one.com/'
    ]
}

Upvotes: 3

Brandon Bailey
Brandon Bailey

Reputation: 811

There are a couple of ways you could do it. One using regex and one without.

if you're not familiar with regex you could try this:

for key, value in mydict.items():
    if val in value:
        mydict.pop(key)

output would be:

    mydict = {'Getting links from: https://www.bar.com/': ['+---OK--- http://www.this.com/',
  '+---OK--- http://www.is.com/',
  '+-BROKEN- http://www.broken.com/'],
 'Getting links from: https://www.boo.com/': ['+---OK--- http://www.this.com/',
  '+---OK--- http://www.is.com/'],
 'Getting links from: https://www.foo.com/': ['+---OK--- http://www.this.com/',
  '+---OK--- http://www.is.com/',
  '+-BROKEN- http://www.broken.com/',
  '+---OK--- http://www.set.com/',
  '+---OK--- http://www.one.com/']}

Upvotes: 0

RomanPerekhrest
RomanPerekhrest

Reputation: 92854

With simple loop:

val = "is"

new_dict = dict()
for k, v in mydict.items():
    values = [i for i in v if val not in i]
    if values: new_dict[k] = values

print(new_dict)

The output:

{'Getting links from: https://www.foo.com/': ['+-BROKEN- http://www.broken.com/', '+---OK--- http://www.set.com/', '+---OK--- http://www.one.com/'], 'Getting links from: https://www.bar.com/': ['+-BROKEN- http://www.broken.com/']}

Upvotes: 4

jpp
jpp

Reputation: 164643

You can use a list comprehension within a dictionary comprehension. You shouldn't change the number of items in a dictionary while you iterate that dictionary.

res = {k: [x for x in v if 'is' not in x] for k, v in mydict.items()}

# {'Getting links from: https://www.foo.com/': ['+-BROKEN- http://www.broken.com/',
#                                               '+---OK--- http://www.set.com/',
#                                               '+---OK--- http://www.one.com/'],
#  'Getting links from: https://www.bar.com/': ['+-BROKEN- http://www.broken.com/'],
#  'Getting links from: https://www.boo.com/': []}

If you wish to remove items with empty list values, you can in a subsequent step:

res = {k: v for k, v in res.items() if v}

Upvotes: 6

YOLO
YOLO

Reputation: 21709

Using dict comprehension, you can try the following:

import re

val = 'is'

# step 1 - remove line having is
mydict = {k:[re.sub(r'.*is*.', '', x) for x in v] for k,v in mydict.items()}

# filtering out keys if there is no value - if needed
mydict = {k:v for k,v in mydict.items() if len(v) > 0}

print(mydict)

{'Getting links from: https://www.foo.com/': ['com/',
  'com/',
  '+-BROKEN- http://www.broken.com/',
  '+---OK--- http://www.set.com/',
  '+---OK--- http://www.one.com/'],
 'Getting links from: https://www.bar.com/': ['com/',
  'com/',
  '+-BROKEN- http://www.broken.com/'],
 'Getting links from: https://www.boo.com/': ['com/', 'com/']}

Upvotes: 0

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