Check if a list of strings contains letters from another list
So I have two lists:
vocabulary = ['a','b','c']
sentences = ['a a b b c c', 'a c b c', 'b c c a b']
I want to calculate how many times the letters in vocabulary appear in the strings in the list sentences.
So I want the output to be:
a = 4
b = 5
c = 6
My program:
counter = Counter()
for word in sentences:
if word in vocabulary:
counter.update(word)
print(counter)
But I keep getting the output:
Counter()
python string python-3.x dictionary counter
edited Nov 13 '18 at 13:31
jpp
97.4k2159109
asked Nov 13 '18 at 13:24
JameshGongJameshGong
486
add a comment |
So I have two lists:
vocabulary = ['a','b','c']
sentences = ['a a b b c c', 'a c b c', 'b c c a b']
I want to calculate how many times the letters in vocabulary appear in the strings in the list sentences.
So I want the output to be:
a = 4
b = 5
c = 6
My program:
counter = Counter()
for word in sentences:
if word in vocabulary:
counter.update(word)
print(counter)
But I keep getting the output:
Counter()
python string python-3.x dictionary counter
edited Nov 13 '18 at 13:31
jpp
97.4k2159109
asked Nov 13 '18 at 13:24
JameshGongJameshGong
486
add a comment |
So I have two lists:
vocabulary = ['a','b','c']
sentences = ['a a b b c c', 'a c b c', 'b c c a b']
I want to calculate how many times the letters in vocabulary appear in the strings in the list sentences.
So I want the output to be:
a = 4
b = 5
c = 6
My program:
counter = Counter()
for word in sentences:
if word in vocabulary:
counter.update(word)
print(counter)
But I keep getting the output:
Counter()
python string python-3.x dictionary counter
edited Nov 13 '18 at 13:31
jpp
97.4k2159109
asked Nov 13 '18 at 13:24
JameshGongJameshGong
486
So I have two lists:
vocabulary = ['a','b','c']
sentences = ['a a b b c c', 'a c b c', 'b c c a b']
I want to calculate how many times the letters in vocabulary appear in the strings in the list sentences.
So I want the output to be:
a = 4
b = 5
c = 6
My program:
counter = Counter()
for word in sentences:
if word in vocabulary:
counter.update(word)
print(counter)
But I keep getting the output:
Counter()
python string python-3.x dictionary counter
python string python-3.x dictionary counter
edited Nov 13 '18 at 13:31
jpp
97.4k2159109
asked Nov 13 '18 at 13:24
JameshGongJameshGong
486
edited Nov 13 '18 at 13:31
jpp
97.4k2159109
asked Nov 13 '18 at 13:24
JameshGongJameshGong
486
edited Nov 13 '18 at 13:31
jpp
97.4k2159109
edited Nov 13 '18 at 13:31
jpp
97.4k2159109
edited Nov 13 '18 at 13:31
jpp
97.4k2159109
97.4k2159109
asked Nov 13 '18 at 13:24
JameshGongJameshGong
486
asked Nov 13 '18 at 13:24
JameshGongJameshGong
486
asked Nov 13 '18 at 13:24
JameshGongJameshGong
486
486
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
Counter is a subclass of dict. dict.update accepts another dictionary or an iterable of pairs. But you're only supplying a single character.
In this case, you can chain your list of strings and pass to Counter, then filter the result via a dictionary comprehension:
from collections import Counter
from itertools import chain
vocabulary = ['a','b','c']
sentences = ['a a b b c c', 'a c b c', 'b c c a b']
vocab_set = set(vocabulary)
c = Counter(chain.from_iterable(sentences))
res = {k: v for k, v in c.items() if k in vocab_set}
{'a': 4, 'b': 5, 'c': 6}
answered Nov 13 '18 at 13:30
jppjpp
97.4k2159109
add a comment |
This will do it, no import needed:
vocabulary = ['a','b','c']
sentences = ['a a b b c c', 'a c b c', 'b c c a b']
data = ''.join(sentences)
for v in vocabulary:
print('{}: {}'.format(v, data.count(v)))
a: 4
b: 5
c: 6
answered Nov 13 '18 at 13:30
zipazipa
16k31637
Worth noting this is at the cost of time complexity O(m * n) vs O(n).
– jpp
Nov 13 '18 at 13:33
add a comment |
An O(n) solution, with no import:
vocabulary = ['a', 'b', 'c']
sentences = ['a a b b c c', 'a c b c', 'b c c a b']
counts = {}
vocab_set = set(vocabulary)
for sentence in sentences:
for ch in sentence:
if ch in vocab_set:
counts[ch] = counts.get(ch, 0) + 1
print(counts)
Output
{'c': 6, 'a': 4, 'b': 5}
answered Nov 13 '18 at 13:43
Daniel MesejoDaniel Mesejo
16.5k21430
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
Counter is a subclass of dict. dict.update accepts another dictionary or an iterable of pairs. But you're only supplying a single character.
In this case, you can chain your list of strings and pass to Counter, then filter the result via a dictionary comprehension:
from collections import Counter
from itertools import chain
vocabulary = ['a','b','c']
sentences = ['a a b b c c', 'a c b c', 'b c c a b']
vocab_set = set(vocabulary)
c = Counter(chain.from_iterable(sentences))
res = {k: v for k, v in c.items() if k in vocab_set}
{'a': 4, 'b': 5, 'c': 6}
answered Nov 13 '18 at 13:30
jppjpp
97.4k2159109
add a comment |
Counter is a subclass of dict. dict.update accepts another dictionary or an iterable of pairs. But you're only supplying a single character.
In this case, you can chain your list of strings and pass to Counter, then filter the result via a dictionary comprehension:
from collections import Counter
from itertools import chain
vocabulary = ['a','b','c']
sentences = ['a a b b c c', 'a c b c', 'b c c a b']
vocab_set = set(vocabulary)
c = Counter(chain.from_iterable(sentences))
res = {k: v for k, v in c.items() if k in vocab_set}
{'a': 4, 'b': 5, 'c': 6}
answered Nov 13 '18 at 13:30
jppjpp
97.4k2159109
add a comment |
Counter is a subclass of dict. dict.update accepts another dictionary or an iterable of pairs. But you're only supplying a single character.
In this case, you can chain your list of strings and pass to Counter, then filter the result via a dictionary comprehension:
from collections import Counter
from itertools import chain
vocabulary = ['a','b','c']
sentences = ['a a b b c c', 'a c b c', 'b c c a b']
vocab_set = set(vocabulary)
c = Counter(chain.from_iterable(sentences))
res = {k: v for k, v in c.items() if k in vocab_set}
{'a': 4, 'b': 5, 'c': 6}
answered Nov 13 '18 at 13:30
jppjpp
97.4k2159109
Counter is a subclass of dict. dict.update accepts another dictionary or an iterable of pairs. But you're only supplying a single character.
In this case, you can chain your list of strings and pass to Counter, then filter the result via a dictionary comprehension:
from collections import Counter
from itertools import chain
vocabulary = ['a','b','c']
sentences = ['a a b b c c', 'a c b c', 'b c c a b']
vocab_set = set(vocabulary)
c = Counter(chain.from_iterable(sentences))
res = {k: v for k, v in c.items() if k in vocab_set}
{'a': 4, 'b': 5, 'c': 6}
answered Nov 13 '18 at 13:30
jppjpp
97.4k2159109
answered Nov 13 '18 at 13:30
jppjpp
97.4k2159109
answered Nov 13 '18 at 13:30
jppjpp
97.4k2159109
answered Nov 13 '18 at 13:30
jppjpp
97.4k2159109
97.4k2159109
add a comment |
add a comment |
This will do it, no import needed:
vocabulary = ['a','b','c']
sentences = ['a a b b c c', 'a c b c', 'b c c a b']
data = ''.join(sentences)
for v in vocabulary:
print('{}: {}'.format(v, data.count(v)))
a: 4
b: 5
c: 6
answered Nov 13 '18 at 13:30
zipazipa
16k31637
Worth noting this is at the cost of time complexity O(m * n) vs O(n).
– jpp
Nov 13 '18 at 13:33
add a comment |
This will do it, no import needed:
vocabulary = ['a','b','c']
sentences = ['a a b b c c', 'a c b c', 'b c c a b']
data = ''.join(sentences)
for v in vocabulary:
print('{}: {}'.format(v, data.count(v)))
a: 4
b: 5
c: 6
answered Nov 13 '18 at 13:30
zipazipa
16k31637
Worth noting this is at the cost of time complexity O(m * n) vs O(n).
– jpp
Nov 13 '18 at 13:33
add a comment |
This will do it, no import needed:
vocabulary = ['a','b','c']
sentences = ['a a b b c c', 'a c b c', 'b c c a b']
data = ''.join(sentences)
for v in vocabulary:
print('{}: {}'.format(v, data.count(v)))
a: 4
b: 5
c: 6
answered Nov 13 '18 at 13:30
zipazipa
16k31637
This will do it, no import needed:
vocabulary = ['a','b','c']
sentences = ['a a b b c c', 'a c b c', 'b c c a b']
data = ''.join(sentences)
for v in vocabulary:
print('{}: {}'.format(v, data.count(v)))
a: 4
b: 5
c: 6
answered Nov 13 '18 at 13:30
zipazipa
16k31637
answered Nov 13 '18 at 13:30
zipazipa
16k31637
answered Nov 13 '18 at 13:30
zipazipa
16k31637
answered Nov 13 '18 at 13:30
zipazipa
16k31637
16k31637
Worth noting this is at the cost of time complexity O(m * n) vs O(n).
– jpp
Nov 13 '18 at 13:33
add a comment |
Worth noting this is at the cost of time complexity O(m * n) vs O(n).
– jpp
Nov 13 '18 at 13:33
Worth noting this is at the cost of time complexity O(m * n) vs O(n).
– jpp
Nov 13 '18 at 13:33
Worth noting this is at the cost of time complexity O(m * n) vs O(n).
– jpp
Nov 13 '18 at 13:33
add a comment |
An O(n) solution, with no import:
vocabulary = ['a', 'b', 'c']
sentences = ['a a b b c c', 'a c b c', 'b c c a b']
counts = {}
vocab_set = set(vocabulary)
for sentence in sentences:
for ch in sentence:
if ch in vocab_set:
counts[ch] = counts.get(ch, 0) + 1
print(counts)
Output
{'c': 6, 'a': 4, 'b': 5}
answered Nov 13 '18 at 13:43
Daniel MesejoDaniel Mesejo
16.5k21430
add a comment |
An O(n) solution, with no import:
vocabulary = ['a', 'b', 'c']
sentences = ['a a b b c c', 'a c b c', 'b c c a b']
counts = {}
vocab_set = set(vocabulary)
for sentence in sentences:
for ch in sentence:
if ch in vocab_set:
counts[ch] = counts.get(ch, 0) + 1
print(counts)
Output
{'c': 6, 'a': 4, 'b': 5}
answered Nov 13 '18 at 13:43
Daniel MesejoDaniel Mesejo
16.5k21430
add a comment |
An O(n) solution, with no import:
vocabulary = ['a', 'b', 'c']
sentences = ['a a b b c c', 'a c b c', 'b c c a b']
counts = {}
vocab_set = set(vocabulary)
for sentence in sentences:
for ch in sentence:
if ch in vocab_set:
counts[ch] = counts.get(ch, 0) + 1
print(counts)
Output
{'c': 6, 'a': 4, 'b': 5}
answered Nov 13 '18 at 13:43
Daniel MesejoDaniel Mesejo
16.5k21430
An O(n) solution, with no import:
vocabulary = ['a', 'b', 'c']
sentences = ['a a b b c c', 'a c b c', 'b c c a b']
counts = {}
vocab_set = set(vocabulary)
for sentence in sentences:
for ch in sentence:
if ch in vocab_set:
counts[ch] = counts.get(ch, 0) + 1
print(counts)
Output
{'c': 6, 'a': 4, 'b': 5}
answered Nov 13 '18 at 13:43
Daniel MesejoDaniel Mesejo
16.5k21430
edited Nov 13 '18 at 13:52
answered Nov 13 '18 at 13:43
Daniel MesejoDaniel Mesejo
16.5k21430
answered Nov 13 '18 at 13:43
Daniel MesejoDaniel Mesejo
16.5k21430
answered Nov 13 '18 at 13:43
Daniel MesejoDaniel Mesejo
16.5k21430
16.5k21430
add a comment |
add a comment |
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