Question

我试图根据感兴趣的单词子集计算单词列中单词存在的次数。

首先我导入我的数据

products = graphlab.SFrame('amazon_baby.gl/')
products['word_count'] = graphlab.text_analytics.count_words(products['review'])
products.head(5)

可以在此处找到数据：https://drive.google.com/open?id=0BzbhZp-qIglxM3VSVWRsVFRhTWc

然后我创建了我感兴趣的单词列表：

words = ['awesome', 'great', 'fantastic']

我想计算产品['word_count']中“单词”中每个单词出现的次数。

我没有使用graphlab结婚。这只是一位同事向我建议的。

Answer 1

好吧，我不太确定'在词典栏中'你的意思。如果是列表：

import operator
dictionary={'texts':['red blue blue','red black','blue white white','red','white','black','blue red']}
words=['red','white','blue']
freqs=dict()
for t in dictionary['texts']:
    for w in words:
        try:
             freqs[w]+=t.count(w)
        except:
            freqs[w]=t.count(w)
top_words = sorted(freqs.items(), key=operator.itemgetter(1),reverse=True)

如果只是一个文字：

import operator
dictionary={'text':'red blue blue red black blue white white red white black blue red'}
words=['red','white','blue']
freqs=dict()
for w in words:
    try:
        freqs[w]+=dictionary['text'].count(w)
    except:
        freqs[w]=dictionary['text'].count(w)
top_words = sorted(freqs.items(), key=operator.itemgetter(1),reverse=True)

Answer 2

如果您想计算单词的出现次数，快速执行此操作的方法是使用collections中的In [3]: from collections import Counter In [4]: c = Counter(['hello', 'world']) In [5]: c Out[5]: Counter({'hello': 1, 'world': 1})对象

例如：

products.head(5)

您能否显示public class Program { public static void Main() { var udp = new Udp("255.255.255.255", 1337); Task.Run(() => { while (true) { Console.WriteLine(udp.Receive()); } }); Task.Run(() => { while (true) { Thread.Sleep(1000); udp.Send("(((1)))"); } }); Console.ReadLine(); } } public class Udp { private readonly UdpClient _sender; private readonly UdpClient _listener; public Udp(string address, int port) { _sender = new UdpClient(address, port); _listener = new UdpClient(); _listener.Client.SetSocketOption(SocketOptionLevel.Socket, SocketOptionName.ReuseAddress, true); _listener.Client.Bind(new IPEndPoint(IPAddress.Any, port)); } public string Receive() { var _ = null as IPEndPoint; return $"{Encoding.Default.GetString(_listener.Receive(ref _))} from {_.Address}:{_.Port}"; } public void Send(string message) { var dataAsBytes = Encoding.ASCII.GetBytes(message); _sender.Send(dataAsBytes, dataAsBytes.Length); } }命令的输出？

Answer 3

如果坚持使用graphlab（或SFrame），请使用SArray.dict_trim_by_keys方法。文档在这里：https://dato.com/products/create/docs/generated/graphlab.SArray.dict_trim_by_keys.html

import graphlab as gl
sf = gl.SFrame({'review': ['what a good book', 'terrible book']})
sf['word_bag'] = gl.text_analytics.count_words(sf['review'])

keywords = ['good', 'book']
sf['key_words'] = sf['word_bag'].dict_trim_by_keys(keywords, exclude=False)
print sf

+------------------+---------------------+---------------------+
|      review      |       word_bag      |      key_words      |
+------------------+---------------------+---------------------+
| what a good book | {'a': 1, 'good':... | {'good': 1, 'boo... |
|  terrible book   | {'book': 1, 'ter... |     {'book': 1}     |
+------------------+---------------------+---------------------+ 
[2 rows x 3 columns]

Answer 4

是否要将每个计数放在单独的列中？在这种情况下，这可能会起作用：

keywords = ['keyword1' , 'keyword2']

def word_counter(dict_cell , word):
if word in dict_cell:
    return dict_cell[word]
else:
    return 0

for words in keywords:
  df[words] = df['word_count'].apply(lambda x:word_counter(x,words))

Answer 5

def count_words(x, w):
    if w in x:
        return x.count(w)
    else:
        return 0   

selected_words = ['awesome', 'great', 'fantastic', 'amazing', 'love', 'horrible', 'bad', 'terrible', 'awful', 'wow', 'hate']

for words in selected_words:
    products[words]=products['review'].apply(lambda x:count_words(x,words))

Python：来自另一个列表的dict中的出现次数

5 个答案: