我有一个如下所示的培训数据，其中所有信息都在一列中。数据集有超过300000个数据。
id features label
1 name=John Matthew;age=25;1.=Post Graduate;2.=Football Player; 1
2 name=Mark clark;age=21;1.=Under Graduate;Interest=Video Games; 1
3 name=David;age=12;1:=High School;2:=Cricketer;native=america; 2
4 name=George;age=11;1:=High School;2:=Carpenter;married=yes 2
.
.
300000 name=Kevin;age=16;1:=High School;2:=Driver;Smoker=No 3
现在我需要转换此培训数据，如下所示
id name age 1 2 Interest married Smoker
1 John Matthew 25 Post Graduate Football Player Nan Nan Nan
2 Mark clark 21 Under Graduate Nan Video Games Nan Nan
.
.
有没有有效的方法来做到这一点。我尝试了下面的代码但是花了3个小时才完成
#Getting the proper features from the features column
cols = {}
for choices in set_label:
collection_list = []
array = train["features"][train["label"] == choices].values
for i in range(1,len(array)):
var_split = array[i].split(";")
try :
d = (dict(s.split('=') for s in var_split))
for x in d.keys():
collection_list.append(x)
except ValueError:
Error = ValueError
count = Counter(collection_list)
for k , v in count.most_common(5):
key = k.replace(":","").replace(" ","_").lower()
cols[key] = v
columns_add = list(cols.keys())
train = train.reindex(columns = np.append( train.columns.values, columns_add))
print (train.columns)
print (train.shape)
#Adding the values for the newly created problem
for row in train.itertuples():
dummy_dic = {}
new_dict={}
value = train.loc[row.Index, 'features']
v_split = value.split(";")
try :
dummy_dict = (dict(s.split('=') for s in v_split))
for k, v in dummy_dict.items():
new_key = k.replace(":","").replace(" ","_").lower()
new_dict[new_key] = v
except ValueError:
Error = ValueError
for k,v in new_dict.items():
if k in train.columns:
train.loc[row.Index, k] = v
是否有任何有用的功能可以在这里应用于有效的特征提取方法？
答案 0 :(得分：1)
创建两个DataFrames（在第一个中，所有功能对于每个数据点都是相同的，第二个是对第一个引入不同功能的某个数据点的修改），符合您的标准：
import pandas as pd
import numpy as np
import random
import time
import itertools
# Create a DataFrame where all the keys for each datapoint in the "features" column are the same.
num = 300000
NAMES = ['John', 'Mark', 'David', 'George', 'Kevin']
AGES = [25, 21, 12, 11, 16]
FEATURES1 = ['Post Graduate', 'Under Graduate', 'High School']
FEATURES2 = ['Football Player', 'Cricketer', 'Carpenter', 'Driver']
LABELS = [1, 2, 3]
df = pd.DataFrame()
df.loc[:num, 0]= ["name={0};age={1};feature1={2};feature2={3}"\
.format(NAMES[np.random.randint(0, len(NAMES))],\
AGES[np.random.randint(0, len(AGES))],\
FEATURES1[np.random.randint(0, len(FEATURES1))],\
FEATURES2[np.random.randint(0, len(FEATURES2))]) for i in xrange(num)]
df['label'] = [LABELS[np.random.randint(0, len(LABELS))] for i in range(num)]
df.rename(columns={0:"features"}, inplace=True)
print df.head(20)
# Create a modified sample DataFrame from the previous one, where not all the keys are the same for each data point.
mod_df = df
random_positions1 = random.sample(xrange(10), 5)
random_positions2 = random.sample(xrange(11, 20), 5)
INTERESTS = ['Basketball', 'Golf', 'Rugby']
SMOKING = ['Yes', 'No']
mod_df.loc[random_positions1, 'features'] = ["name={0};age={1};interest={2}"\
.format(NAMES[np.random.randint(0, len(NAMES))],\
AGES[np.random.randint(0, len(AGES))],\
INTERESTS[np.random.randint(0, len(INTERESTS))]) for i in xrange(len(random_positions1))]
mod_df.loc[random_positions2, 'features'] = ["name={0};age={1};smoking={2}"\
.format(NAMES[np.random.randint(0, len(NAMES))],\
AGES[np.random.randint(0, len(AGES))],\
SMOKING[np.random.randint(0, len(SMOKING))]) for i in xrange(len(random_positions2))]
print mod_df.head(20)
假设您的原始数据存储在名为df的数据文件中。
def func2(y):
lista = y.split('=')
value = lista[1]
return value
def function(x):
lista = x.split(';')
array = [func2(i) for i in lista]
return array
# Calculate the execution time
start = time.time()
array = pd.Series(df.features.apply(function)).tolist()
new_df = df.from_records(array, columns=['name', 'age', '1', '2'])
end = time.time()
new_df
print 'Total time:', end - start
Total time: 1.80923295021
编辑：您需要做的一件事是相应地编辑columns列表。
import pandas as pd
import numpy as np
import time
import itertools
# The following functions are meant to extract the keys from each row, which are going to be used as columns.
def extract_key(x):
return x.split('=')[0]
def def_columns(x):
lista = x.split(';')
keys = [extract_key(i) for i in lista]
return keys
df = mod_df
columns = pd.Series(df.features.apply(def_columns)).tolist()
flattened_columns = list(itertools.chain(*columns))
flattened_columns = np.unique(np.array(flattened_columns)).tolist()
flattened_columns
# This function turns each row from the original dataframe into a dictionary.
def function(x):
lista = x.split(';')
dict_ = {}
for i in lista:
key, val = i.split('=')
dict_[key ] = val
return dict_
df.features.apply(function)
arr = pd.Series(df.features.apply(function)).tolist()
pd.DataFrame.from_dict(arr)
答案 1 :(得分：0)
假设您的数据是这样的：
features= ["name=John Matthew;age=25;1:=Post Graduate;2:=Football Player;",
'name=Mark clark;age=21;1:=Under Graduate;2:=Football Player;',
"name=David;age=12;1:=High School;2:=Cricketer;",
"name=George;age=11;1:=High School;2:=Carpenter;",
'name=Kevin;age=16;1:=High School;2:=Driver; ']
df = pd.DataFrame({'features': features})
我将开始 by this 回答并尝试将所有分隔符（name，age，1：=，2：=）替换为;
使用此功能
def replace_feature(x):
for r in (("name=", ";"), (";age=", ";"), (';1:=', ';'), (';2:=', ";")):
x = x.replace(*r)
x = x.split(';')
return x
df = df.assign(features= df.features.apply(replace_feature))
将该功能应用到您的df后，所有值都将显示功能列表。你可以通过索引获得每一个 然后我使用4个海关功能来获取每个属性名称，年龄，等级;工作， 注意：只使用一个函数
可以有更好的方法def get_name(df):
return df['features'][1]
def get_age(df):
return df['features'][2]
def get_grade(df):
return df['features'][3]
def get_job(df):
return df['features'][4]
最终将该功能应用于您的数据框：
df = df.assign(name = df.apply(get_name, axis=1),
age = df.apply(get_age, axis=1),
grade = df.apply(get_grade, axis=1),
job = df.apply(get_job, axis=1))
希望这会快速而快速
答案 2 :(得分：0)
据我了解您的代码，糟糕的表现来自于您按元素创建dataframe元素的事实。最好用一个字典列表一次创建整个数据帧。
让我们重新创建输入数据框：
from StringIO import StringIO
data=StringIO("""id features label
1 name=John Matthew;age=25;1.=Post Graduate;2.=Football Player; 1
2 name=Mark clark;age=21;1.=Under Graduate;2.=Football Player; 1
3 name=David;age=12;1:=High School;2:=Cricketer; 2
4 name=George;age=11;1:=High School;2:=Carpenter; 2""")
df=pd.read_table(data,sep=r'\s{3,}',engine='python')
我们可以查看：
print df
id features label
0 1 name=John Matthew;age=25;1.=Post Graduate;2.=F... 1
1 2 name=Mark clark;age=21;1.=Under Graduate;2.=Fo... 1
2 3 name=David;age=12;1:=High School;2:=Cricketer; 2
3 4 name=George;age=11;1:=High School;2:=Carpenter; 2
现在我们可以使用以下代码创建所需的字典列表：
feat=[]
for line in df['features']:
line=line.replace(':','.')
lsp=line.split(';')[:-1]
feat.append(dict([elt.split('=') for elt in lsp]))
结果数据框：
print pd.DataFrame(feat)
1. 2. age name
0 Post Graduate Football Player 25 John Matthew
1 Under Graduate Football Player 21 Mark clark
2 High School Cricketer 12 David
3 High School Carpenter 11 George