假如我们要构建新特征b
目的是从a中筛选出数值在4~6之间的数据,如果符合就是True,否则就是False。
那么代码如下
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import pandas as pd
lists = pd.DataFrame({ 'a' :[ 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 ]})
lists[ 'b' ] = (lists[ 'a' ]< 6 ).mul(lists[ 'a' ]> 4 )
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补充:dataframe求两列的相乘,再将输出为新的一列
看代码吧~
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df[ "new" ] = df3[ "rate" ] * df3[ "duration" ]
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new为新的一列的列名
rate和duration为需要相乘的列
加,减,乘,除都适用!
补充:DataFrame衍生新特征操作
1.DataFrame中某一列的值衍生为新的特征
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#将LBL1特征的值衍生为one-hot形式的新特征
piao = df_train_log.LBL1.value_counts().index
#先构造一个临时的df
df_tmp = pd.DataFrame({ 'USRID' :df_train_log.drop_duplicates( 'USRID' ).USRID.values})
#将所有的新特征列都置为0
for i in piao:
df_tmp[ 'PIAO_' + i] = 0
#进行分组便利,有这个特征就置为1,原数据每个USRID有多条记录,所以分组统计
group = df_train_log.groupby([ 'USRID' ])
for k in group.groups.keys():
t = group.get_group(k)
id = t.USRID.value_counts().index[ 0 ]
tmp_list = t.LBL1.value_counts().index
for j in tmp_list:
df_tmp[ 'PIAO_' + j].loc[df_tmp.USRID = = id ] = 1
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2.分组统计,选出同一USRID下该变量中出现次数最多的值项
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group = df_train_log.groupby([ 'USRID' ])
lt = []
list_max_lbl1 = []
list_max_lbl2 = []
list_max_lbl3 = []
for k in group.groups.keys():
t = group.get_group(k)
#通过value_counts找出出现次数最多的项
argmx = np.argmax(t[ 'EVT_LBL' ].value_counts())
lbl1_max = np.argmax(t[ 'LBL1' ].value_counts())
lbl2_max = np.argmax(t[ 'LBL2' ].value_counts())
lbl3_max = np.argmax(t[ 'LBL3' ].value_counts())
list_max_lbl1.append(lbl1_max)
list_max_lbl2.append(lbl2_max)
list_max_lbl3.append(lbl3_max)
#只留下出现次数最多的项
c = t[t[ 'EVT_LBL' ] = = argmx].drop_duplicates( 'EVT_LBL' )
#放入list中
lt.append(c)
#构造一个新的df
df_train_log_new = pd.concat(lt)
#另外又构造了三个特征,LBL1-LBL3分别出现次数最多的项
df_train_log_new[ 'LBL1_MAX' ] = list_max_lbl1
df_train_log_new[ 'LBL2_MAX' ] = list_max_lbl2
df_train_log_new[ 'LBL3_MAX' ] = list_max_lbl3
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3.衍生出某天是否发生的ont-hot新特征
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#创造临时df,星期三,星期六,星期七,都默认置为0
df_day = pd.DataFrame({ 'USRID' :df_train_log.drop_duplicates( 'USRID' ).USRID.values})
df_day[ 'weekday_3' ] = 0
df_day[ 'weekday_6' ] = 0
df_day[ 'weekday_7' ] = 0
#分组统计,有就置为1,没有置为0
group = df_train_log.groupby([ 'USRID' ])
for k in group.groups.keys():
t = group.get_group(k)
id = t.USRID.value_counts().index[ 0 ]
tmp_list = t.occ_dayofweek.value_counts().index
for j in tmp_list:
if j = = 3 :
df_day[ 'weekday_3' ].loc[df_tmp.USRID = = id ] = 1
elif j = = 6 :
df_day[ 'weekday_6' ].loc[df_tmp.USRID = = id ] = 1
elif j = = 7 :
df_day[ 'weekday_7' ].loc[df_tmp.USRID = = id ] = 1
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4.查看用户一共停留在APP上多少秒,共有几天看了APP
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#首先将日期转化为时间戳,并赋予一个新特征
tmp_list = []
for i in df_train_log.OCC_TIM:
d = datetime.datetime.strptime( str (i), "%Y-%m-%d %H:%M:%S" )
evt_time = time.mktime(d.timetuple())
tmp_list.append(evt_time)
df_train_log[ 'time' ] = tmp_list
#每下一行减去上一行,得到app停留时间
df_train_log[ 'diff_time' ] = df_train_log.time - df_train_log.time.shift( 1 )
#构造一个新的dataFrame,分组得到查看app的天数
df_time = pd.DataFrame({ 'USRID' :df_train_log.drop_duplicates( 'USRID' ).USRID.values})
#有几天查看
df_time[ 'days' ] = 0
group = df_train_log.groupby([ 'USRID' ])
for k in group.groups.keys():
t = group.get_group(k)
id = set (t.USRID).pop()
df_time[ 'days' ].loc[df_time.USRID = = id ] = len (t.occ_day.value_counts().index)
#去掉一些异常时间戳,比如间隔两天的相减,肯定不合适,na的也去掉了
df_train_log = df_train_log[(df_train_log.diff_time> 0 )&(df_train_log.diff_time< 8000 )]
#累计停留时间
group_stayTime = df_train_log[ 'diff_time' ].groupby(df_train_log[ 'USRID' ]). sum ()
#创造新的df
df_tmp = pd.DataFrame({ 'USRID' : list (group_stayTime.index.values), 'stay_time' : list (group_stayTime.values)})
#合并成一个新的df
df = pd.merge(df_time,df_tmp,on = [ 'USRID' ],how = 'left' ) #合并后,缺失的停留时间,置为0df.fillna(0,axis=1,inplace=True)
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以上为个人经验,希望能给大家一个参考,也希望大家多多支持服务器之家。
原文链接:https://yuchi.blog.csdn.net/article/details/101059922