客户需求+查看销售人员的非空行为+数据存储情况如下:+代码实现+import pandas as pddata=pd.read_excel('test.xlsx', sheet_name='Sheet1')data_not_na=data[data['销售人员'].notna()]print(data_not_na)+输出结果+D:
客户需求
查看销售人员不为空值的行
数据存储情况如图:
代码实现
import pandas as pd
data = pd.read_excel('test.xlsx',sheet_name='Sheet1')
datanota = data[data['销售人员'].notna()]
print(datanota)
import pandas as pd
data = pd.read_csv('my_data.csv', sep=',')
data.head()
它的输出如下:
id city department sms category
01 khi revenue NaN 0
02 lhr revenue good 1
03 lhr revenue NaN 0
我想删除sms列为空/ NaN的所有行.什么是有效的方法呢?
解决方法:
将dropna与参数子集一起使用以指定用于检查NaN的列:
data = data.dropna(subset=['sms'])
print (data)
id city department sms category
1 2 lhr revenue good 1
boolean indexing和notnull的另一个解决方案:
data = data[data['sms'].notnull()]
print (data)
id city department sms category
1 2 lhr revenue good 1
替代query:
print (data.query("sms == sms"))
id city department sms category
1 2 lhr revenue good 1
计时
#[300000 rows x 5 columns]
data = pd.concat([data]*100000).reset_index(drop=True)
In [123]: %timeit (data.dropna(subset=['sms']))
100 loops, best of 3: 19.5 ms per loop
In [124]: %timeit (data[data['sms'].notnull()])
100 loops, best of 3: 13.8 ms per loop
In [125]: %timeit (data.query("sms == sms"))
10 loops, best of 3: 23.6 ms per loop
客户需求+查看销售人员的非空行为+数据存储情况如下:+代码实现+import pandas as pddata=pd.read_excel('test.xlsx', sheet_name='Sheet1')data_not_na=data[data['销售人员'].notna()]print(data_not_na)+输出结果+D:
客户需求
查看销售人员不为空值的行
数据存储情况如图:
代码实现
import pandas as pd
data = pd.read_excel('test.xlsx',sheet_name='Sheet1')
datanota = data[data['销售人员'].notna()]
print(datanota)
import pandas as pd
data = pd.read_csv('my_data.csv', sep=',')
data.head()
它的输出如下:
id city department sms category
01 khi revenue NaN 0
02 lhr revenue good 1
03 lhr revenue NaN 0
我想删除sms列为空/ NaN的所有行.什么是有效的方法呢?
解决方法:
将dropna与参数子集一起使用以指定用于检查NaN的列:
data = data.dropna(subset=['sms'])
print (data)
id city department sms category
1 2 lhr revenue good 1
boolean indexing和notnull的另一个解决方案:
data = data[data['sms'].notnull()]
print (data)
id city department sms category
1 2 lhr revenue good 1
替代query:
print (data.query("sms == sms"))
id city department sms category
1 2 lhr revenue good 1
计时
#[300000 rows x 5 columns]
data = pd.concat([data]*100000).reset_index(drop=True)
In [123]: %timeit (data.dropna(subset=['sms']))
100 loops, best of 3: 19.5 ms per loop
In [124]: %timeit (data[data['sms'].notnull()])
100 loops, best of 3: 13.8 ms per loop
In [125]: %timeit (data.query("sms == sms"))
10 loops, best of 3: 23.6 ms per loop