标签：search df boston 数据库 CHAS 00.5736 SQL 筛选 pandas

我们都知道SQL数据库的语句很简洁，python中的pandas库也很好用，但是如何将两者给关联起来？？？

本文将通过pandas来实现类似于SQL中的【过滤】、【排序】、【关联】、【合并】、【更新】、【删除】等操作。

用到的数据集

from sklearn.datasets import load_boston
import pandas as pd

boston = load_boston()
df = pd.DataFrame(boston.data,columns = boston.feature_names)
df['target'] =pd.Series(boston.target) 

df

	CRIM	ZN	INDUS	CHAS	NOX	RM	AGE	DIS	RAD	TAX	PTRATIO	B	LSTAT	target
0	0.00632	18.0	2.31	0.0	0.538	6.575	65.2	4.0900	1.0	296.0	15.3	396.90	4.98	24.0
1	0.02731	0.0	7.07	0.0	0.469	6.421	78.9	4.9671	2.0	242.0	17.8	396.90	9.14	21.6
2	0.02729	0.0	7.07	0.0	0.469	7.185	61.1	4.9671	2.0	242.0	17.8	392.83	4.03	34.7
3	0.03237	0.0	2.18	0.0	0.458	6.998	45.8	6.0622	3.0	222.0	18.7	394.63	2.94	33.4
4	0.06905	0.0	2.18	0.0	0.458	7.147	54.2	6.0622	3.0	222.0	18.7	396.90	5.33	36.2
...	...	...	...	...	...	...	...	...	...	...	...	...	...	...
501	0.06263	0.0	11.93	0.0	0.573	6.593	69.1	2.4786	1.0	273.0	21.0	391.99	9.67	22.4
502	0.04527	0.0	11.93	0.0	0.573	6.120	76.7	2.2875	1.0	273.0	21.0	396.90	9.08	20.6
503	0.06076	0.0	11.93	0.0	0.573	6.976	91.0	2.1675	1.0	273.0	21.0	396.90	5.64	23.9
504	0.10959	0.0	11.93	0.0	0.573	6.794	89.3	2.3889	1.0	273.0	21.0	393.45	6.48	22.0
505	0.04741	0.0	11.93	0.0	0.573	6.030	80.8	2.5050	1.0	273.0	21.0	396.90	7.88	11.9

506 rows × 14 columns

简单的字段查询

‘’‘SQL语句’’’

• SELECT CRIM,ZN,CHAS,NOX,RM,RAD FROM boston LIMIT 2;

实现返回每行记录的CRIM,ZN,CHAS,NOX,RM,RAD字段，返回2行。

search = df[["CRIM","ZN","CHAS","NOX","RM","RAD"]].head(2)
search

	CRIM	ZN	CHAS	NOX	RM	RAD
0	0.00632	18.0	0.0	0.538	6.575	1.0
1	0.02731	0.0	0.0	0.469	6.421	2.0

简单的条件过滤 WHERE

‘’‘SQL语句’’’

• SELECT * FROM boston WHERE CHAS=1 LIMIT 2;

set(df['CHAS']) # 我们将CHAS特征作为分类标签

{0.0, 1.0}

search = df[df['CHAS']==1].head(2)
search

	CRIM	ZN	INDUS	CHAS	NOX	RM	AGE	DIS	RAD	TAX	PTRATIO	B	LSTAT	target
142	3.32105	0.0	19.58	1.0	0.871	5.403	100.0	1.3216	5.0	403.0	14.7	396.90	26.82	13.4
152	1.12658	0.0	19.58	1.0	0.871	5.012	88.0	1.6102	5.0	403.0	14.7	343.28	12.12	15.3

多条件与或查询WHERE AND|OR

如上满足SQL实现查询同时满足CHAS=1和CRIM>1.0两个条件的记录，返回两行

与关系&

‘’‘SQL语句’’’

• SELECT * FROM boston WHERE CHAS=1 AND CRIM>=1.0 LIMIT 2;

search = df[(df['CHAS']==1) & (df['CRIM'] >= 1.0)] .head(2) # 注意这里条件是用的圆括号！！！不是方括号
search

	CRIM	ZN	INDUS	CHAS	NOX	RM	AGE	DIS	RAD	TAX	PTRATIO	B	LSTAT	target
142	3.32105	0.0	19.58	1.0	0.871	5.403	100.0	1.3216	5.0	403.0	14.7	396.90	26.82	13.4
152	1.12658	0.0	19.58	1.0	0.871	5.012	88.0	1.6102	5.0	403.0	14.7	343.28	12.12	15.3

或关系|

‘’‘SQL语句’’’

• SELECT * FROM boston WHERE CHAS=1 OR CRIM>=1.0 LIMIT 2;

search = df[(df['CHAS']==1) | (df['CRIM'] >= 1.0)] .head(2) # 注意这里条件是用的圆括号！！！不是方括号
search

	CRIM	ZN	INDUS	CHAS	NOX	RM	AGE	DIS	RAD	TAX	PTRATIO	B	LSTAT	target
16	1.05393	0.0	8.14	0.0	0.538	5.935	29.3	4.4986	4.0	307.0	21.0	386.85	6.58	23.1
20	1.25179	0.0	8.14	0.0	0.538	5.570	98.1	3.7979	4.0	307.0	21.0	376.57	21.02	13.6

条件过滤 控制判断

空判断 查询null的记录

‘’‘SQL语句’’’

• SELECT * FROM boston WHERE CHAS IS NULL;

search = df[df['CHAS'].isna()]# 注意这里条件是用的圆括号！！！不是方括号
search

	CRIM	ZN	INDUS	CHAS	NOX	RM	AGE	DIS	RAD	TAX	PTRATIO	B	LSTAT	target

非空判断is not null

查询不为null的记录

‘’‘SQL语句’’’

• SELECT * FROM boston WHERE CHAS IS NOT NULL;

search = df[df['CHAS'].notna()]# 注意这里条件是用的圆括号！！！不是方括号
search

	CRIM	ZN	INDUS	CHAS	NOX	RM	AGE	DIS	RAD	TAX	PTRATIO	B	LSTAT	target
0	0.00632	18.0	2.31	0.0	0.538	6.575	65.2	4.0900	1.0	296.0	15.3	396.90	4.98	24.0
1	0.02731	0.0	7.07	0.0	0.469	6.421	78.9	4.9671	2.0	242.0	17.8	396.90	9.14	21.6
2	0.02729	0.0	7.07	0.0	0.469	7.185	61.1	4.9671	2.0	242.0	17.8	392.83	4.03	34.7
3	0.03237	0.0	2.18	0.0	0.458	6.998	45.8	6.0622	3.0	222.0	18.7	394.63	2.94	33.4
4	0.06905	0.0	2.18	0.0	0.458	7.147	54.2	6.0622	3.0	222.0	18.7	396.90	5.33	36.2
...	...	...	...	...	...	...	...	...	...	...	...	...	...	...
501	0.06263	0.0	11.93	0.0	0.573	6.593	69.1	2.4786	1.0	273.0	21.0	391.99	9.67	22.4
502	0.04527	0.0	11.93	0.0	0.573	6.120	76.7	2.2875	1.0	273.0	21.0	396.90	9.08	20.6
503	0.06076	0.0	11.93	0.0	0.573	6.976	91.0	2.1675	1.0	273.0	21.0	396.90	5.64	23.9
504	0.10959	0.0	11.93	0.0	0.573	6.794	89.3	2.3889	1.0	273.0	21.0	393.45	6.48	22.0
505	0.04741	0.0	11.93	0.0	0.573	6.030	80.8	2.5050	1.0	273.0	21.0	396.90	7.88	11.9

506 rows × 14 columns

排序 ORDER BY ASC|DESC

‘’‘SQL语句’’’

• SELECT * FROM boston WHERE INDUS>10 ORDER BY DESC CHAS;

满足INDUS>10的值按照CHAS进行降序排列

search = df[(df['INDUS']>10)].sort_values(by='CHAS',ascending=False)
search

	CRIM	ZN	INDUS	CHAS	NOX	RM	AGE	DIS	RAD	TAX	PTRATIO	B	LSTAT	target
162	1.83377	0.0	19.58	1.0	0.605	7.802	98.2	2.0407	5.0	403.0	14.7	389.61	1.92	50.0
212	0.21719	0.0	10.59	1.0	0.489	5.807	53.8	3.6526	4.0	277.0	18.6	390.94	16.03	22.4
364	3.47428	0.0	18.10	1.0	0.718	8.780	82.9	1.9047	24.0	666.0	20.2	354.55	5.29	21.9
363	4.22239	0.0	18.10	1.0	0.770	5.803	89.0	1.9047	24.0	666.0	20.2	353.04	14.64	16.8
357	3.84970	0.0	18.10	1.0	0.770	6.395	91.0	2.5052	24.0	666.0	20.2	391.34	13.27	21.7
...	...	...	...	...	...	...	...	...	...	...	...	...	...	...
293	0.08265	0.0	13.92	0.0	0.437	6.127	18.4	5.5027	4.0	289.0	16.0	396.90	8.58	23.9
294	0.08199	0.0	13.92	0.0	0.437	6.009	42.3	5.5027	4.0	289.0	16.0	396.90	10.40	21.7
295	0.12932	0.0	13.92	0.0	0.437	6.678	31.1	5.9604	4.0	289.0	16.0	396.90	6.27	28.6
296	0.05372	0.0	13.92	0.0	0.437	6.549	51.0	5.9604	4.0	289.0	16.0	392.85	7.39	27.1
505	0.04741	0.0	11.93	0.0	0.573	6.030	80.8	2.5050	1.0	273.0	21.0	396.90	7.88	11.9

236 rows × 14 columns

更新UPDATE

‘’‘SQL语句’’’

• UPDATE boston SET CHAS = 1 WHERE INDUS >10 AND ZN = 0

将满足INDUS >10 和ZN > 0的CHAS字段更新为1

df.loc[(df['INDUS']>10) & (df['ZN']==0),'CHAS']=1
df[df['INDUS']>=10]

	CRIM	ZN	INDUS	CHAS	NOX	RM	AGE	DIS	RAD	TAX	PTRATIO	B	LSTAT	target
70	0.08826	0.0	10.81	1.0	0.413	6.417	6.6	5.2873	4.0	305.0	19.2	383.73	6.72	24.2
71	0.15876	0.0	10.81	1.0	0.413	5.961	17.5	5.2873	4.0	305.0	19.2	376.94	9.88	21.7
72	0.09164	0.0	10.81	1.0	0.413	6.065	7.8	5.2873	4.0	305.0	19.2	390.91	5.52	22.8
73	0.19539	0.0	10.81	1.0	0.413	6.245	6.2	5.2873	4.0	305.0	19.2	377.17	7.54	23.4
74	0.07896	0.0	12.83	1.0	0.437	6.273	6.0	4.2515	5.0	398.0	18.7	394.92	6.78	24.1
...	...	...	...	...	...	...	...	...	...	...	...	...	...	...
501	0.06263	0.0	11.93	1.0	0.573	6.593	69.1	2.4786	1.0	273.0	21.0	391.99	9.67	22.4
502	0.04527	0.0	11.93	1.0	0.573	6.120	76.7	2.2875	1.0	273.0	21.0	396.90	9.08	20.6
503	0.06076	0.0	11.93	1.0	0.573	6.976	91.0	2.1675	1.0	273.0	21.0	396.90	5.64	23.9
504	0.10959	0.0	11.93	1.0	0.573	6.794	89.3	2.3889	1.0	273.0	21.0	393.45	6.48	22.0
505	0.04741	0.0	11.93	1.0	0.573	6.030	80.8	2.5050	1.0	273.0	21.0	396.90	7.88	11.9

236 rows × 14 columns

分组统计

根据CHAS进行分组，返回CHAS和每组的数量

‘’‘SQL语句’’’

• SELECT CHAS，COUNT(*) FROM boston GROUP BY CHAS;

search = df.groupby('CHAS').size()
search

CHAS
0.0    261
1.0    245
dtype: int64

分组统计 聚合输出

根据CHAS进行分组，返回CHAS，每个组的INDUS和NOX的平均值、最大值

‘’‘SQL语句’’’

• SELECT CHAS,avg(INDUS),max(INDUS),avg(NOX),max(NOX) FROM boston GROUP BY CHAS

import numpy as np
search = df.groupby('CHAS').agg({'INDUS':[np.mean,np.max],'NOX':[np.mean,np.max]})
search

	INDUS		NOX
	mean	amax	mean	amax
CHAS
0.0	5.485479	15.04	0.481091	0.647
1.0	17.157143	27.74	0.633106	0.871

删除

‘’‘SQL语句’’’

• DELETE FROM boston WHERE WHERE TAX=305 AND RAD = 4;

drop = df.drop(df[(df['TAX']=305) & (df['RAD']=4)].index)

  File "<ipython-input-90-fff36bdbf4c7>", line 1
    drop = df.drop(df[(df['TAX']=305) & (df['RAD']=4)].index)
                                ^
SyntaxError: invalid syntax

标签：search,df,boston,数据库,CHAS,00.5736,SQL,筛选,pandas
来源： https://blog.csdn.net/weixin_43213884/article/details/121653284

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