(데이터 분석) 파이썬 - Pandas DataFrame SQL처럼 활용하기(CRUD)

 

 

 

 

 

 

 

 

 

EMP 파일 실습

 

In [1]:

import pandas as pd
import numpy as np

 

 

1. Data Load

 

 

• df = read_csv()
• df.info()
• df.head()
• df.shape

In [51]:

df=pd.read_csv("./emp2.csv")  #csv파일 불러서 df변수에 저장
df

Out[51]:

	EMPNO	ENAME	JOB	MGR	HIRDDATE	SAL	COMM	DEPTNO
0	7369	SMITH	CLERK	7902.0	1980-12-17	1800	NaN	30
1	7499	ALLEN	SALESMAN	7698.0	1981-02-20	1600	300.0	30
2	7521	WARD	SALESMAN	7698.0	1981-02-22	1250	500.0	30
3	7566	JONES	MANAGER	7839.0	1981-04-02	2975	NaN	20
4	7654	MARTIN	SALESMAN	7698.0	1981-09-28	1250	1400.0	30
5	7698	BLAKE	MANAGER	7839.0	1981-05-01	2850	NaN	30
6	7782	CLARK	MANAGER	7839.0	1981-06-09	2450	NaN	10
7	7788	SCOTT	ANALYST	7566.0	1987-07-13	3000	NaN	20
8	7839	KING	PRESIDENT	NaN	1981-11-17	5000	NaN	10
9	7844	TURNER	SALESMAN	7698.0	1981-09-08	1500	0.0	30
10	7876	ADAMS	CLERK	7788.0	1987-07-13	1100	NaN	20
11	7900	JAMES	CLERK	7698.0	1981-12-03	950	NaN	30
12	7902	FORD	ANALYST	7566.0	1981-12-03	6000	NaN	20
13	7934	MILLER	CLERK	7782.0	1982-01-23	1300	NaN	10

 

• object(글자)타입 X
• 결측(NaN) X

In [52]:

df.info() #데이터 구조

 

<class 'pandas.core.frame.DataFrame'>
RangeIndex: 14 entries, 0 to 13
Data columns (total 8 columns):
 #   Column    Non-Null Count  Dtype  
---  ------    --------------  -----  
 0   EMPNO     14 non-null     int64  
 1   ENAME     14 non-null     object 
 2   JOB       14 non-null     object 
 3   MGR       13 non-null     float64
 4   HIRDDATE  14 non-null     object 
 5   SAL       14 non-null     int64  
 6   COMM      4 non-null      float64
 7   DEPTNO    14 non-null     int64  
dtypes: float64(2), int64(3), object(3)
memory usage: 1.0+ KB

In [53]:

df.isna() #결측 확인

Out[53]:

	EMPNO	ENAME	JOB	MGR	HIRDDATE	SAL	COMM	DEPTNO
0	False	False	False	False	False	False	True	False
1	False	False	False	False	False	False	False	False
2	False	False	False	False	False	False	False	False
3	False	False	False	False	False	False	True	False
4	False	False	False	False	False	False	False	False
5	False	False	False	False	False	False	True	False
6	False	False	False	False	False	False	True	False
7	False	False	False	False	False	False	True	False
8	False	False	False	True	False	False	True	False
9	False	False	False	False	False	False	False	False
10	False	False	False	False	False	False	True	False
11	False	False	False	False	False	False	True	False
12	False	False	False	False	False	False	True	False
13	False	False	False	False	False	False	True	False

In [54]:

df.isna().sum() #결측 수 세기

Out[54]:

EMPNO        0
ENAME        0
JOB          0
MGR          1
HIRDDATE     0
SAL          0
COMM        10
DEPTNO       0
dtype: int64

In [55]:

df.head()

Out[55]:

	EMPNO	ENAME	JOB	MGR	HIRDDATE	SAL	COMM	DEPTNO
0	7369	SMITH	CLERK	7902.0	1980-12-17	1800	NaN	30
1	7499	ALLEN	SALESMAN	7698.0	1981-02-20	1600	300.0	30
2	7521	WARD	SALESMAN	7698.0	1981-02-22	1250	500.0	30
3	7566	JONES	MANAGER	7839.0	1981-04-02	2975	NaN	20
4	7654	MARTIN	SALESMAN	7698.0	1981-09-28	1250	1400.0	30

In [56]:

df.shape #(행,열) 수

Out[56]:

(14, 8)

 

컬럼명 대문자를 소문자로 바꾸기

 

In [58]:

lists = df.columns.tolist()
for i, col in enumerate(lists):
    list[i] = col.lower()
print(list)

 

['empno', 'ename', 'job', 'mgr', 'hirddate', 'sal', 'comm', 'deptno']

In [59]:

list=[col.lower() for col in df.columns.tolist()]
print(list)

 

['empno', 'ename', 'job', 'mgr', 'hirddate', 'sal', 'comm', 'deptno']

In [60]:

df.columns=list

 

2. SELECT

 

 

• 칼럼별로 호출 df[칼럼명]
• array처럼 [행,열]로 호출 iloc()
• 로우명, 칼럼명으로 호출 loc()

 

df[] - 칼럼별로 호출 df[칼럼명]

In [61]:

df[['empno','ename','sal']] #select empno, enmae, sal from emp

Out[61]:

	empno	ename	sal
0	7369	SMITH	1800
1	7499	ALLEN	1600
2	7521	WARD	1250
3	7566	JONES	2975
4	7654	MARTIN	1250
5	7698	BLAKE	2850
6	7782	CLARK	2450
7	7788	SCOTT	3000
8	7839	KING	5000
9	7844	TURNER	1500
10	7876	ADAMS	1100
11	7900	JAMES	950
12	7902	FORD	6000
13	7934	MILLER	1300

 

df.iloc() -array처럼 [행,열]로 호출

In [62]:

df.iloc[[1,2,3],[0,1,2]] #[[1,2,3행],[0,1,2열]]

Out[62]:

	empno	ename	job
1	7499	ALLEN	SALESMAN
2	7521	WARD	SALESMAN
3	7566	JONES	MANAGER

In [49]:

df.iloc[[0,1,2,3],0:3] #[0,1,2,3행],0~3열] **마지막 번호까지 출력함 ==> 통상 사용하는 슬라이싱이랑 다름

Out[49]:

	empno	ename	job
0	7369	smith999	CLERK
1	7499	ALLEN	SALESMAN
2	7521	WARD	SALESMAN
3	7566	JONES	MANAGER

 

df.loc() -로우명, 칼럼명으로 호출 loc()

In [63]:

df.loc[[0,1,2,3],['empno','ename','job']] #[[행이름][열이름]]

Out[63]:

	empno	ename	job
0	7369	SMITH	CLERK
1	7499	ALLEN	SALESMAN
2	7521	WARD	SALESMAN
3	7566	JONES	MANAGER

In [64]:

df.loc[[0,1,2,3],'empno':'job']

Out[64]:

	empno	ename	job
0	7369	SMITH	CLERK
1	7499	ALLEN	SALESMAN
2	7521	WARD	SALESMAN
3	7566	JONES	MANAGER

In [69]:

df #전체 데이터 출력

Out[69]:

	empno	ename	job	mgr	hirddate	sal	comm	deptno
0	7369	SMITH	CLERK	7902.0	1980-12-17	1800	NaN	30
1	7499	ALLEN	SALESMAN	7698.0	1981-02-20	1600	300.0	30
2	7521	WARD	SALESMAN	7698.0	1981-02-22	1250	500.0	30
3	7566	JONES	MANAGER	7839.0	1981-04-02	2975	NaN	20
4	7654	MARTIN	SALESMAN	7698.0	1981-09-28	1250	1400.0	30
5	7698	BLAKE	MANAGER	7839.0	1981-05-01	2850	NaN	30
6	7782	CLARK	MANAGER	7839.0	1981-06-09	2450	NaN	10
7	7788	SCOTT	ANALYST	7566.0	1987-07-13	3000	NaN	20
8	7839	KING	PRESIDENT	NaN	1981-11-17	5000	NaN	10
9	7844	TURNER	SALESMAN	7698.0	1981-09-08	1500	0.0	30
10	7876	ADAMS	CLERK	7788.0	1987-07-13	1100	NaN	20
11	7900	JAMES	CLERK	7698.0	1981-12-03	950	NaN	30
12	7902	FORD	ANALYST	7566.0	1981-12-03	6000	NaN	20
13	7934	MILLER	CLERK	7782.0	1982-01-23	1300	NaN	10

 

3. 조건절 where

 

df[조건][칼럼]

df[칼럼][조건]

In [72]:

#select empno, ename, sal from emp where deptno=10;
df[['empno','ename','sal']][df['deptno']==10]

Out[72]:

	empno	ename	sal
6	7782	CLARK	2450
8	7839	KING	5000
13	7934	MILLER	1300

In [73]:

df[df['deptno']==10][['empno','ename','sal']]

Out[73]:

	empno	ename	sal
6	7782	CLARK	2450
8	7839	KING	5000
13	7934	MILLER	1300

 

df에서는 and=&, or=|

df[(df[]조건) &/| (df[]조건)] ==> ()처리 반드시

In [81]:

df[(df['deptno']==30) & (df['job']=='SALESMAN')]

Out[81]:

	empno	ename	job	mgr	hirddate	sal	comm	deptno
1	7499	ALLEN	SALESMAN	7698.0	1981-02-20	1600	300.0	30
2	7521	WARD	SALESMAN	7698.0	1981-02-22	1250	500.0	30
4	7654	MARTIN	SALESMAN	7698.0	1981-09-28	1250	1400.0	30
9	7844	TURNER	SALESMAN	7698.0	1981-09-08	1500	0.0	30

In [83]:

df[(df['job']=='MANAGER') | (df['job']=='SALESMAN')]

Out[83]:

	empno	ename	job	mgr	hirddate	sal	comm	deptno
1	7499	ALLEN	SALESMAN	7698.0	1981-02-20	1600	300.0	30
2	7521	WARD	SALESMAN	7698.0	1981-02-22	1250	500.0	30
3	7566	JONES	MANAGER	7839.0	1981-04-02	2975	NaN	20
4	7654	MARTIN	SALESMAN	7698.0	1981-09-28	1250	1400.0	30
5	7698	BLAKE	MANAGER	7839.0	1981-05-01	2850	NaN	30
6	7782	CLARK	MANAGER	7839.0	1981-06-09	2450	NaN	10
9	7844	TURNER	SALESMAN	7698.0	1981-09-08	1500	0.0	30

In [85]:

df[((df['deptno']==30) & (df['job']=='SALESMAN'))|(df['deptno']==10)][['job','deptno']]

Out[85]:

	job	deptno
1	SALESMAN	30
2	SALESMAN	30
4	SALESMAN	30
6	MANAGER	10
8	PRESIDENT	10
9	SALESMAN	30
13	CLERK	10

 

4. 서브쿼리 사용하는 법 (비추) - 가능하면 db에서 해결하는게 빠름

 

단일행 서브쿼리

 

Value 값을 변수에 넣어서 비교

In [113]:

x=df[df['empno']==7782]['job'].values[0]
x
df[df['job']==x][['ename','job']]

Out[113]:

	ename	job
3	JONES	MANAGER
5	BLAKE	MANAGER
6	CLARK	MANAGER

 

다중행 서브쿼리

 

in -> df[].isin() 함수 사용

In [125]:

x_list=df[df['deptno']==10]['job'].values
x_list

Out[125]:

array(['MANAGER', 'PRESIDENT', 'CLERK'], dtype=object)

In [127]:

df[df['job'].isin(x_list)][['ename','job']]

Out[127]:

	ename	job
0	SMITH	CLERK
3	JONES	MANAGER
5	BLAKE	MANAGER
6	CLARK	MANAGER
8	KING	PRESIDENT
10	ADAMS	CLERK
11	JAMES	CLERK
13	MILLER	CLERK

 

5. SQL 연산 ||, +-*/ 사용하는 법

 

 

|| 사용법

In [133]:

new_col=df['ename']+df['job']
df['new_col']=new_col
df

Out[133]:

	empno	ename	job	mgr	hirddate	sal	comm	deptno	new_col
0	7369	SMITH	CLERK	7902.0	1980-12-17	1800	NaN	30	SMITHCLERK
1	7499	ALLEN	SALESMAN	7698.0	1981-02-20	1600	300.0	30	ALLENSALESMAN
2	7521	WARD	SALESMAN	7698.0	1981-02-22	1250	500.0	30	WARDSALESMAN
3	7566	JONES	MANAGER	7839.0	1981-04-02	2975	NaN	20	JONESMANAGER
4	7654	MARTIN	SALESMAN	7698.0	1981-09-28	1250	1400.0	30	MARTINSALESMAN
5	7698	BLAKE	MANAGER	7839.0	1981-05-01	2850	NaN	30	BLAKEMANAGER
6	7782	CLARK	MANAGER	7839.0	1981-06-09	2450	NaN	10	CLARKMANAGER
7	7788	SCOTT	ANALYST	7566.0	1987-07-13	3000	NaN	20	SCOTTANALYST
8	7839	KING	PRESIDENT	NaN	1981-11-17	5000	NaN	10	KINGPRESIDENT
9	7844	TURNER	SALESMAN	7698.0	1981-09-08	1500	0.0	30	TURNERSALESMAN
10	7876	ADAMS	CLERK	7788.0	1987-07-13	1100	NaN	20	ADAMSCLERK
11	7900	JAMES	CLERK	7698.0	1981-12-03	950	NaN	30	JAMESCLERK
12	7902	FORD	ANALYST	7566.0	1981-12-03	6000	NaN	20	FORDANALYST
13	7934	MILLER	CLERK	7782.0	1982-01-23	1300	NaN	10	MILLERCLERK

 

수치(db: number, python: int,float, df: int64, float64) 컬럼, 컬럼간의 연산

 

• NaN에는 어떤 연산을 해도 NaN이다
• db: nvl(col,0)
• python : df.fillna(0) df[컬럼].fillna(0)
• db: 컬럼 is not null, 컬럼 is null
• df.isna() df.notna()
• df.isnull() df.notnull()
• df.astype() 캐스팅

In [152]:

new_col2=df['sal']+df['comm']
df['sal+comm']=new_col2
df

Out[152]:

	empno	ename	job	mgr	hirddate	sal	comm	deptno	new_col	sal+comm
0	7369	SMITH	CLERK	7902.0	1980-12-17	1800	NaN	30	SMITHCLERK	NaN
1	7499	ALLEN	SALESMAN	7698.0	1981-02-20	1600	300.0	30	ALLENSALESMAN	1900.0
2	7521	WARD	SALESMAN	7698.0	1981-02-22	1250	500.0	30	WARDSALESMAN	1750.0
3	7566	JONES	MANAGER	7839.0	1981-04-02	2975	NaN	20	JONESMANAGER	NaN
4	7654	MARTIN	SALESMAN	7698.0	1981-09-28	1250	1400.0	30	MARTINSALESMAN	2650.0
5	7698	BLAKE	MANAGER	7839.0	1981-05-01	2850	NaN	30	BLAKEMANAGER	NaN
6	7782	CLARK	MANAGER	7839.0	1981-06-09	2450	NaN	10	CLARKMANAGER	NaN
7	7788	SCOTT	ANALYST	7566.0	1987-07-13	3000	NaN	20	SCOTTANALYST	NaN
8	7839	KING	PRESIDENT	NaN	1981-11-17	5000	NaN	10	KINGPRESIDENT	NaN
9	7844	TURNER	SALESMAN	7698.0	1981-09-08	1500	0.0	30	TURNERSALESMAN	1500.0
10	7876	ADAMS	CLERK	7788.0	1987-07-13	1100	NaN	20	ADAMSCLERK	NaN
11	7900	JAMES	CLERK	7698.0	1981-12-03	950	NaN	30	JAMESCLERK	NaN
12	7902	FORD	ANALYST	7566.0	1981-12-03	6000	NaN	20	FORDANALYST	NaN
13	7934	MILLER	CLERK	7782.0	1982-01-23	1300	NaN	10	MILLERCLERK	NaN

In [154]:

new_col2=df['sal']+df['comm'].fillna(0)
df['sal+comm']=new_col2.astype(int) #int로 캐스팅
df

Out[154]:

	empno	ename	job	mgr	hirddate	sal	comm	deptno	new_col	sal+comm
0	7369	SMITH	CLERK	7902.0	1980-12-17	1800	NaN	30	SMITHCLERK	1800
1	7499	ALLEN	SALESMAN	7698.0	1981-02-20	1600	300.0	30	ALLENSALESMAN	1900
2	7521	WARD	SALESMAN	7698.0	1981-02-22	1250	500.0	30	WARDSALESMAN	1750
3	7566	JONES	MANAGER	7839.0	1981-04-02	2975	NaN	20	JONESMANAGER	2975
4	7654	MARTIN	SALESMAN	7698.0	1981-09-28	1250	1400.0	30	MARTINSALESMAN	2650
5	7698	BLAKE	MANAGER	7839.0	1981-05-01	2850	NaN	30	BLAKEMANAGER	2850
6	7782	CLARK	MANAGER	7839.0	1981-06-09	2450	NaN	10	CLARKMANAGER	2450
7	7788	SCOTT	ANALYST	7566.0	1987-07-13	3000	NaN	20	SCOTTANALYST	3000
8	7839	KING	PRESIDENT	NaN	1981-11-17	5000	NaN	10	KINGPRESIDENT	5000
9	7844	TURNER	SALESMAN	7698.0	1981-09-08	1500	0.0	30	TURNERSALESMAN	1500
10	7876	ADAMS	CLERK	7788.0	1987-07-13	1100	NaN	20	ADAMSCLERK	1100
11	7900	JAMES	CLERK	7698.0	1981-12-03	950	NaN	30	JAMESCLERK	950
12	7902	FORD	ANALYST	7566.0	1981-12-03	6000	NaN	20	FORDANALYST	6000
13	7934	MILLER	CLERK	7782.0	1982-01-23	1300	NaN	10	MILLERCLERK	1300

In [157]:

df.isnull() #null인 값 True 논리값으로 반환

Out[157]:

	empno	ename	job	mgr	hirddate	sal	comm	deptno	new_col	sal+comm
0	False	False	False	False	False	False	True	False	False	False
1	False	False	False	False	False	False	False	False	False	False
2	False	False	False	False	False	False	False	False	False	False
3	False	False	False	False	False	False	True	False	False	False
4	False	False	False	False	False	False	False	False	False	False
5	False	False	False	False	False	False	True	False	False	False
6	False	False	False	False	False	False	True	False	False	False
7	False	False	False	False	False	False	True	False	False	False
8	False	False	False	True	False	False	True	False	False	False
9	False	False	False	False	False	False	False	False	False	False
10	False	False	False	False	False	False	True	False	False	False
11	False	False	False	False	False	False	True	False	False	False
12	False	False	False	False	False	False	True	False	False	False
13	False	False	False	False	False	False	True	False	False	False

In [159]:

df.notnull() #null이 아닌 값 True 논리값으로 반환

Out[159]:

	empno	ename	job	mgr	hirddate	sal	comm	deptno	new_col	sal+comm
0	True	True	True	True	True	True	False	True	True	True
1	True	True	True	True	True	True	True	True	True	True
2	True	True	True	True	True	True	True	True	True	True
3	True	True	True	True	True	True	False	True	True	True
4	True	True	True	True	True	True	True	True	True	True
5	True	True	True	True	True	True	False	True	True	True
6	True	True	True	True	True	True	False	True	True	True
7	True	True	True	True	True	True	False	True	True	True
8	True	True	True	False	True	True	False	True	True	True
9	True	True	True	True	True	True	True	True	True	True
10	True	True	True	True	True	True	False	True	True	True
11	True	True	True	True	True	True	False	True	True	True
12	True	True	True	True	True	True	False	True	True	True
13	True	True	True	True	True	True	False	True	True	True

In [158]:

df.isnull().sum() #null 개수

Out[158]:

empno        0
ename        0
job          0
mgr          1
hirddate     0
sal          0
comm        10
deptno       0
new_col      0
sal+comm     0
dtype: int64

 

6. 그룹함수

 

 

Group 함수

• db : count() min() max() sum() avg()
• df : value_counts() min() max() sum() mean() ... std() var()

In [184]:

#부서별 사원수 출력
#select deptno, count(1) from emp group by deptno;
df.groupby("deptno")["empno"].count()

Out[184]:

deptno
10    3
20    4
30    7
Name: empno, dtype: int64

In [182]:

df.groupby("deptno")["sal"].mean()

Out[182]:

deptno
10    2916.666667
20    3268.750000
30    1600.000000
Name: sal, dtype: float64

In [174]:

#select deptno, min(sal), max(sal), sum(sal), avg(sal) from emp group by deptno 
df.groupby("deptno")["sal"].agg(["min","max","sum","mean"])

Out[174]:

	min	max	sum	mean
deptno
10	1300	5000	8750	2916.666667
20	1100	6000	13075	3268.750000
30	950	2850	11200	1600.000000

In [180]:

#select deptno, max(sal), min(comm) from emp group by deptno 
mydict = {"sal":"max", "comm":"min"} #컬럼이름(키 값)이 같으면 안됨
df.groupby("deptno").agg(mydict)

Out[180]:

	sal	comm
deptno
10	5000	NaN
20	6000	NaN
30	2850	0.0

 

• 조건문 + 그룹바이
• df[df[]조건].groupby().....

In [195]:

#select deptno, min(sal), max(sal), sum(sal), avg(sal) from emp 
#where deptno =10
#group by deptno
df[df['deptno']==10].groupby("deptno")[["sal"]].agg(["min","max","sum","mean"])

Out[195]:

	sal
	min	max	sum	mean
deptno
10	1300	5000	8750	2916.666667

In [202]:

df.describe()

Out[202]:

	empno	mgr	sal	comm	deptno	sal+comm
count	14.000000	13.000000	14.000000	4.000000	14.000000	14.000000
mean	7726.571429	7739.307692	2358.928571	550.000000	22.857143	2516.071429
std	178.294361	103.714660	1520.362025	602.771377	8.254203	1450.053286
min	7369.000000	7566.000000	950.000000	0.000000	10.000000	950.000000
25%	7588.000000	7698.000000	1262.500000	225.000000	20.000000	1562.500000
50%	7785.000000	7698.000000	1700.000000	400.000000	25.000000	2175.000000
75%	7868.000000	7839.000000	2943.750000	725.000000	30.000000	2943.750000
max	7934.000000	7902.000000	6000.000000	1400.000000	30.000000	6000.000000

In [199]:

df['sal'].describe()

Out[199]:

count      14.000000
mean     2358.928571
std      1520.362025
min       950.000000
25%      1262.500000
50%      1700.000000
75%      2943.750000
max      6000.000000
Name: sal, dtype: float64

In [205]:

gdf=df.describe()
gdf.loc[['count','mean','max'],['sal','mgr','comm']]

Out[205]:

	sal	mgr	comm
count	14.000000	13.000000	4.0
mean	2358.928571	7739.307692	550.0
max	6000.000000	7902.000000	1400.0

 

7. UPDATE

 

In [211]:

# create table empcp as (select empno, ename, sal, job from emp where deptno=30)

empcp = df[df['deptno']==30][['empno','ename','sal','job']].copy() ##카피
empcp

Out[211]:

	empno	ename	sal	job
0	7369	SMITH	1800	CLERK
1	7499	ALLEN	1600	SALESMAN
2	7521	WARD	1250	SALESMAN
4	7654	MARTIN	1250	SALESMAN
5	7698	BLAKE	2850	MANAGER
9	7844	TURNER	1500	SALESMAN
11	7900	JAMES	950	CLERK

In [213]:

empcp.iloc[1] = [7499,'ALLEN22',1800,'MANAGER']
empcp

Out[213]:

	empno	ename	sal	job
0	7369	SMITH	1800	CLERK
1	7499	ALLEN22	1800	MANAGER
2	7521	WARD	1250	SALESMAN
4	7654	MARTIN	1250	SALESMAN
5	7698	BLAKE	2850	MANAGER
9	7844	TURNER	1500	SALESMAN
11	7900	JAMES	950	CLERK

 

 

8. DELETE

 

• 행 한줄 지우기
• empcp.drop(index = '행', axis=0)

In [211]:

# create table empcp as (select empno, ename, sal, job from emp where deptno=30)

empcp = df[df['deptno']==30][['empno','ename','sal','job']].copy() ##카피
empcp

Out[211]:

	empno	ename	sal	job
0	7369	SMITH	1800	CLERK
1	7499	ALLEN	1600	SALESMAN
2	7521	WARD	1250	SALESMAN
4	7654	MARTIN	1250	SALESMAN
5	7698	BLAKE	2850	MANAGER
9	7844	TURNER	1500	SALESMAN
11	7900	JAMES	950	CLERK

In [216]:

empcp.drop(index=11, axis=0)  #-- inplace=True를 사용해야 실제로 지워짐 -> 사용안하면 삭제한 견본만 보임
#empcp.drop(index=11, axis=0, inplace=True)
#empcp = empcp.drop(index=11, axis=0) 이렇게도 지우기 가능

Out[216]:

	empno	ename	sal	job
0	7369	SMITH	1800	CLERK
1	7499	ALLEN22	1800	MANAGER
2	7521	WARD	1250	SALESMAN
4	7654	MARTIN	1250	SALESMAN
5	7698	BLAKE	2850	MANAGER
9	7844	TURNER	1500	SALESMAN

 

• 열(컬럼) 지우기
• empcp.drop(['컬럼','컬럼'], axis=1)

In [218]:

empcp

Out[218]:

	empno	ename	sal	job
0	7369	SMITH	1800	CLERK
1	7499	ALLEN22	1800	MANAGER
2	7521	WARD	1250	SALESMAN
4	7654	MARTIN	1250	SALESMAN
5	7698	BLAKE	2850	MANAGER
9	7844	TURNER	1500	SALESMAN
11	7900	JAMES	950	CLERK

In [221]:

empcp.drop('job', axis=1) ##칼럼 지우기 (axis = 1)
#-- inplace=True를 사용해야 실제로 지워짐 -> 사용안하면 삭제한 견본만 보임
#empcp.drop('job', axis=1, inplace=True)
#empcp=empcp.drop(['job','sal','ename'], axis=1)

Out[221]:

	empno	ename	sal
0	7369	SMITH	1800
1	7499	ALLEN22	1800
2	7521	WARD	1250
4	7654	MARTIN	1250
5	7698	BLAKE	2850
9	7844	TURNER	1500
11	7900	JAMES	950

 

9. INSERT