R, Python 분석과 프로그래밍의 친구 (by R Friend)

> rm(list=ls()) # clean all

# Sample vector

> set.seed(123) # for reproducibility

> vec_raw <- sample(c(0, 1), size=30, replace=TRUE, prob=(c(0.8, 0.2)))

> vec_raw

 [1] 0 0 0 1 1 0 0 1 0 0 1 0 0 0 0 1 0 0 0 1 1 0 0 1 0 0 0 0 0 0

> vec <- vec_raw # copy for manipulation

##---------------------------------------------

# Vector Bin Split by successive '0's criteria

##---------------------------------------------

# Setting

vec_tmp <- c() # null vector

bin_range <- 3 # successive '0's criterion

vec_idx <- 1

vec_num <- 1

# Zero_Bin_Splitter

while (length(vec) > 0){

  if (sum(vec[1:bin_range], na.rm=T) != 0){

    vec_tmp[vec_idx] <- vec[1]

    vec <- vec[-1]

    vec_idx <- vec_idx + 1

  } else if (is.null(vec_tmp)){

    assign(paste0("vec_split_", vec_num), vec[1:bin_range])

    vec <- vec[-(1:bin_range)]

    vec_idx <- 1 # initialization

    vec_num <- vec_num + 1

  } else {

    assign(paste0("vec_split_", vec_num), vec_tmp)

    vec_tmp <- c() # initialization

    vec_num <- vec_num + 1

    assign(paste0("vec_split_", vec_num), vec[1:bin_range])

    vec <- vec[-(1:bin_range)]

    vec_idx <- 1 # initialization

    vec_num <- vec_num + 1

  }

}

# delete temp vector and arguments

rm(vec, vec_tmp, bin_range, vec_idx, vec_num)

# by June Young Lee

library(data.table)

# 0이 세번 연속으로 들어 있는 아무 벡터 생성

vec <- c(0,0,0,3,10,2,3,0,4,0,0,0,1,2,50,4,0,0,32,1,0,0,0,1,1)

# data.table함수인 rleid이용하여 연속인 행들을 구분하는 idx1 컬럼 생성 

dt <- data.table(v1=vec, idx1=rleid(vec))

# idx1 기준으로 갯수를 세어 N 컬럼 생성 

dt[,N:=.N, by=idx1]

# 0이 세번 연속으로 나온 그룹(=N컬럼변수가 3인)을 idx2로 체크하기 

dt[v1==0&N==3,idx2:=1L]

# split 하기 위한 기준 벡터를 마찬가지로 rleid함수를 이용하여 idx3로 생성

dt[,idx3:=rleid(idx2)]

# data.table의 split 함수 적용하고, lapply 적용하여 v1 칼럼만 추출 

res <- lapply(split(dt, by="idx3"), function(x) x[,v1])

res

# $`1`

# [1] 0 0 0

# 

# $`2`

# [1] 3 10 2 3 0 4

# 

# $`3`

# [1] 0 0 0

# 

# $`4`

# [1] 1 2 50 4 0 0 32 1

# 

# $`5`

# [1] 0 0 0

# 

# $`6`

# [1] 1 1

# 0이 3번 연속으로 나온 놈들을 남겨두도록 작성한 이유는, 

# 위치확인 & 확인용입니다. 

# 필요없으면, 아래 정도의 코드를 추가하면 되겠네요.

res2 <- res[!unlist(lapply(res, function(x) length(x)==3&sum(x)==0))]

res2

# $`2`

# [1] 3 10 2 3 0 4

# 

# $`4`

# [1] 1 2 50 4 0 0 32 1

# 

# $`6`

# [1] 1 1

 (1) while: 조건이 참인 동안 반복 (loop while the condition is True)

while condition:

    conditional code block 

while 반복문

실행 결과 (result)

count = 0

while count < 5:

    print('count : ', count)

    count = count + 1

print('The end')

count : 0

count : 1

count : 2

count : 3

count : 4

The end

while 반복문 

실행 결과 (result) 

count = 0

while count < 5:

    print(count, ' is less than 5')

    count = count + 1

else:

    print(count, ' is not less 5 so stop here') 

0  is less than 5

1  is less than 5

2  is less than 5

3  is less than 5

4  is less than 5

5  is not less 5 so stop here

while 반복문 (infinite loop)

실행 결과 (result) 

 # Infinite loop

while(True):

    print('Infinite loop goes on and on ...')

Infinite loop goes on and on ...

Infinite loop goes on and on ...

Infinite loop goes on and on ...

File "zmq/backend/cython/checkrc.pxd", line 12, in zmq.backend.cython.checkrc._check_rc (zmq\backend\cython\socket.c:9621)

KeyboardInterrupt

 while 반복문 (with if, elif, else & break)

실행 결과 (result) 

while True:

    print('continue or stop? [continue/stop]:')

    answer = input()

    if answer == 'continue':

        print('>> continue again!')

    elif answer == 'stop':

        print('>> stop here!')

        break

    else:

        print('>> wrong answer!')

continue or stop? [continue/stop]:

continue

>> continue again!

continue or stop? [continue/stop]:

repeate

>> wrong answer!

continue or stop? [continue/stop]:

continue

>> continue again!

continue or stop? [continue/stop]:

stop

>> stop here!

 (2) for:  순서열의 처음부터 끝까지 반복 (loop until the end of the sequence)

for iterating_var in sequence:

    code block 

 for i in range(start, end, interval)

실행 결과 (result) 

for i in range(0, 5, 1): # (start, end, interval)

    print(i)

0

1

2

3

4

for i in range(3, 10, 2):

    print(i)

3

5

7

9

for i in range(0, 5): # interval default : 1

    print(i)

0

1

2

3

4

for i in range(5): # start default : 0

    print(i)

0

1

2

3

4

for i in range(5, 5):

    print(i)

(no result)

# TypeError

for i in range(4.9, 5):

    print(i)

Traceback (most recent call last):

  File "<ipython-input-11-bb2a792a3440>", line 1, in <module>

    for i in range(4.9, 5):

TypeError: 'float' object cannot be interpreted as an integer

for i in range() 복수 반복문

실행 결과 (result) 

for i in range(1, 6):

    for j in range(i+1, 6):

        print(i, '+', j, '=', i+j)

1 + 2 = 3

1 + 3 = 4

1 + 4 = 5

1 + 5 = 6

2 + 3 = 5

2 + 4 = 6

2 + 5 = 7

3 + 4 = 7

3 + 5 = 8

4 + 5 = 9

for i in tuple

실행 결과 (result) 

for i in ('Python', 'is', 'easy'):

    print(i)

print('Really?')

Python

is

easy

Really?

for loop with range() & tuple index

실행 결과 (result) 

# iterating by sequence index

mytuple = ('Python', 'is', 'easy')

for index in range(len(mytuple)):

    print(mytuple[index])

print('Really?')

Python

is

easy

Really? 

for i in list

실행 결과 (result) 

for mylist in ['Python', 'is', 'easy']:

    print(mylist)

print('Are you sure?') 

Python

is

easy

Are you sure? 

for i in string

실행 결과 (result) 

for mystr in ('Python is easy'):

    print(mystr)

print('--The end of string character--')

P

y

t

h

o

n

i

s

e

a

s

y

--The end of string character--

for i in dictionary

실행 결과 (result) 

mydic = {'cust_01': 'kim', 

            'cust_02': 'lee', 

            'cust_03': 'choi'}

for key, val in mydic.items(): # tuple unpacking

    print("ID is '{0}' and NAME is '{1}'".format(key, val))

ID is 'cust_01' and NAME is 'kim'

ID is 'cust_02' and NAME is 'lee'

ID is 'cust_03' and NAME is 'choi'

3. continue: 조건이 참이면 반복문 계속 (continue loop if the condition is True)

     break: 조건이 참이면 반복문 중단 (stop loop if the condition is True)

for 반복문 with continue 제어문

실행 결과 (result) 

# print only odd numbers b/w 1 ~ 5

for i in range(1, 6):

    if i % 2 == 0:

        continue

    print(i)

1

3

5

for 반복문 with continue 제어문

실행 결과 (result) 

# if i+j == 5 then print i, j combination

for i in range(1, 6):

    for j in range(i+1, 6):

        if i+j != 5:

            continue        

        print(i, '+', j, '=', i+j)

1 + 4 = 5

2 + 3 = 5

for 반복문 with break 제어문

실행 결과 (result) 

for i in range(1, 6):

    print(i)

    if i == 4:

        break

1

2

3

4

for 반복문 with break, if, else

실행 결과 (result) 

for i in range(3, 21):

    for j in range(2, i):

        if i % j == 0:

            k = i / j

            print("%d is %d * %d" %(i, j, k))

            break

        else:

            if j != i - 1:

                continue

            else:

                print(i, "is prime number.")

                break

print("The End")

3 is prime number.

4 is 2 * 2

5 is prime number.

6 is 2 * 3

7 is prime number.

8 is 2 * 4

9 is 3 * 3

10 is 2 * 5

11 is prime number.

12 is 2 * 6

13 is prime number.

14 is 2 * 7

15 is 3 * 5

16 is 2 * 8

17 is prime number.

18 is 2 * 9

19 is prime number.

20 is 2 * 10

The End

nested while loops with break, if

실행 결과 (result) 

i = 3

while(i < 21):

    j = 2

    while(j <= (i/j)):

        if not(i%j): 

            print('%d is %d * %d' %(i, j, i/j))

            break

        j = j + 1

    if (j > i/j):

        print(i, " is prime")

    i = i + 1

print('The end')

# True or False?

# not(4%2) # True

# not(3%2) # False

3  is prime

4 is 2 * 2

5  is prime

6 is 2 * 3

7  is prime

8 is 2 * 4

9 is 3 * 3

10 is 2 * 5

11  is prime

12 is 2 * 6

13  is prime

14 is 2 * 7

15 is 3 * 5

16 is 2 * 8

17  is prime

18 is 2 * 9

19  is prime

20 is 2 * 10

The end