python 字串處理

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s="0123456789"

print(s[:5])

print(s[:5:2])

s.find("2") #  找字

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# 印出三角*

for i in range(0,10):

    for j in range(0,i+1):

        print("*",end="") # 不換行

    print("")

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#Longest Common SubSequence

import numpy as np

S1="abcdefg"

S2="abzzcdzzefg"

L1 = list(S1)

L2 = list(S2)

#補零作為起始

L1.insert(0,"0")

L2.insert(0,"0")

A = np.zeros((len(L1), len(L2)), dtype=np.int)

#初始值

for i in range(0,len(L1)):

    A[i,0] = -i

for j in range(0,len(L2)):

    A[0,j] = -j

#計算距離矩陣

for i in range(1,len(L1)):

    for j in range(1,len(L2)):

        if L1[i] == L2[j]:

            A[i,j] = A[i-1,j-1] + 2

        else:

            A[i,j] = max(A[i-1,j-1],A[i,j-1],A[i-1,j]) - 1

#回溯字串

i = len(L1)-1

j = len(L2)-1

LCS1 = ""

LCS2 = ""

while i > 0 and j > 0:

    #左上

    if A[i-1,j-1] >= max(A[i,j-1],A[i-1,j]):

        LCS1 = L1[i] + LCS1

        LCS2 = L2[j] + LCS2

        i = i - 1

        j = j - 1

    #左

    elif A[i,j-1] >= A[i-1,j]:

        LCS1 = "-" + LCS1

        LCS2 = L2[j] + LCS2

        j = j - 1

    #上

    else:

        LCS1 = L1[i] + LCS1

        LCS2 = "-" + LCS2

        i = i - 1

while i > 0:

    LCS1 = L1[i] + LCS1

    LCS2 = "-" + LCS2

    i = i - 1

print("輸入的字串")

print(S1)

print(S2)

print("輸出的字串")

print(LCS1)

print(LCS2)

print("共同的字串")

for i in range(0,len(LCS1)):

    if LCS1[i] == LCS2[i]:

        print(LCS1[i],end="")