如何用颜色显示两个字符串序列的差异?
How to show diff of two string sequences in colors?
我正在尝试寻找 Python 区分字符串的方法。我知道 difflib
,但我一直没能找到一种内联模式,其功能类似于 this JS library(绿色插入,红色删除):
one_string = "beep boop"
other_string = "beep boob blah"
有办法实现吗?
您可以使用 ndiff.
示例...
import difflib
cases=[('afrykanerskojęzyczny', 'afrykanerskojęzycznym'),
('afrykanerskojęzyczni', 'nieafrykanerskojęzyczni'),
('afrykanerskojęzycznym', 'afrykanerskojęzyczny'),
('nieafrykanerskojęzyczni', 'afrykanerskojęzyczni'),
('nieafrynerskojęzyczni', 'afrykanerskojzyczni'),
('abcdefg','xac')]
for a,b in cases:
print('{} => {}'.format(a,b))
for i,s in enumerate(difflib.ndiff(a, b)):
if s[0]==' ': continue
elif s[0]=='-':
print(u'Delete "{}" from position {}'.format(s[-1],i))
elif s[0]=='+':
print(u'Add "{}" to position {}'.format(s[-1],i))
print()
Returns.......
afrykanerskojęzyczny => afrykanerskojęzycznym
Add "m" to position 20
afrykanerskojęzyczni => nieafrykanerskojęzyczni
Add "n" to position 0
Add "i" to position 1
Add "e" to position 2
afrykanerskojęzycznym => afrykanerskojęzyczny
Delete "m" from position 20
nieafrykanerskojęzyczni => afrykanerskojęzyczni
Delete "n" from position 0
Delete "i" from position 1
Delete "e" from position 2
nieafrynerskojęzyczni => afrykanerskojzyczni
Delete "n" from position 0
Delete "i" from position 1
Delete "e" from position 2
Add "k" to position 7
Add "a" to position 8
Delete "ę" from position 16
abcdefg => xac
Add "x" to position 0
Delete "b" from position 2
Delete "d" from position 4
Delete "e" from position 5
Delete "f" from position 6
Delete "g" from position 7
查看此post了解更多信息..
Python - difference between two strings
试试这个基于 Minimum Edit Distance, in this case I use this algorithm 的解决方案来计算距离矩阵。之后,矩阵上的迭代从前到后以确定字符串中包含或删除了哪些字符,因为我需要反转结果。
我使用 colorama 模块为终端着色。
#!/bin/python
import sys
from colorama import *
from numpy import zeros
init()
inv_WHITE = Fore.WHITE[::-1]
inv_RED = Fore.RED[::-1]
inv_GREEN = Fore.GREEN[::-1]
def edDistDp(y, x):
res = inv_WHITE
D = zeros((len(x)+1, len(y)+1), dtype=int)
D[0, 1:] = range(1, len(y)+1)
D[1:, 0] = range(1, len(x)+1)
for i in xrange(1, len(x)+1):
for j in xrange(1, len(y)+1):
delt = 1 if x[i-1] != y[j-1] else 0
D[i, j] = min(D[i-1, j-1]+delt, D[i-1, j]+1, D[i, j-1]+1)
#print D
# iterate the matrix's values from back to forward
i = len(x)
j = len(y)
while i > 0 and j > 0:
diagonal = D[i-1, j-1]
upper = D[i, j-1]
left = D[i-1, j]
# check back direction
direction = "\" if diagonal <= upper and diagonal <= left else "<-" if left < diagonal and left <= upper else "^"
#print "(",i,j,")",diagonal, upper, left, direction
i = i-1 if direction == "<-" or direction == "\" else i
j = j-1 if direction == "^" or direction == "\" else j
# Colorize caracters
if (direction == "\"):
if D[i+1, j+1] == diagonal:
res += x[i] + inv_WHITE
elif D[i+1, j+1] > diagonal:
res += y[j] + inv_RED
res += x[i] + inv_GREEN
else:
res += x[i] + inv_GREEN
res += y[j] + inv_RED
elif (direction == "<-"):
res += x[i] + inv_GREEN
elif (direction == "^"):
res += y[j] + inv_RED
return res[::-1]
one_string = "beep boop"
other_string = "beep boob blah"
print ("'%s'-'%s'='%s'" % (one_string, other_string, edDistDp(one_string, other_string)))
print ("'%s'-'%s'='%s'" % (other_string, one_string, edDistDp(other_string, one_string)))
other_string = "hola nacho"
one_string = "hola naco"
print ("'%s'-'%s'='%s'" % (one_string, other_string, edDistDp(one_string, other_string)))
print ("'%s'-'%s'='%s'" % (other_string, one_string, edDistDp(other_string, one_string)))
使用@interjay 的评论,我得到了
import difflib
red = lambda text: f"3[38;2;255;0;0m{text}3[38;2;255;255;255m"
green = lambda text: f"3[38;2;0;255;0m{text}3[38;2;255;255;255m"
blue = lambda text: f"3[38;2;0;0;255m{text}3[38;2;255;255;255m"
white = lambda text: f"3[38;2;255;255;255m{text}3[38;2;255;255;255m"
def get_edits_string(old, new):
result = ""
codes = difflib.SequenceMatcher(a=old, b=new).get_opcodes()
for code in codes:
if code[0] == "equal":
result += white(old[code[1]:code[2]])
elif code[0] == "delete":
result += red(old[code[1]:code[2]])
elif code[0] == "insert":
result += green(new[code[3]:code[4]])
elif code[0] == "replace":
result += (red(old[code[1]:code[2]]) + green(new[code[3]:code[4]]))
return result
这仅取决于 difflib
,并且可以使用
进行测试
one_string = "beep boop"
other_string = "beep boob blah"
print(get_edits_string(one_string, other_string))
我正在尝试寻找 Python 区分字符串的方法。我知道 difflib
,但我一直没能找到一种内联模式,其功能类似于 this JS library(绿色插入,红色删除):
one_string = "beep boop"
other_string = "beep boob blah"
有办法实现吗?
您可以使用 ndiff.
示例...
import difflib
cases=[('afrykanerskojęzyczny', 'afrykanerskojęzycznym'),
('afrykanerskojęzyczni', 'nieafrykanerskojęzyczni'),
('afrykanerskojęzycznym', 'afrykanerskojęzyczny'),
('nieafrykanerskojęzyczni', 'afrykanerskojęzyczni'),
('nieafrynerskojęzyczni', 'afrykanerskojzyczni'),
('abcdefg','xac')]
for a,b in cases:
print('{} => {}'.format(a,b))
for i,s in enumerate(difflib.ndiff(a, b)):
if s[0]==' ': continue
elif s[0]=='-':
print(u'Delete "{}" from position {}'.format(s[-1],i))
elif s[0]=='+':
print(u'Add "{}" to position {}'.format(s[-1],i))
print()
Returns.......
afrykanerskojęzyczny => afrykanerskojęzycznym
Add "m" to position 20
afrykanerskojęzyczni => nieafrykanerskojęzyczni
Add "n" to position 0
Add "i" to position 1
Add "e" to position 2
afrykanerskojęzycznym => afrykanerskojęzyczny
Delete "m" from position 20
nieafrykanerskojęzyczni => afrykanerskojęzyczni
Delete "n" from position 0
Delete "i" from position 1
Delete "e" from position 2
nieafrynerskojęzyczni => afrykanerskojzyczni
Delete "n" from position 0
Delete "i" from position 1
Delete "e" from position 2
Add "k" to position 7
Add "a" to position 8
Delete "ę" from position 16
abcdefg => xac
Add "x" to position 0
Delete "b" from position 2
Delete "d" from position 4
Delete "e" from position 5
Delete "f" from position 6
Delete "g" from position 7
查看此post了解更多信息..
Python - difference between two strings
试试这个基于 Minimum Edit Distance, in this case I use this algorithm 的解决方案来计算距离矩阵。之后,矩阵上的迭代从前到后以确定字符串中包含或删除了哪些字符,因为我需要反转结果。
我使用 colorama 模块为终端着色。
#!/bin/python
import sys
from colorama import *
from numpy import zeros
init()
inv_WHITE = Fore.WHITE[::-1]
inv_RED = Fore.RED[::-1]
inv_GREEN = Fore.GREEN[::-1]
def edDistDp(y, x):
res = inv_WHITE
D = zeros((len(x)+1, len(y)+1), dtype=int)
D[0, 1:] = range(1, len(y)+1)
D[1:, 0] = range(1, len(x)+1)
for i in xrange(1, len(x)+1):
for j in xrange(1, len(y)+1):
delt = 1 if x[i-1] != y[j-1] else 0
D[i, j] = min(D[i-1, j-1]+delt, D[i-1, j]+1, D[i, j-1]+1)
#print D
# iterate the matrix's values from back to forward
i = len(x)
j = len(y)
while i > 0 and j > 0:
diagonal = D[i-1, j-1]
upper = D[i, j-1]
left = D[i-1, j]
# check back direction
direction = "\" if diagonal <= upper and diagonal <= left else "<-" if left < diagonal and left <= upper else "^"
#print "(",i,j,")",diagonal, upper, left, direction
i = i-1 if direction == "<-" or direction == "\" else i
j = j-1 if direction == "^" or direction == "\" else j
# Colorize caracters
if (direction == "\"):
if D[i+1, j+1] == diagonal:
res += x[i] + inv_WHITE
elif D[i+1, j+1] > diagonal:
res += y[j] + inv_RED
res += x[i] + inv_GREEN
else:
res += x[i] + inv_GREEN
res += y[j] + inv_RED
elif (direction == "<-"):
res += x[i] + inv_GREEN
elif (direction == "^"):
res += y[j] + inv_RED
return res[::-1]
one_string = "beep boop"
other_string = "beep boob blah"
print ("'%s'-'%s'='%s'" % (one_string, other_string, edDistDp(one_string, other_string)))
print ("'%s'-'%s'='%s'" % (other_string, one_string, edDistDp(other_string, one_string)))
other_string = "hola nacho"
one_string = "hola naco"
print ("'%s'-'%s'='%s'" % (one_string, other_string, edDistDp(one_string, other_string)))
print ("'%s'-'%s'='%s'" % (other_string, one_string, edDistDp(other_string, one_string)))
使用@interjay 的评论,我得到了
import difflib
red = lambda text: f"3[38;2;255;0;0m{text}3[38;2;255;255;255m"
green = lambda text: f"3[38;2;0;255;0m{text}3[38;2;255;255;255m"
blue = lambda text: f"3[38;2;0;0;255m{text}3[38;2;255;255;255m"
white = lambda text: f"3[38;2;255;255;255m{text}3[38;2;255;255;255m"
def get_edits_string(old, new):
result = ""
codes = difflib.SequenceMatcher(a=old, b=new).get_opcodes()
for code in codes:
if code[0] == "equal":
result += white(old[code[1]:code[2]])
elif code[0] == "delete":
result += red(old[code[1]:code[2]])
elif code[0] == "insert":
result += green(new[code[3]:code[4]])
elif code[0] == "replace":
result += (red(old[code[1]:code[2]]) + green(new[code[3]:code[4]]))
return result
这仅取决于 difflib
,并且可以使用
one_string = "beep boop"
other_string = "beep boob blah"
print(get_edits_string(one_string, other_string))