Python 如何更有效地将字符串中的多个字符映射为单个字符?
我正在寻找一种将字符组映射为单个字符的有效方法 目前,我的代码与以下代码类似:Python 如何更有效地将字符串中的多个字符映射为单个字符?,python,python-3.x,string,dictionary,replace,Python,Python 3.x,String,Dictionary,Replace,我正在寻找一种将字符组映射为单个字符的有效方法 目前,我的代码与以下代码类似: example = 'Accomodation' VOWELS = 'aeiou' CONSONANTS = 'bcdfghjklmnpqrstvwxyz' output = '' for char in example: if char in VOWELS: output += 'v' elif char in VOWELS.upper(): output +=
example = 'Accomodation'
VOWELS = 'aeiou'
CONSONANTS = 'bcdfghjklmnpqrstvwxyz'
output = ''
for char in example:
if char in VOWELS:
output += 'v'
elif char in VOWELS.upper():
output += 'V'
elif char in CONSONANTS:
....
在本例中,它最终将返回vcvvc
我想让这部分更有效率:
for char in example:
if char in VOWELS:
output += 'v'
elif char in VOWELS.upper():
output += 'V'
elif char in CONSONANTS:
....
理想情况下,该解决方案将允许映射到的字符字典作为键,其值为选项列表。例如
replace_dict = {'v': VOWELS,
'V': VOWELS.upper(),
'c': CONSONANTS,
...
我不太熟悉map
,但我希望解决方案能以某种方式利用它
研究
我在这里发现了一个类似的问题:
该问题的解决方案表明,我需要以下内容:
target = 'Accomodation'
charset = 'aeioubcdfghjklmnpqrstvwxyzAEIOUBCDFGHJKLMNPQRSTVWXYZ'
key = 'vvvvvcccccccccccccccccccccVVVVVCCCCCCCCCCCCCCCCCCCCC'
然而,我并不认为这些赋值看起来特别清晰——尽管它节省了一块if
/else
语句。此外,如果我想添加更多字符集,则赋值的可读性甚至更低,例如,对于不同的外来字符集
任何人,也许对内置函数有更好的了解,是否能够制作一个比上述两个例子更高效/干净的例子 我也愿意接受其他不需要使用字典的想法
解决方案应该在
python3
中这是一种使用dict
的方法
Ex:
example = 'Accomodation'
VOWELS = 'aeiou'
CONSONANTS = 'bcdfghjklmnpqrstvwxyz'
replace_dict = {'v': VOWELS,
"V": VOWELS.upper(),
"c": CONSONANTS
}
print("".join(k for i in example
for k, v in replace_dict.items() if i in v
)
)
Vccvcvcvcvvc
输出:
example = 'Accomodation'
VOWELS = 'aeiou'
CONSONANTS = 'bcdfghjklmnpqrstvwxyz'
replace_dict = {'v': VOWELS,
"V": VOWELS.upper(),
"c": CONSONANTS
}
print("".join(k for i in example
for k, v in replace_dict.items() if i in v
)
)
Vccvcvcvcvvc
有一种更有效的方法可以创建这样一个目录:
example = 'Accomodation'
VOWELS = 'aeiou'
CONSONANTS = 'bcdfghjklmnpqrstvwxyz'
replace_dict = {
**{v: 'v' for v in VOWELS},
**{V: 'V' for V in VOWELS.upper()},
**{c: 'c' for c in CONSONANTS}
}
print(''.join(replace_dict[s] for s in example))
# Vccvcvcvcvvc
您的
replace_dict
想法很接近,但最好“翻转”dict“由内而外”,即将其从{v':'aei',c':'bc'}
转换为{a':'v','e':'v','b':'c',…}
def从目录(replace目录)获取目录(replace目录):
替换_map={}
对于cls,替换目录项()中的字符:
替换映射更新(dict.fromkeys(chars,cls))
返回替换映射
def用_图替换_(s,替换_图):
return.''join(替换映射get(c,c)表示s中的c)
元音=“aeiou”
辅音=“bcdfghjklmnpnpqrstvwxyz”
replace\u map=从目录中获取\u replace\u map\u(
{“v”:元音,“v”:元音.upper(),“c”:辅音}
)
打印(用地图替换地图(“住宿,谢谢!”,替换地图))
上面的replace_with_map
函数保留了所有未映射的字符(但您可以使用第二个参数将其更改为.get()
),因此输出是
vccccvccvcc,ccvccc
对你正在做的事情进行反向查找怎么样?应该是可伸缩的
VOWELS = 'aeiou'
CONSONANTS = 'bcdfghjklmnpqrstvwxyz'
example = "Accomodation"
lookup_dict = {k: "v" for k in VOWELS}
lookup_dict.update({k: "c" for k in CONSONANTS})
lookup_dict.update({k: "V" for k in VOWELS.upper()})
lookup_dict.update({k: "C" for k in CONSONANTS.upper()})
''.join([lookup_dict[i] for i in example])
试试这个。不需要辅音,不仅适用于英语,也适用于俄语字母(我很惊讶): vcvcvvvvvvvvvcvvc
我是Python新手,玩Python很有趣。让我们看看这些字典有多好。此处建议的四种算法:
地名时间总计
1.AKX 0.6777 16.5018金牌得主!!!
2.Sanyash 0.8874 21.5725下降31%
3.Alex 0.9573 23.2569下降41%
4.亚当0.9584 23.2210减速41%
然后我对代码做了一些小的改进:
VOWELS_UP = VOWELS.upper()
def vowels_consonants0(example):
output = ''
for char in example:
if char.isalpha():
if char.islower():
output += 'v' if char in VOWELS else 'c'
else:
output += 'V' if char in VOWELS_UP else 'C'
return output
这让我获得了第二名:
地名时间总计
1.AKX 0.6825 16.5331金牌得主!!!
2.亚历克斯0.7026 17.1036下跌3%
3.Sanyash 0.8557 20.8817下降25%
4.亚当0.9631 23.3327下跌41%
现在我需要剃掉这3%的胡子,获得第一名。我用列夫·托尔斯泰小说的文本进行了测试
原始源代码:
import time
import itertools
VOWELS = 'eaiouу' # in order of letter frequency
CONSONANTS = 'bcdfghjklmnpqrstvwxyz'
def vowels_consonants0(example):
output = ''
for char in example:
if char.isalpha():
x = 'v' if char.lower() in VOWELS else 'c'
output += x if char.islower() else x.upper()
return output
def vowels_consonants1(example):
output = ''
for char in example:
if char in VOWELS:
output += 'v'
elif char in VOWELS.upper():
output += 'V'
elif char in CONSONANTS:
output += 'c'
elif char in CONSONANTS.upper():
output += 'C'
return output
def vowels_consonants2(example):
replace_dict = {
**{v: 'v' for v in VOWELS},
**{V: 'V' for V in VOWELS.upper()},
**{c: 'c' for c in CONSONANTS},
**{c: 'c' for c in CONSONANTS.upper()}
}
return ''.join(replace_dict[s] if s in replace_dict else '' for s in example)
def get_replace_map_from_dict(replace_dict):
replace_map = {}
for cls, chars in replace_dict.items():
replace_map.update(dict.fromkeys(chars, cls))
return replace_map
def replace_with_map(s, replace_map):
return "".join(replace_map.get(c, c) for c in s)
replace_map = get_replace_map_from_dict(
{"v": VOWELS, "V": VOWELS.upper(), "c": CONSONANTS, "C": CONSONANTS.upper()}
)
def vowels_consonants3(example):
output = ''
for char in example:
if char in replace_map:
output += char
output = replace_with_map(output, replace_map)
return output
def test(function, name):
text = open(name, encoding='utf-8')
t0 = time.perf_counter()
line_number = 0
char_number = 0
vc_number = 0 # vowels and consonants
while True:
line_number += 1
line = text.readline()
if not line:
break
char_number += len(line)
vc_line = function(line)
vc_number += len(vc_line)
t0 = time.perf_counter() - t0
text.close()
return t0, line_number, char_number, vc_number
tests = [vowels_consonants0, vowels_consonants1, vowels_consonants2, vowels_consonants3]
names = ["Alex", "Adam", "Sanyash", "AKX"]
best_time = float('inf')
run_times = [best_time for _ in tests]
sum_times = [0.0 for _ in tests]
show_result = [True for _ in tests]
print("\n!!! Start the race by permutation with no repetitions now ...\n")
print(" * - best time in race so far")
print(" + - personal best time\n")
print("Note Name Time (Permutation)")
products = itertools.permutations([0, 1, 2, 3])
for p in list(products):
print(p)
for n in p:
clock, lines, chars, vcs = test(tests[n], 'war_peace.txt')
sum_times[n] += clock
note = " "
if clock < run_times[n]:
run_times[n] = clock
note = "+" # Improved personal best time
if clock < best_time:
best_time = clock
note = "*" # Improved total best time
print("%s %8s %6.4f" % (note, names[n], clock), end="")
if show_result[n]:
show_result[n] = False
print(" Lines:", lines, "Characters:", chars, "Letters:", vcs)
else:
print()
print("\n!!! Finish !!! and the winner by the best run time is ...\n")
print("Place Name Time Total")
i = 0
for n in sorted(range(len(run_times)), key=run_times.__getitem__):
i += 1
t = run_times[n]
print("%d. %8s %.4f %.4f " % (i, names[n], t, sum_times[n]), end="")
if i == 1:
print("The winner of Gold medal!!!")
else:
print("Slower by %2d%%" % (round(100.0 * (t - best_time)/best_time)))
导入时间
进口itertools
元音='eaiou#'#按字母频率顺序排列
辅音='bcdfghjklmnpnpqrstvwxyz'
def元音辅音S0(示例):
输出=“”
对于示例中的char:
如果char.isalpha():
x='v'如果元音中的char.lower()为'c'
如果char.islower()或x.upper(),则输出+=x
返回输出
def元音和辅音1(示例):
输出=“”
对于示例中的char:
如果元音中有字符:
输出+='v'
元音中的elif char.upper():
输出+='V'
辅音中的elif字符:
输出+='c'
辅音中的elif char.upper():
输出+='C'
返回输出
def元音和辅音2(示例):
替换[u dict={
**{v:'v'代表元音中的v},
**{V:'V'代表元音中的V.upper()},
**{c:'c'表示辅音中的c},
**{c:'c'表示辅音中的c.upper()}
}
返回“”。join(如果示例中的s在replace_dict else“”中为s,则为s替换_dict)
def从目录中获取目录替换映射目录(替换目录):
替换_map={}
对于cls,替换目录项()中的字符:
替换映射更新(dict.fromkeys(chars,cls))
返回替换映射
def用_图替换_(s,替换_图):
return.''join(替换映射get(c,c)表示s中的c)
replace\u map=从目录中获取\u replace\u map\u(
{“v”:元音,“v”:元音.upper(),“c”:辅音,“c”:辅音.upper()}
)
def元音和辅音3(示例):
输出=“”
对于示例中的char:
如果替换映射中的字符:
输出+=字符
输出=用图替换图(输出,替换图)
返回输出
def测试(功能、名称):
text=open(名称,编码='utf-8')
t0=时间。性能计数器()
线号=0
字符数=0
vc_数=0#元音和辅音
尽管如此:
行号+=1
line=text.readline()
如果不是直线:
打破
字符数+=len(行)
vc_线=功能(线)
vc_编号+=len(vc_线)
t0=时间。性能计数器()-t0