从python中的字符串集中删除不需要的字符

从python中的字符串集中删除不需要的字符,python,string,re,Python,String,Re,我正在尝试清理一组字符串以删除不需要的字符 输入 想要的产出 Lethal Lunch Muscika Typhoon Ten Wentworth Falls One Night Stand Dancinginthewoods Case Key 我试过这个 re.findall('([a-zA-Z ]*)\d*.*',final_df.loc[index, 'Horse']) 这将删除数字后的所有内容,但在第一个条目上保留t。我想知道是否有更好的方法?我会使用re.split: for d

我正在尝试清理一组字符串以删除不需要的字符

输入

想要的产出

Lethal Lunch
Muscika
Typhoon Ten
Wentworth Falls
One Night Stand
Dancinginthewoods 
Case Key
我试过这个

re.findall('([a-zA-Z ]*)\d*.*',final_df.loc[index, 'Horse'])

这将删除数字后的所有内容,但在第一个条目上保留t。我想知道是否有更好的方法?

我会使用
re.split

for d in data.splitlines():
    print(re.split(r'\s+t?[0-9]\+?', d)[0])
结果 说明:它在指定模式匹配的位置拆分字符串,然后获取第一部分。您可能想要调整它,以便其他模式也匹配

熊猫 我刚刚注意到你似乎在使用熊猫——假设你的df看起来是这样的:

                                               Horse
0  Lethal Lunch t5+ 0 0 D 10 t5+ Michael Bell . A...
1  Muscika 1 v5+ W5+ 0 0 D 5 v5+ W5+ D O'Meara . ...
2  Typhoon Ten 1 0 0 D 13 R Hannon . Luke Catton7 .
3  Wentworth Falls 1 cp5+ 0 0 C D 45 cp5+ G Harke...
4  One Night Stand 0 0 D 34 W Jarvis . Silvestre ...
5  Dancinginthewoods 1 0 0 D 24 D Ivory . 14 Jami...
6  Case Key 1 v3 0 0 D 13 v3 M Appleby . Andrew M...
你能行

from operator import itemgetter

df["name"] = df.Horse.str.split('\s+t?[0-9]\+?').map(itemgetter(0))
要获得此信息:

                                               Horse               name
0  Lethal Lunch t5+ 0 0 D 10 t5+ Michael Bell . A...       Lethal Lunch
1  Muscika 1 v5+ W5+ 0 0 D 5 v5+ W5+ D O'Meara . ...            Muscika
2  Typhoon Ten 1 0 0 D 13 R Hannon . Luke Catton7 .        Typhoon Ten
3  Wentworth Falls 1 cp5+ 0 0 C D 45 cp5+ G Harke...    Wentworth Falls
4  One Night Stand 0 0 D 34 W Jarvis . Silvestre ...    One Night Stand
5  Dancinginthewoods 1 0 0 D 24 D Ivory . 14 Jami...  Dancinginthewoods
6  Case Key 1 v3 0 0 D 13 v3 M Appleby . Andrew M...           Case Key


像这样的方法应该会奏效:

filtered_text = list()

for line in text:
    part = ""
    for word in text.split(" "):
        if len(word) <= 3:
            break
        else:
            part = str(part) + " " + str(word)

    part = part[1:] # skip first space
    filtered_text.append(part)
filtered_text=list()
对于文本中的行:
part=“”
对于文本中的单词。拆分(“”):

如果len(word)像这样的东西就足够了吗

input = [
    "Lethal Lunch t5+ 0 0 D 10 t5+ Michael Bell . Alex Jary7 .",
    "Muscika 1 v5+ W5+ 0 0 D 5 v5+ W5+ D O'Meara . Cam Hardie . C5",
    "Typhoon Ten 1 0 0 D 13 R Hannon . Luke Catton7 .",
    "Wentworth Falls 1 cp5+ 0 0 C D 45 cp5+ G Harker . Connor Beasley .",
    "One Night Stand 0 0 D 34 W Jarvis . Silvestre De Sousa . 30 C1 C5",
    "Dancinginthewoods 1 0 0 D 24 D Ivory . 14 Jamie Spencer . 30",
    "Case Key 1 v3 0 0 D 13 v3 M Appleby . Andrew Mullen . 14",
]

for inp in input:
    print(re.findall(r'\b[a-zA-Z ]+\b', inp)[0])
我们基本上忽略了带有数字或奇怪符号的单词。 输出:

Lethal Lunch 
Muscika 
Typhoon Ten 
Wentworth Falls 
One Night Stand 
Dancinginthewoods 
Case Key 

太棒了,谢谢你加上熊猫片,真的很有帮助。非常感谢。
input = [
    "Lethal Lunch t5+ 0 0 D 10 t5+ Michael Bell . Alex Jary7 .",
    "Muscika 1 v5+ W5+ 0 0 D 5 v5+ W5+ D O'Meara . Cam Hardie . C5",
    "Typhoon Ten 1 0 0 D 13 R Hannon . Luke Catton7 .",
    "Wentworth Falls 1 cp5+ 0 0 C D 45 cp5+ G Harker . Connor Beasley .",
    "One Night Stand 0 0 D 34 W Jarvis . Silvestre De Sousa . 30 C1 C5",
    "Dancinginthewoods 1 0 0 D 24 D Ivory . 14 Jamie Spencer . 30",
    "Case Key 1 v3 0 0 D 13 v3 M Appleby . Andrew Mullen . 14",
]

for inp in input:
    print(re.findall(r'\b[a-zA-Z ]+\b', inp)[0])
Lethal Lunch 
Muscika 
Typhoon Ten 
Wentworth Falls 
One Night Stand 
Dancinginthewoods 
Case Key