python文本文件到行和列

Question

所以我已经尝试了一段时间，似乎遇到了障碍并希望得到帮助。

我有几个文本文件。没有把它全部写出来，这是一个例子：

2020
Grum Grum
Stamina: 20
Agility: 23
Strength: 20.5%
Resistances: 20-21-30

2020
Mondo Silo
Stamina: 23
Agility: 13
Strength: 10.5%
Resistances: 20-21-20

等等等等。有些像这样每 6 行开始一个新的统计文件，有些文本文件有它，所以每 10 行就有一个新的统计表。

我的目标是每次统计表结束时，将其放入一行和列中。我认为这在电子表格术语中被称为转置，但我不知道我做错了什么。或者即使这样说是正确的..

例如，我希望文件在完成后看起来像这样。

Year | Name | Stamina | Agility | Str | Res
2020 | Grum Grum | Stamina: 20 | Agility: 23 | Strength: 20.5% | Resistances: 20-21-30

我已经尝试过 Numpy、Pandas 和 idk 我做错了什么，老实说不知道要搜索什么才能找到正确的答案。

如果我能得到任何帮助，我将不胜感激，这些文件非常大，我希望能够具体说明我需要统计表填写的列数。

如果您能提供帮助，请提前致谢。

Answer 1

你可以试试这个来获得所需的数据框：

with open(r'test1.txt','r') as file:
    data=file.read().split('\n\n')
data=[i.split('\n') for i in data]
df=pd.DataFrame(data,columns=['Year','Name','Stamina','Agility','Str','Res'])

print(df)

输出：

   Year        Name  ...              Str                    Res
0  2020   Grum Grum  ...  Strength: 20.5%  Resistances: 20-21-30
1  2020  Mondo Silo  ...  Strength: 10.5%  Resistances: 20-21-20
2  2020   Grum Grum  ...  Strength: 20.5%  Resistances: 20-21-30
3  2020  Mondo Silo  ...  Strength: 10.5%  Resistances: 20-21-20

并编写.txt具有不同行数和相同结构的文件列表的数据帧，您可以尝试：

选项1

import pandas as pd

files=['test1.txt','test2.txt']                     #list of files

df=pd.DataFrame(columns=['Year','Name','Stamina','Agility','Str','Res'])  #create the dataframe

for file in files:                                  #we open each file
    with open(r'path_of_files'+file,'r') as file_r:   
        data=file_r.read().strip().split('\n\n')
        data=[i.split('\n') for i in data if i!=''] #get the rows
        print(data)
        s = pd.DataFrame(data, columns=df.columns)  
        df =pd.concat([df, s], ignore_index=True)   #we append the new rows to the dataframe
        
        
print(df)
df.to_csv(r'test3.txt', sep='|', index=False)       #write the final dataframe to the output file('test3.txt'), with '|' as separator 

选项 2

import pandas as pd

files=['test1.txt','test2.txt']                      #list of files

for file in files:                                   #we open each file
    with open(r'path_of_files'+file,'r') as file_r, open(r'test3.txt', 'a') as fout:
        data=file_r.read().strip().split('\n\n')
        data=[i.split('\n') for i in data if i!='']
        df=pd.DataFrame(data,columns=['Year','Name','Stamina','Agility','Str','Res'])   #create a dataframe with the data of the current file
        if files.index(file)==0:
            fout.write(df.to_string( index = False)) #we let header=true to the first iteration to write the columns, and also write the data
        else:
            fout.write(df.to_string(header = False, index = False))  #we write the dataframe without the index and the columns names
        fout.write('\n')                             #a newline to place correctly the next rows

示例
使用一些像下面 ( test1.txt,test2.txt) 的虚拟文件，您可以看到test3.txt带有两个选项的结果 ( )：

测试1.txt

2020
Grum Grum
Stamina: 20
Agility: 23
Strength: 20.5%
Resistances: 20-21-30

2020
Mondo Silo
Stamina: 23
Agility: 13
Strength: 10.5%
Resistances: 20-21-20

测试2.txt

2020
Grum Grum
Stamina: 20
Agility: 23
Strength: 20.5%
Resistances: 20-21-30

2020
Mondo Silo
Stamina: 23
Agility: 13
Strength: 10.5%
Resistances: 20-21-20

2020
Mondo Silo
Stamina: 23
Agility: 13
Strength: 10.5%
Resistances: 20-21-20

2020
Mondo Silo
Stamina: 23
Agility: 13
Strength: 10.5%
Resistances: 20-21-20

test3.txt（输出文件）与选项 1

Year|Name|Stamina|Agility|Str|Res
2020|Grum Grum|Stamina: 20|Agility: 23|Strength: 20.5%|Resistances: 20-21-30
2020|Mondo Silo|Stamina: 23|Agility: 13|Strength: 10.5%|Resistances: 20-21-20
2020|Grum Grum|Stamina: 20|Agility: 23|Strength: 20.5%|Resistances: 20-21-30
2020|Mondo Silo|Stamina: 23|Agility: 13|Strength: 10.5%|Resistances: 20-21-20
2020|Mondo Silo|Stamina: 23|Agility: 13|Strength: 10.5%|Resistances: 20-21-20
2020|Mondo Silo|Stamina: 23|Agility: 13|Strength: 10.5%|Resistances: 20-21-20

带有选项 2 的test3.txt（输出文件）

 Year        Name      Stamina      Agility              Str                    Res
 2020   Grum Grum  Stamina: 20  Agility: 23  Strength: 20.5%  Resistances: 20-21-30
 2020  Mondo Silo  Stamina: 23  Agility: 13  Strength: 10.5%  Resistances: 20-21-20
 2020   Grum Grum  Stamina: 20  Agility: 23  Strength: 20.5%  Resistances: 20-21-30
 2020  Mondo Silo  Stamina: 23  Agility: 13  Strength: 10.5%  Resistances: 20-21-20
 2020  Mondo Silo  Stamina: 23  Agility: 13  Strength: 10.5%  Resistances: 20-21-20
 2020  Mondo Silo  Stamina: 23  Agility: 13  Strength: 10.5%  Resistances: 20-21-20