在python中重新排列表格

如何解决在python中重新排列表格

我有一个表(example_table.txt),其中包含700多个行。每行包含对应于17个不同类的值。我想以以下方式(Desired_output.text)重新排列我的表

Example_table.txt链接(https://drive.google.com/file/d/1sz9XkPzMqCZItUBN-QugQKq39X0buIoX/view?usp=sharing

Desired_output.txt链接(https://drive.google.com/file/d/1OXm2b4VMbuQ1GqBzBf48bDE_gPyzRpnU/view?usp=sharing

输入表

ID  Class 1 Class 2 Class 3 Class 4 Class 5 Class 6 Class 7 Class 8 Class 9 Class 10    Class 11    Class 12    Class 13
1   0   0.0013865   0   0   0.0005675   0.00317325  0.00008725  0   0.0000925   0   0   0   0
2   0   0.02396475  0   0   0.00045075  0.008391    0.00161075  0   0.00033725  0   0   0   0
3   0   0.0260415   0   0   0   0.0210125   0.011682    0   0.00092125  0   0   0   0
4   0   0.01287525  0   0.00007425  0   0.02698525  0.02130875  0   0.0012565   0   0   0   0
5   0   0.008697    0.00012475  0   0.012641    0.00643825  0.0332455   0   0.00116475  0   0.00018875  0   0

所需的输出

Id  No of class and Class Name  Area
1   5   
    2   0.0013865
    5   0.0005675
    6   0.00317325
    7   0.00008725
    9   0.0000925
2   5   
    2   0.02396475
    5   0.00045075
    6   0.008391
    7   0.00161075
    9   0.00033725
3   4   
    2   0.0260415
    6   0.0210125
    7   0.011682
    9   0.00092125
4   5   
    2   0.01287525
    4   0.00007425
    6   0.02698525
    7   0.02130875
    9   0.0012565
5   7   
    2   0.008697
    3   0.00012475
    5   0.012641
    6   0.00643825
    7   0.0332455
    9   0.00116475
    11  0.00018875

如何使用python以所需的方式重新排列这些数据

感谢前进

解决方法

您可以使用熊猫df.melt

df = pd.read_csv(r'C:\Users\XXX\Downloads\example_table.txt',delimiter='\t')
ExpectedOutput = df.melt(id_vars = 'ID',var_name='No of class and Class Name',value_name='Area')
ExpectedOutput = ExpectedOutput[ExpectedOutput.Area != 0] #remove record with 0 values
ExpectedOutput.sort_values(by=['ID'],inplace=True,ascending=True) #sort the data
ExpectedOutput['No of class and Class Name'] = ExpectedOutput['No of class and Class Name'].str.split(' ').str[1] #split and get class name from "class x" string


 ID No of class and Class Name  Area
1   5   0.000567
1   2   0.001386
1   7   0.000087
1   6   0.003173
1   9   0.000092
... ... ... ...
783 7   0.005627
783 4   0.000896
783 3   0.001235
783 2   0.045130
783 12  0.006651
,

将使用ID和classes数组复制该数组:

classes = np.tile(np.arange(len(df.columns)),(df.shape[0],1)).flatten()
IDs = np.tile(df.index.values.reshape([-1,1]),(1,df.shape[1])).flatten()

X = df.values.flatten()

删除零

mask = X!=0


dfNew = pd.DataFrame()

dfNew['ID'] = IDs[mask]
dfNew['Classes'] = classes[mask]
dfNew['area'] = X[mask]

然后是groupby: dfNew.groupby(['ID','Classes'])['area']。sum()

,

这里是转换数据的一种方法。

from io import StringIO
import pandas as pd

# copy data from original post into triple-quoted string
data='''ID   Class 1  Class 2  Class 3  Class 4  Class 5  Class 6  Class 7  Class 8  Class 9  Class 10     Class 11     Class 12     Class 13
1   0   0.0013865   0   0   0.0005675   0.00317325  0.00008725  0   0.0000925   0   0   0   0
2   0   0.02396475  0   0   0.00045075  0.008391    0.00161075  0   0.00033725  0   0   0   0
3   0   0.0260415   0   0   0   0.0210125   0.011682    0   0.00092125  0   0   0   0
4   0   0.01287525  0   0.00007425  0   0.02698525  0.02130875  0   0.0012565   0   0   0   0
5   0   0.008697    0.00012475  0   0.012641    0.00643825  0.0332455   0   0.00116475  0   0.00018875  0   0
'''

现在分三步处理数据:

# create data frame
df = pd.read_csv(StringIO(data),sep='\s\s+',engine='python',index_col='ID')

# convert 'Class n' to 'n' (with type integer)
df.columns = df.columns.str.replace('Class ','').astype(int).rename('class_num')

# re-shape,filter,sort,rename
df = df.stack().loc[lambda x: x > 0].sort_index().rename('area')

# UPDATE: count of IDs with non-zero area
t = df.groupby(level=0).transform('count').rename('non-zero-count')
df = pd.concat([df,t],axis=1)

# show first 10 rows
df.head(10)

                  area  non-zero-count
ID class_num                          
1  2          0.001386               5
   5          0.000567               5
   6          0.003173               5
   7          0.000087               5
   9          0.000092               5
2  2          0.023965               5
   5          0.000451               5
   6          0.008391               5
   7          0.001611               5
   9          0.000337               5

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