Python 如何从文件中进行计算？_Python

Python 如何从文件中进行计算？

python

Python 如何从文件中进行计算？,python,Python,我有一个包含两列的文件，需要对它们应用这个等式，就像 x y 1.2 6.8 2.5 7.0 3 8 4 9 5 10 方程式是 de = sqrt((xi-xj)^2-(yi-yj)^2) 这意味着结果将是一列 row1 = sqrt((x1-x2)^2-(y1-y2)^2) row2 = sqrt((x1-x3)^2-(y1-y3)^2) 对每个点x1和其他点做这个方程，对其他点做y1，直到完成，然后开始计算 row 6 = sqrt((x2-x3)

我有一个包含两列的文件，需要对它们应用这个等式，就像

方程式是

de = sqrt((xi-xj)^2-(yi-yj)^2)

这意味着结果将是一列

row1 = sqrt((x1-x2)^2-(y1-y2)^2)

row2 = sqrt((x1-x3)^2-(y1-y3)^2)

对每个点x1和其他点做这个方程，对其他点做y1，直到完成，然后开始计算

row 6 = sqrt((x2-x3)^2-(y2-y3)^2)

row 7 = sqrt((x2-x4)^2-(y2-y4)^2)

对每个点做这个等式，x2到其他点，y2到其他点，直到完成，依此类推，直到完成所有的x和y，并将结果存储在一个文件中

我尝试使用2个数组来实现这一点，并将数字存储在其中，然后进行计算，但数据太大，数组将是错误的选择。。如何在python中实现这一点。。从文件中读取每个值的I和j

我的努力，如果太糟糕，我会感到抱歉

import math
with open('columnss.txt', 'r', encoding='utf-8') as f:
      for line in f: 
           [x, y] = (int(n) for n in line.split())
           d = math.sqrt(((x[0] - y[0])**2) + ((x[1] - y[1])** 2)) 
           with open('result.txt', 'w', encoding='utf-8') as f1:
                  f1.write( str(d) + '\n')

我得到

我在excel中进行了计算，但也尝试使用python进行计算我应该把每一列放在一个单独的文件中，以便更容易捕捉数字，还是可以用同一个文件来实现这一点

*您需要在输入文件中循环两次。第二个循环可以跳过第一个循环中该行之前的所有行

如果可以将文件内容加载到列表或数组中，则可以通过迭代索引而不是跳过行来更轻松地完成此操作

此外，您应该只打开输出文件一次。每次通过循环时都会覆盖它

import cmath

with open('columnss.txt', 'r', encoding='utf-8') as f1, open('columnss.txt', 'r', encoding='utf-8') as f2, open('result.txt', 'w', encoding='utf-8') as outfile:
    for i1, line in enumerate(f1):
        x1, y1 = (float(n) for n in line.split())
        f2.seek(0)
        for i2, line in enumerate(f2):
            if i1 < i2:
                x2, y2 = (float(n) for n in line.split())
                print(cmath.sqrt((x1-x2)**2-(y1-y2)**2), file=outfile)

导入cmath
将open（'columnss.txt'，'r'，encoding='utf-8'）作为f1，将open（'columnss.txt'，'r'，encoding='utf-8'）作为f2，将open（'result.txt'，'w'，encoding='utf-8'）作为输出文件：
对于i1，枚举（f1）中的行：
x1，y1=（在line.split（）中n的浮点（n）
f2.寻道（0）
对于i2，枚举（f2）中的行：
如果i1

试试这个：
import math
with open('result.txt', 'w', encoding='utf-8') as f1:
    with open('columnss.txt', 'r', encoding='utf-8') as f:
           while True :
              line=f.readline()
              [x, y] = (int(float(n)) for n in line.split())
              if ((x[0] - y[0])**2) + ((x[1] - y[1])** 2)> 0:
                d = math.sqrt(((x[0] - y[0])**2) + ((x[1] - y[1])** 2)) 
                f1.write(line +':' str(d) + '\n')
              if not line :
                 break 
f.close()
f.close()

每当有一个问题看起来像是可以在excel工作表中完成的，并且想要启用python方法来完成它时，我就使用pandas
我想你也可以使用熊猫
下面是读取“columns.txt”文件并输出为“output.csv”的代码
import pandas as pd
import cmath
df = pd.read_csv('columns.txt', sep=r"\s+") # read columns.txt into a dataframe, using space as deliimter
df.dropna(inplace=True,axis=1)                 # multiple whitespaces create NA columns. Better to use csv file
df = df.astype(float)                          # specify the columsn as type float
print("-"*20 + "Input" + "-"*20)
print(df)                                      # 
print("-"*50)

for index, row in df.iterrows():
    origin=row                              # specify current row as origin

    '''
    Adding  equation column
    Here we are using a lambda function (same as de used in the question)
    and creating a new column called equation
    '''
    df["equation from row {}".format(index)]=df.apply(lambda row_lambda: cmath.sqrt((origin.x-row_lambda.x)**2 - (origin.y-row_lambda.y)**2), axis=1)

print("-"*20 + "Output" + "-"*20)
print(df)
print("-"*50)

# Save this output as csv file (even excel is possible)
df.to_csv('Output.csv')```


The output will look like:

    --------------------Input--------------------
             x         y
    0 -99.9580 -28.84930
    1 -71.5378 -26.77280
    2 -91.6913 -40.90390
    3 -69.0989 -12.95010
    4 -79.6443  -9.20575
    5 -92.1975 -20.02760
    6 -99.7732 -14.26070
    7 -80.3767 -18.16040
    --------------------------------------------------
    --------------------Output--------------------
             x         y      distance from row 0      distance from row 1  \
    0 -99.9580 -28.84930                       0j  (28.344239552155912+0j)   
    1 -71.5378 -26.77280  (28.344239552155912+0j)                       0j   
    2 -91.6913 -40.90390       8.773542743384796j  (14.369257985017867+0j)   
    3 -69.0989 -12.95010  (26.448052710360358+0j)      13.605837059144871j   
    4 -79.6443  -9.20575   (5.174683670283624+0j)      15.584797189970107j   
    5 -92.1975 -20.02760       4.194881481043308j  (19.527556965734348+0j)   
    6 -99.7732 -14.26070      14.587429482948666j   (25.31175945583396+0j)   
    7 -80.3767 -18.16040   (16.40654523292457+0j)  (1.9881447256173002+0j)   
    
           distance from row 2      distance from row 3      distance from row 4  \
    0       8.773542743384796j  (26.448052710360358+0j)   (5.174683670283624+0j)   
    1  (14.369257985017867+0j)      13.605837059144871j     -15.584797189970107j   
    2                       0j      16.462028935705348j      29.319660714655278j   
    3      16.462028935705348j                       0j   (9.858260710566546-0j)   
    4      29.319660714655278j   (9.858260710566546+0j)                       0j   
    5       20.87016203219323j  (21.987594586720945+0j)   (6.361634445447185+0j)   
    6      25.387851398454337j  (30.646288651809048+0j)  (19.483841913429192+0j)   
    7       19.72933397482034j  (10.002077121778257+0j)       8.924648276682952j   
    
           distance from row 5      distance from row 6      distance from row 7  
    0       4.194881481043308j      14.587429482948666j   (16.40654523292457+0j)  
    1  (19.527556965734348-0j)   (25.31175945583396-0j)  (1.9881447256173002-0j)  
    2      -20.87016203219323j     -25.387851398454337j       19.72933397482034j  
    3  (21.987594586720945+0j)  (30.646288651809048+0j)  (10.002077121778257+0j)  
    4   (6.361634445447185+0j)  (19.483841913429192+0j)       8.924648276682952j  
    5                       0j   (4.912646423263124-0j)  (11.672398074089152+0j)  
    6   (4.912646423263124+0j)                       0j  (19.000435578165046+0j)  
    7  (11.672398074089152+0j)  (19.000435578165046-0j)                       0j  
    --------------------------------------------------

To know more about pandas:
[https://pandas.pydata.org/docs/][1]

Stackoverflow itself is an excellent resource for gathering all way of using pandas.


  [1]: https://pandas.pydata.org/docs/



Here column names are defined as 'x' and 'y' in the header.
If the column names are not specified you can add a new header by:
df.columns=['x','y'] 
after reading the csv file (or text file).

If it already has a header and want to use that name just specify that in the lambdas formula.

Please see: 
https://stackoverflow.com/questions/14365542/import-csv-file-as-a-pandas-dataframe

Hope this helps

每当有一个问题看起来像是可以在excel工作表中完成的，并且想要启用python方法来完成它时，我就使用pandas
我想你也可以使用熊猫
下面是读取“columns.txt”文件并输出为“output.csv”的代码，该文件查找每行与其他行的距离，并添加一个新列
import pandas as pd
import cmath
df = pd.read_csv('columns.txt', sep=r"\s+") # read columns.txt into a dataframe, using space as deliimter
df.dropna(inplace=True,axis=1)                 # multiple whitespaces create NA columns. Better to use csv file
df = df.astype(float)                          # specify the columsn as type float
print("-"*20 + "Input" + "-"*20)
print(df)                                      # 
print("-"*50)

for index, row in df.iterrows():
    origin=row                              # specify first row as origin

    '''
    Adding distance column
    Here we are using a lambda function (same as de used in the question)
    and creating a new column called distance
    '''
    df["distance from row {}".format(index)]=df.apply(lambda row_lambda: cmath.sqrt((origin.x-row_lambda.x)**2 - (origin.y-row_lambda.y)**2), axis=1)

print("-"*20 + "Output" + "-"*20)
print(df)
print("-"*50)

# Save this output as csv file (even excel is possible)
df.to_csv('Output.csv')```


The output will look like:

--------------------Input--------------------
         x         y
0 -99.9580 -28.84930
1 -71.5378 -26.77280
2 -91.6913 -40.90390
3 -69.0989 -12.95010
4 -79.6443  -9.20575
5 -92.1975 -20.02760
6 -99.7732 -14.26070
7 -80.3767 -18.16040
--------------------------------------------------
--------------------Output--------------------
         x         y      distance from row 0      distance from row 1  \
0 -99.9580 -28.84930                       0j  (28.344239552155912+0j)   
1 -71.5378 -26.77280  (28.344239552155912+0j)                       0j   
2 -91.6913 -40.90390       8.773542743384796j  (14.369257985017867+0j)   
3 -69.0989 -12.95010  (26.448052710360358+0j)      13.605837059144871j   
4 -79.6443  -9.20575   (5.174683670283624+0j)      15.584797189970107j   
5 -92.1975 -20.02760       4.194881481043308j  (19.527556965734348+0j)   
6 -99.7732 -14.26070      14.587429482948666j   (25.31175945583396+0j)   
7 -80.3767 -18.16040   (16.40654523292457+0j)  (1.9881447256173002+0j)   

       distance from row 2      distance from row 3      distance from row 4  \
0       8.773542743384796j  (26.448052710360358+0j)   (5.174683670283624+0j)   
1  (14.369257985017867+0j)      13.605837059144871j     -15.584797189970107j   
2                       0j      16.462028935705348j      29.319660714655278j   
3      16.462028935705348j                       0j   (9.858260710566546-0j)   
4      29.319660714655278j   (9.858260710566546+0j)                       0j   
5       20.87016203219323j  (21.987594586720945+0j)   (6.361634445447185+0j)   
6      25.387851398454337j  (30.646288651809048+0j)  (19.483841913429192+0j)   
7       19.72933397482034j  (10.002077121778257+0j)       8.924648276682952j   

       distance from row 5      distance from row 6      distance from row 7  
0       4.194881481043308j      14.587429482948666j   (16.40654523292457+0j)  
1  (19.527556965734348-0j)   (25.31175945583396-0j)  (1.9881447256173002-0j)  
2      -20.87016203219323j     -25.387851398454337j       19.72933397482034j  
3  (21.987594586720945+0j)  (30.646288651809048+0j)  (10.002077121778257+0j)  
4   (6.361634445447185+0j)  (19.483841913429192+0j)       8.924648276682952j  
5                       0j   (4.912646423263124-0j)  (11.672398074089152+0j)  
6   (4.912646423263124+0j)                       0j  (19.000435578165046+0j)  
7  (11.672398074089152+0j)  (19.000435578165046-0j)                       0j  
--------------------------------------------------
To know more about pandas:
[https://pandas.pydata.org/docs/][1]

Stackoverflow itself is an excellent resource for gathering all way of using pandas.


  [1]: https://pandas.pydata.org/docs/

仍然得到相同的错误[x，y]=（int（n）表示line.split（）中的n）ValueError:int（）的文本无效，以10为基数：'-9.0检查更新的解决方案感谢您的帮助我检查并获得了ValueError:math domain error问题在于您的值有一些CA，您试图对负值进行sqr，尝试我的更新我从计算中删除了负值我对你的答案进行了投票，因为你帮助了我。。谢谢但是我仍然有一个问题，那就是当我得到错误数字的结果时，将要计算的点序列。这不是一个距离公式。这是减去差异的平方，而不是相加。我的错误-我认为cmath将非常容易地在这个基础上工作df[“距行{}.format（index）]=df.apply（lambda row_lambda:cmath.sqrt（（origin.x-row_lambda.x）**2-（origin y-row_lambda.y）**2），axis=1）什么是origin？问题是他想用每一行计算公式，而不仅仅是从第一行开始。我想我已经回答了两次——我要删除这个版本。也许通过标题工作会有所帮助。我已经看到，一旦你开始使用熊猫，它就会变得非常高效和直观。谢谢你的努力。我尝试了代码，但得到了第2行，在df=pd.read_csv（'columnss.txt'，delimiter=“”）pandas.errors.parserror:Error标记化数据。C错误：第1792行应为1个字段，SAW3您可以将columns.txt保存为csv文件并直接读取。我没有看到使用示例输入时出现的错误。我现在再试一次，我想可能是因为您的数据文件中有空行。我使用了增量变量，但它不会影响代码。。今天我从你那里学到了很多。。谢谢
import pandas as pd
import cmath
df = pd.read_csv('columns.txt', sep=r"\s+") # read columns.txt into a dataframe, using space as deliimter
df.dropna(inplace=True,axis=1)                 # multiple whitespaces create NA columns. Better to use csv file
df = df.astype(float)                          # specify the columsn as type float
print("-"*20 + "Input" + "-"*20)
print(df)                                      # 
print("-"*50)

for index, row in df.iterrows():
    origin=row                              # specify first row as origin

    '''
    Adding distance column
    Here we are using a lambda function (same as de used in the question)
    and creating a new column called distance
    '''
    df["distance from row {}".format(index)]=df.apply(lambda row_lambda: cmath.sqrt((origin.x-row_lambda.x)**2 - (origin.y-row_lambda.y)**2), axis=1)

print("-"*20 + "Output" + "-"*20)
print(df)
print("-"*50)

# Save this output as csv file (even excel is possible)
df.to_csv('Output.csv')```


The output will look like:

--------------------Input--------------------
         x         y
0 -99.9580 -28.84930
1 -71.5378 -26.77280
2 -91.6913 -40.90390
3 -69.0989 -12.95010
4 -79.6443  -9.20575
5 -92.1975 -20.02760
6 -99.7732 -14.26070
7 -80.3767 -18.16040
--------------------------------------------------
--------------------Output--------------------
         x         y      distance from row 0      distance from row 1  \
0 -99.9580 -28.84930                       0j  (28.344239552155912+0j)   
1 -71.5378 -26.77280  (28.344239552155912+0j)                       0j   
2 -91.6913 -40.90390       8.773542743384796j  (14.369257985017867+0j)   
3 -69.0989 -12.95010  (26.448052710360358+0j)      13.605837059144871j   
4 -79.6443  -9.20575   (5.174683670283624+0j)      15.584797189970107j   
5 -92.1975 -20.02760       4.194881481043308j  (19.527556965734348+0j)   
6 -99.7732 -14.26070      14.587429482948666j   (25.31175945583396+0j)   
7 -80.3767 -18.16040   (16.40654523292457+0j)  (1.9881447256173002+0j)   

       distance from row 2      distance from row 3      distance from row 4  \
0       8.773542743384796j  (26.448052710360358+0j)   (5.174683670283624+0j)   
1  (14.369257985017867+0j)      13.605837059144871j     -15.584797189970107j   
2                       0j      16.462028935705348j      29.319660714655278j   
3      16.462028935705348j                       0j   (9.858260710566546-0j)   
4      29.319660714655278j   (9.858260710566546+0j)                       0j   
5       20.87016203219323j  (21.987594586720945+0j)   (6.361634445447185+0j)   
6      25.387851398454337j  (30.646288651809048+0j)  (19.483841913429192+0j)   
7       19.72933397482034j  (10.002077121778257+0j)       8.924648276682952j   

       distance from row 5      distance from row 6      distance from row 7  
0       4.194881481043308j      14.587429482948666j   (16.40654523292457+0j)  
1  (19.527556965734348-0j)   (25.31175945583396-0j)  (1.9881447256173002-0j)  
2      -20.87016203219323j     -25.387851398454337j       19.72933397482034j  
3  (21.987594586720945+0j)  (30.646288651809048+0j)  (10.002077121778257+0j)  
4   (6.361634445447185+0j)  (19.483841913429192+0j)       8.924648276682952j  
5                       0j   (4.912646423263124-0j)  (11.672398074089152+0j)  
6   (4.912646423263124+0j)                       0j  (19.000435578165046+0j)  
7  (11.672398074089152+0j)  (19.000435578165046-0j)                       0j  
--------------------------------------------------
To know more about pandas:
[https://pandas.pydata.org/docs/][1]

Stackoverflow itself is an excellent resource for gathering all way of using pandas.


  [1]: https://pandas.pydata.org/docs/