python-3.x - 使用 pandas 转换文件中的数据时出现 KeyError5

Question

如果总和小于或等于五，我需要从 csv 读取数据并添加两行或多行，然后将其保存到新的 csv 文件中。

预期的输出模式：

 Application  time_taken
  Serv6, Serv2         2.2
         Serv1         2.3
         Serv9         3.3
         Serv4         4.4
         Serv8         4.6
        Serv10         5.0
         Serv7         5.5
         Serv3         6.0
         Serv5         8.0

错误：

Traceback (most recent call last):

  File "<ipython-input-24-fc02128770c1>", line 1, in <module>
    runfile('C:/Users/prasad.namala/Documents/Python Scripts/Programs/New_Csv.py', wdir='C:/Users/prasad.namala/Documents/Python Scripts/Programs')

  File "C:\Users\prasad.namala\AppData\Local\Continuum\anaconda3\lib\site-packages\spyder_kernels\customize\spydercustomize.py", line 827, in runfile
    execfile(filename, namespace)

  File "C:\Users\prasad.namala\AppData\Local\Continuum\anaconda3\lib\site-packages\spyder_kernels\customize\spydercustomize.py", line 110, in execfile
    exec(compile(f.read(), filename, 'exec'), namespace)

  File "C:/Users/prasad.namala/Documents/Python Scripts/Programs/New_Csv.py", line 36, in <module>
    final_csv = value_checker(input_data)

  File "C:/Users/prasad.namala/Documents/Python Scripts/Programs/New_Csv.py", line 32, in value_checker
    return value_checker(new_df)

  File "C:/Users/prasad.namala/Documents/Python Scripts/Programs/New_Csv.py", line 20, in value_checker
    print(f"{df.loc[i,['Application']].item()}, {df.loc[i+1,['Application']].item()}")

  File "C:\Users\prasad.namala\AppData\Local\Continuum\anaconda3\lib\site-packages\pandas\core\indexing.py", line 1494, in __getitem__
    return self._getitem_tuple(key)

  File "C:\Users\prasad.namala\AppData\Local\Continuum\anaconda3\lib\site-packages\pandas\core\indexing.py", line 868, in _getitem_tuple
    return self._getitem_lowerdim(tup)

  File "C:\Users\prasad.namala\AppData\Local\Continuum\anaconda3\lib\site-packages\pandas\core\indexing.py", line 988, in _getitem_lowerdim
    section = self._getitem_axis(key, axis=i)

  File "C:\Users\prasad.namala\AppData\Local\Continuum\anaconda3\lib\site-packages\pandas\core\indexing.py", line 1913, in _getitem_axis
    return self._get_label(key, axis=axis)

  File "C:\Users\prasad.namala\AppData\Local\Continuum\anaconda3\lib\site-packages\pandas\core\indexing.py", line 141, in _get_label
    return self.obj._xs(label, axis=axis)

  File "C:\Users\prasad.namala\AppData\Local\Continuum\anaconda3\lib\site-packages\pandas\core\generic.py", line 3585, in xs
    loc = self.index.get_loc(key)

  File "C:\Users\prasad.namala\AppData\Local\Continuum\anaconda3\lib\site-packages\pandas\core\indexes\base.py", line 2659, in get_loc
    return self._engine.get_loc(self._maybe_cast_indexer(key))

  File "pandas\_libs\index.pyx", line 108, in pandas._libs.index.IndexEngine.get_loc

  File "pandas\_libs\index.pyx", line 132, in pandas._libs.index.IndexEngine.get_loc

  File "pandas\_libs\hashtable_class_helper.pxi", line 987, in pandas._libs.hashtable.Int64HashTable.get_item

  File "pandas\_libs\hashtable_class_helper.pxi", line 993, in pandas._libs.hashtable.Int64HashTable.get_item

KeyError: 5

**Data in input csv file:**

Application  time_taken
       Serv6         1.0
       Serv2         1.2
       Serv1         2.3
       Serv9         3.3
       Serv4         4.4
       Serv8         4.6
      Serv10         5.0
       Serv7         5.5
       Serv3         6.0
       Serv5         8.0

**Here is the code:**

import pandas as pd

def value_checker(sf):
    col = []
    col1 = []
    sf.sort_values(by='time_taken', inplace= True)
    sf.reset_index(drop = True, inplace = True)
    df = sf[sf['time_taken']<5]
    print(sf)
    if df.shape[0]>2:
        for i in range(0, df.shape[0], 2):
            if df.loc[i:i+1,['time_taken']].sum().item() <= 5:
                print(f"{df.loc[i,['Application']].item()}, {df.loc[i+1,['Application']].item()}")
                col.append(f"{df.loc[i,['Application']].item()}, {df.loc[i+1,['Application']].item()}")
#                print(df.loc[i,['time_taken']].item() + df.loc[i+1,['time_taken']].item())
                col1.append(df.loc[i,['time_taken']].item() + df.loc[i+1,['time_taken']].item())
                print(col1)
            else:
                print(f"in else block {i}th iter ")
                col.extend([df.loc[i,['Application']].item(), df.loc[i+1,['Application']].item()])
                col1.extend([df.loc[i,['time_taken']].item(), df.loc[i+1,['time_taken']].item()])
        new_df = pd.DataFrame(data={'Application':col, 'time_taken': col1})
        new_df = pd.concat([new_df, sf[sf['time_taken']>=5]])
        if new_df[new_df['time_taken']<5].shape[0]>=2:
            return value_checker(new_df)
    return sf

input_data = pd.read_csv('inputfile.csv')
final_csv = value_checker(input_data)
final_csv.to_csv("Outputfile.csv",index=False)
print("Task Completed")

注意：忽略打印语句

预期的输出应该是上面提到的模式。如果任何两行的总和小于或等于 5，则再次添加它们，依此类推。

标签： python-3.xpandasdataframe

Answer 1

新行会添加到数据框的底部吗？还是插入满足您条件的两个值之间？

如果满足条件，第一列中第 i 行的值应该是多少？

在等待您的回复时，开始，我认为您必须在列表中进行二进制搜索。

现在编辑我的答案:)）。尝试这个：

# Placing your 2nd column's values in a list.

time_taken_vals = df.iloc[:,1].values.tolist() 

# Doing the search for each entry and creating a new list of values for the result to be used as your 3rd column's values.

sum_result = []
for i in range(0, len(time_taken_vals)):
    if time_taken_vals[i] + time_taken_vals[i+1] <= 5 and i != len(time_list)-1:
        sum_result.append(time_taken_vals[i] + time_taken_vals[i+1])
    else:
        sum_result.append('')
sum_result.append('') # to equate the length of the two lists

# Now all you need to do is use the sum_result as value of your new column

df['Sum of time taken'] = sum_result

所以从你最初的DataFrame：

Application  time_taken
   Serv6         1.0
   Serv2         1.2
   Serv1         2.3
   Serv9         3.3
   Serv4         4.4
   Serv8         4.6
  Serv10         5.0
   Serv7         5.5
   Serv3         6.0
   Serv5         8.0

结果将是：

Application  time_taken  Sum of time taken
   Serv6         1.0           2.2
   Serv2         1.2           3.5
   Serv1         2.3           
   Serv9         3.3           
   Serv4         4.4           
   Serv8         4.6           
  Serv10         5.0           
   Serv7         5.5           
   Serv3         6.0           
   Serv5         8.0           

因为只有1.0+1.2和1.2+2.3满足你的条件。

这是您正在寻找的解决方案吗？无论如何，我希望这能给你一个解决问题的想法，祝你好运，别忘了给我的答案投票，谢谢:))