将多个Python行转换为并发数据框架，并与源数据合并

Question

如果这是一个基本问题，我深表歉意。我觉得这应该很容易，但我无法弄清楚。我有下面列出的代码，这些代码本质上查看了CSV文件中的两个列，并匹配了具有0.7相似性的作业标题。为此，我使用difflib.get_close_matches。但是，输出是多个单行，每当我尝试将其转换为数据框架时，每条线都是其自己的数据框架，我无法弄清楚如何合并/concat。所有代码以及当前和所需的输出都在下面。任何帮助都将不胜感激。

当前代码为：

import pandas as pd
import difflib

df = pd.read_csv('name.csv')

aLists = list(df['JTs'])

bLists = list(df['JT'])

n=3
cutoff = 0.7

for aList in aLists:
    best = difflib.get_close_matches(aList, bLists, n, cutoff)
    print(best)

当前输出为：

['SW Engineer']
['Manu Engineer']
[]
['IT Help']

所需的输出为：

          Output
0    SW Engineer
1  Manu Engineer
2        (blank)
3        IT Help

我试图执行此操作的表是：

< img src =“ https://i.sstatic.net/8va9u.png” alt =“必需的表格格式快照”>

任何帮助都将不胜感激！

Answer 1

这是一种实现此目的的简单方法。我首先转换为字符串。然后将第一个和最后一个括号从该字符串中删除，然后将其附加到全局列表中。

import pandas as pd
import difflib
import numpy as np

df = pd.read_csv('name.csv')

aLists = list(df['JTs'])

bLists = list(df['JT'])

n = 3
cutoff = 0.7

best = []

for aList in aLists:
    temp = difflib.get_close_matches(aList, bLists, n, cutoff)
    temp = str(temp)
    strippedString = temp.lstrip("[").rstrip("]")
    # print(temp)    
    best.append(strippedString)

print(best)

输出

[ 
  "'SW Engineer'", 
  "'Manu Engineer'", 
  '', 
  "'IT Help'"
]

是实现这一目标的另一种更好的方法。
您只需使用numpy将多个数组加入单个阵列。然后，可以在需要的情况下将其转换为普通数组。

import pandas as pd
import difflib
import numpy as np

df = pd.read_csv('name.csv')
aLists = list(df['JTs'])
bLists = list(df['JT'])
n = 3
cutoff = 0.7
best = []
for aList in aLists:
    temp = difflib.get_close_matches(aList, bLists, n, cutoff)
    best.append(temp)

# print(best)
# Use concatenate() to join two arrays
combinedNumpyArray = np.concatenate(best)
#Converting numpy array to normal array
normalArray = combinedNumpyArray.tolist()
print(normalArray)

输出

['SW Engineer', 'Manu Engineer', 'IT Help']

谢谢

Answer 2

您可以使用熊猫的.apply（）在每个条目上运行您的功能。然后，可以将其添加为新列或创建的新数据框架。

例如：

import pandas as pd
import difflib


def get_best_match(word):
    matches = difflib.get_close_matches(word, JT, n, cutoff)
    return matches[0] if matches else None

df = pd.read_csv('name.csv')
JT = df['JT']

n = 3
cutoff = 0.7

df['Output'] = df['JTs'].apply(get_best_match)

或用于新的数据帧：

df_output = pd.DataFrame({'Output' : df['JTs'].apply(get_best_match)})

给您：

                      JTs             JT         Output
0       Software Engineer  Manu Engineer    SW Engineer
1  Manufacturing Engineer    SW Engineer  Manu Engineer
2  Human Resource Manager        IT Help           None
3            IT Help Desk              f        IT Help

或：：

          Output
0    SW Engineer
1  Manu Engineer
2           None
3        IT Help