Question

我有一些包含拼写错误的数据。例如：

# Define the correct spellings:
Li_A = ["potato", "tomato", "squash", "apple", "pear"]

# Define the data that contains spelling errors:
B    = {'one' : pd.Series(["potat0", "toma3o", "s5uash", "ap8le", "pea7"], index=['a', 'b', 'c', 'd', 'e']),
        'two' : pd.Series(["po1ato", "2omato", "squ0sh", "2pple", "p3ar"], index=['a', 'b', 'c', 'd', 'e'])}

df_B = pd.DataFrame(B)

我正在尝试使用以下代码更正它们：

import pandas as pd
import difflib

# Define the function that corrects the spelling:

def Spelling(ask):
    difflib.get_close_matches(ask, Li_A, n=1, cutoff=0.5)

# Apply the function that corrects the spelling:

for index,row in df_B.iterrows():
    df_B.loc[index,'Correct one'] = Spelling(df_B['one'])

for index,row in df_B.iterrows():
    df_B.loc[index,'Correct two'] = Spelling(df_B['two'])

df_B

但我得到的只是：

      one     two  Correct one  Correct two
a  potat0  po1ato          NaN          NaN
b  toma3o  2omato          NaN          NaN
c  s5uash  squ0sh          NaN          NaN
d   ap8le   2pple          NaN          NaN
e    pea7    p3ar          NaN          NaN

如何在我的数据框架上添加正确的拼写作为新列添加到目前所说的＆＃34; Nan＆＃34;请？

当我一次只运行一个单词时，它确实有效：

import difflib
Li_A = ["potato", "tomato", "squash", "apple", "pear"]
B    = 'potat0'
C    = difflib.get_close_matches(B, Li_A, n=1, cutoff=0.5)
C

Out: ['potato']

Answer 1

您忘了功能(([\s\S]*?)(production|public))(?P<app>\2)\g和return使用iterrows每个循环的选择值，row只使用一次：

iterrows

但更简单的是使用applymap：

def Spelling(ask):
    return difflib.get_close_matches(ask, Li_A, n=1, cutoff=0.5)

# Apply the function that corrects the spelling:

for index,row in df_B.iterrows():
    df_B.loc[index,'Correct one'] = Spelling(row['one'])
    df_B.loc[index,'Correct two'] = Spelling(row['two'])

print (df_B)
      one     two Correct one Correct two
a  potat0  po1ato    [potato]    [potato]
b  toma3o  2omato    [tomato]    [tomato]
c  s5uash  squ0sh    [squash]    [squash]
d   ap8le   2pple     [apple]     [apple]
e    pea7    p3ar      [pear]      [pear]

将拼写检查功能应用于Dataframe

1 个答案: