我正在研究一项要求,有两个CSV如下-
CSV1.csv
Short Description Category
Device is DOWN! Server Down
CPU Warning Monitoron XSSXSXSXSXSX.com CPU Utilization
CPU Warning Monitoron XSSXSXSXSXSX.com CPU Utilization
CPU Warning Monitoron XSSXSXSXSXSX.com CPU Utilization
CPU Warning Monitoron XSSXSXSXSXSX.com CPU Utilization
Device Performance Alerts was triggered on Physical memory Memory Utilization
Device Performance Alerts was triggered on Physical memory Memory Utilization
Device Performance Alerts was triggered on Physical memory Memory Utilization
Disk Space Is Lowon ;E: Disk Space Utilization
Disk Space Is Lowon;C: Disk Space Utilization
Network Interface Down Interface Down
Active Directory 和reference.csv
Category Complexity
Server Down Simple
Network Interface down Complex
Drive Cleanup Windows Medium
CPU Utilization Medium
Memory Utilization Medium
Disk Space Utilization Unix Simple
Windows Service Restart Medium
UNIX Service Restart Medium
Web Tomcat Instance Restart Simple
Expected Output
Short Description Category Complexity
Device is DOWN! Server Down Simple
CPU Warning Monitoron XSSXSXSXSXSX.com CPU Utilization Medium
CPU Warning Monitoron XSSXSXSXSXSX.com CPU Utilization Medium
CPU Warning Monitoron XSSXSXSXSXSX.com CPU Utilization Medium
CPU Warning Monitoron XSSXSXSXSXSX.com CPU Utilization Medium
Device Performance Alerts was triggered on Physical memory Memory Utilization Medium
Device Performance Alerts was triggered on Physical memory Memory Utilization Medium
Device Performance Alerts was triggered on Physical memory Memory Utilization Medium
Disk Space Is Lowon ;E: Disk Space Utilization Medium
Disk Space Is Lowon;C: Disk Space Utilization Medium
Network Interface Down Interface Down Complex我尝试了下面的代码--但是在输出数据中,我可以看到空白的[],不确定我遗漏了什么。在输出复杂性列中,我只能看到每一行的[]。我试图得到完全匹配,但我需要得到所有可能的组合,所以我使用get_close_matches。如何在下面的代码中传递数据中的可能性参数,我不知道如何传递这个可能性。
我尝试了很少其他的可能性,如精确,但没有给出预期的结果,因为我正在寻找所有可能的组合,同时比较列和字符串。
import pandas as pd
import difflib
df1 = pd.read_csv('csv1.csv')
df1 = df1.fillna('')
df2 = pd.read_csv('reference.csv')
my_dict = dict(zip(df2['Category'].values, df2['Complexity'].values))
def match_key(key, default_value):
if not key:
return default_value
for d_key in my_dict.keys():
if key in d_key or d_key in key:
return my_dict[d_key]
return default_value
df1['Complexity'] = df1['Category'].apply(lambda x: difflib.get_close_matches(x, list(my_dict.keys(), n=1)))
df1 = df1.explode('Complexity')
df1['Complexity'] = df1['Complexity'].map(my_dict)
print(df1)发布于 2021-03-29 16:13:15
difflib.get_close_matches期望第一个参数是'word',在您的情况下是x,第二个参数是‘可能性’。你提供了一个空字符串。这就是为什么你的函数不能工作,它试图匹配一个词,基本上什么都没有。
my_dict包含作为键的有效选项,因此我们可以将它们用作“可能性”列表。
# Use n=1, so only tries to get 1 match
df1['Complexity'] = df1['Category'].apply(lambda x: difflib.get_close_matches(x, list(my_dict.keys()), n=1))
# The output of get_close_matches is a list, we use explode to convert it to a string
df1 = df1.explode('Complexity')
# We can now apply our map, to the *Complexity* column,
# which is technically the best match *Category*, via get_close_matches
df1['Complexity'] = df1['Complexity'].map(my_dict)原始坏答案
但是,与其继续使用difflib,我认为您可以更改您的方法。您希望将my_dict应用于df1的Category列。这在传统上被称为应用map。pandas已经通过pandas.Series.map准备好了这个实现。
df1['Complexity'] = df1['Category'].map(my_dict)https://stackoverflow.com/questions/66857040
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