Question

SO SO社区，

在逐行突出显示df中的错误时，我一直遇到问题。

reference_dict = {'jobclass' : ['A','B'], 'Jobs' : ['Teacher','Plumber']}
dict = {'jobclass': ['A','C','A'], 'Jobs': ['Teacher', 'Plumber','Policeman']}
df = pd.DataFrame(data=dict) 

def highlight_rows(df):
  for i in df.index:
    if df.jobclass[i] in reference_dict['jobclass']:
      print(df.jobclass[i])
      return 'background-color: green'

df.style.apply(highlight_rows, axis = 1)

我遇到了错误： TypeError ：（“字符串索引必须是整数”，“发生在索引0”）

我希望得到的df突出显示了我reference_dict中找不到的值。

任何帮助将不胜感激。。干杯！

编辑：

x = {'jobclass' : ['A','B'], 'Jobs' : ['Teacher','Plumber']}
d = {'jobclass': ['A','C','A'], 'Jobs': ['Teacher', 'Plumber','Policeman']}
df = pd.DataFrame(data=d) 
print(df)
def highlight_rows(s):
  ret = ["" for i in s.index]
  for i in df.index:
    if df.jobclass[i] not in x['jobclass']:
      ret[s.index.get_loc('Jobs')] = "background-color: yellow"
      return ret
df.style.apply(highlight_rows, axis = 1)

尝试过此操作，并突出显示了整个列，而不是我想要的特定行值。.= /

Answer 1

也祝您有美好的一天！

What i hope to get is my df with values not found in my reference_dict being highlighted.

如果您要查找在reference_dict中突出显示的 not 值，则表示该函数具有以下含义？

def highlight_rows(df):
  for i in df.index:
    if df.jobclass[i] not in reference_dict['jobclass']:
      print(df.jobclass[i])
      return 'background-color: green'

无论哪种方式，为什么当您可以隔离行时突出显示行？似乎您想查看 df 中的所有作业类，而 reference_dict 中没有一个。

import pandas as pd


reference_dict = {'jobclass' : ['A','B'], 'Jobs' : ['Teacher','Plumber']}

data_dict = {'jobclass': ['A','C','A'], 'Jobs': ['Teacher', 'Plumber','Policeman']}



ref_df = pd.DataFrame(reference_dict)
df = pd.DataFrame(data_dict)

outliers = df.merge(ref_df, how='outer', on='jobclass') # merge the two tables together, how='outer' includes jobclasses which the DataFrames do not have in common. Will automatically generate columns Jobs_x and Jobs_y once joined together because the columns have the same name
outliers = outliers[ outliers['Jobs_y'].isnull() ] # Jobs_y is null when there is no matching jobclass in the reference DataFrame, so we can take advantage of that by filtering
outliers = outliers.drop('Jobs_y', axis=1) # let's drop the junk column after we used it to filter for what we wanted

print("The reference DataFrame is:")
print(ref_df,'\n')

print("The input DataFrame is:")
print(df,'\n')

print("The result is a list of all the jobclasses not in the reference DataFrame and what job is with it:")
print(outliers)

结果是：

The reference DataFrame is:
  jobclass     Jobs
0        A  Teacher
1        B  Plumber 

The input DataFrame is:
  jobclass       Jobs
0        A    Teacher
1        C    Plumber
2        A  Policeman 

The result is a list of all the jobclasses not in the reference DataFrame and what job is with it:
  jobclass   Jobs_x
2        C  Plumber

这可能是切线，但这是我要做的。我不知道您根本无法突出显示大熊猫中的行，很酷的技巧。

Answer 2

您可以将merge与参数indicator一起用于发现不匹配的值，然后创建样式的DataFrame：

x = {'jobclass' : ['A','B'], 'Jobs' : ['Teacher','Plumber']}
d = {'jobclass': ['A','C','A'], 'Jobs': ['Teacher', 'Plumber','Policeman']}
df = pd.DataFrame(data=d) 
print (df)
  jobclass       Jobs
0        A    Teacher
1        C    Plumber
2        A  Policeman

详细信息：

print (df.merge(pd.DataFrame(x) , on='jobclass', how='left', indicator=True))
  jobclass     Jobs_x   Jobs_y     _merge
0        A    Teacher  Teacher       both
1        C    Plumber      NaN  left_only
2        A  Policeman  Teacher       both

def highlight_rows(s):
    c1 = 'background-color: yellow'
    c2 = '' 

    df1 = pd.DataFrame(x)
    m = s.merge(df1, on='jobclass', how='left', indicator=True)['_merge'] == 'left_only'
    df2 = pd.DataFrame(c2, index=s.index, columns=s.columns)
    df2.loc[m, 'Jobs'] = c1
    return df2

df.style.apply(highlight_rows, axis = None)

根据条件突出显示熊猫df错误

2 个答案: