根据条件突出显示 panda df 错误

Posted 2023-03-06

【中文标题】根据条件突出显示 panda df 错误

【英文标题】：Highlight panda df errors based on conditions

【发布时间】：2019-03-23 22:46:58

【问题描述】：

美好的一天 SO 社区，

我在尝试逐行突出显示我的 df 中的错误时遇到了问题。

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)

我收到错误： TypeError: ('字符串索引必须是整数', '发生在索引 0')

我希望得到的是我的 df，其中突出显示了在我的 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)

试过这个并突出显示整个列而不是我想要的特定行值.. =/

【参考方案1】：

您可以使用merge 和参数indicator 查找未匹配的值，然后创建DataFrame 的样式：

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

详情：

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)

【讨论】：

嗨耶兹瑞尔，谢谢！我基于您的解决方案已被接受。

@hakkonen - 欢迎您！也可以免费投票给我的解决方案。谢谢。

【参考方案2】：

祝你也有美好的一天！

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

如果您正在查找要突出显示的 reference_dict 中的 not 值，您的意思是函数如下吗？

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'

不管怎样，既然可以隔离行，为什么还要突出显示它们呢？您似乎想查看 df 中的所有作业类，而 reference_dict 中没有。

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

这可能是一个切线，但这是我会做的。我根本不知道您可以突出显示 pandas 中的行，很酷的技巧。

【讨论】：

谢谢 Jess，它很有用，但我的任务是创建一个突出显示值的函数。但是非常感谢您的回答！=))

啊，好吧！很高兴你发现它很有用，希望你能找出突出显示的地方。