我试图突出显示两个数据帧之间发生了什么变化。
假设我有两个Python Pandas数据框架:
"StudentRoster Jan-1":
id Name score isEnrolled Comment
111 Jack 2.17 True He was late to class
112 Nick 1.11 False Graduated
113 Zoe 4.12 True
"StudentRoster Jan-2":
id Name score isEnrolled Comment
111 Jack 2.17 True He was late to class
112 Nick 1.21 False Graduated
113 Zoe 4.12 False On vacation
我的目标是输出一个HTML表，它:
标识已更改的行(可以是int, float, boolean，字符串)
输出具有相同的OLD和NEW值的行(理想情况下是HTML表)，以便消费者可以清楚地看到两个数据框架之间发生了什么变化:
“StudentRoster差异Jan-1 - Jan-2”:
id名称分数isregistered评论
尼克是1.11|现在1.21假毕业
113佐伊4.12是真的|现在是假的|现在“度假”
我想我可以逐行逐列比较，但有没有更简单的方法?
pandas >= 1.1: DataFrame.compare
使用pandas 1.1，基本上可以用一个函数调用复制Ted Petrou的输出。例子摘自文档:
pd.__version__
# '1.1.0'
df1.compare(df2)
score isEnrolled Comment
self other self other self other
1 1.11 1.21 NaN NaN NaN NaN
2 NaN NaN 1.0 0.0 NaN On vacation
这里，“self”指的是LHS数据帧，而“other”指的是RHS数据帧。默认情况下，相等的值将被nan替换，因此您可以只关注差异。如果您想显示相同的值，请使用
df1.compare(df2, keep_equal=True, keep_shape=True)
score isEnrolled Comment
self other self other self other
1 1.11 1.21 False False Graduated Graduated
2 4.12 4.12 True False NaN On vacation
你也可以使用align_axis改变比较轴:
df1.compare(df2, align_axis='index')
score isEnrolled Comment
1 self 1.11 NaN NaN
other 1.21 NaN NaN
2 self NaN 1.0 NaN
other NaN 0.0 On vacation
这是逐行比较值，而不是逐列比较值。
下面是另一种使用选择和合并的方法:
In [6]: # first lets create some dummy dataframes with some column(s) different
...: df1 = pd.DataFrame({'a': range(-5,0), 'b': range(10,15), 'c': range(20,25)})
...: df2 = pd.DataFrame({'a': range(-5,0), 'b': range(10,15), 'c': [20] + list(range(101,105))})
In [7]: df1
Out[7]:
a b c
0 -5 10 20
1 -4 11 21
2 -3 12 22
3 -2 13 23
4 -1 14 24
In [8]: df2
Out[8]:
a b c
0 -5 10 20
1 -4 11 101
2 -3 12 102
3 -2 13 103
4 -1 14 104
In [10]: # make condition over the columns you want to comapre
...: condition = df1['c'] != df2['c']
...:
...: # select rows from each dataframe where the condition holds
...: diff1 = df1[condition]
...: diff2 = df2[condition]
In [11]: # merge the selected rows (dataframes) with some suffixes (optional)
...: diff1.merge(diff2, on=['a','b'], suffixes=('_before', '_after'))
Out[11]:
a b c_before c_after
0 -4 11 21 101
1 -3 12 22 102
2 -2 13 23 103
3 -1 14 24 104
以下是来自Jupyter的截图:
突出显示两个数据框架之间的差异
可以使用DataFrame样式属性来突出显示有差异的单元格的背景颜色。
使用原始问题中的示例数据
第一步是用concat函数水平连接dataframe，并用keys参数区分每一帧:
df_all = pd.concat([df.set_index('id'), df2.set_index('id')],
axis='columns', keys=['First', 'Second'])
df_all
交换列级别并将相同的列名放在彼此旁边可能更容易:
df_final = df_all.swaplevel(axis='columns')[df.columns[1:]]
df_final
现在，很容易看出不同的框架。但是，我们可以进一步使用style属性来突出显示不同的单元格。我们定义了一个自定义函数来实现这一点，您可以在本部分文档中看到。
def highlight_diff(data, color='yellow'):
attr = 'background-color: {}'.format(color)
other = data.xs('First', axis='columns', level=-1)
return pd.DataFrame(np.where(data.ne(other, level=0), attr, ''),
index=data.index, columns=data.columns)
df_final.style.apply(highlight_diff, axis=None)
这将突出显示两者都有缺失值的单元格。您可以填充它们或提供额外的逻辑，这样它们就不会被突出显示。
这个答案只是扩展了@Andy Hayden的答案，使其能够适应数值字段为nan的情况，并将其包装成一个函数。
import pandas as pd
import numpy as np
def diff_pd(df1, df2):
"""Identify differences between two pandas DataFrames"""
assert (df1.columns == df2.columns).all(), \
"DataFrame column names are different"
if any(df1.dtypes != df2.dtypes):
"Data Types are different, trying to convert"
df2 = df2.astype(df1.dtypes)
if df1.equals(df2):
return None
else:
# need to account for np.nan != np.nan returning True
diff_mask = (df1 != df2) & ~(df1.isnull() & df2.isnull())
ne_stacked = diff_mask.stack()
changed = ne_stacked[ne_stacked]
changed.index.names = ['id', 'col']
difference_locations = np.where(diff_mask)
changed_from = df1.values[difference_locations]
changed_to = df2.values[difference_locations]
return pd.DataFrame({'from': changed_from, 'to': changed_to},
index=changed.index)
所以对于你的数据(稍微编辑一下，在分数列中有一个NaN):
import sys
if sys.version_info[0] < 3:
from StringIO import StringIO
else:
from io import StringIO
DF1 = StringIO("""id Name score isEnrolled Comment
111 Jack 2.17 True "He was late to class"
112 Nick 1.11 False "Graduated"
113 Zoe NaN True " "
""")
DF2 = StringIO("""id Name score isEnrolled Comment
111 Jack 2.17 True "He was late to class"
112 Nick 1.21 False "Graduated"
113 Zoe NaN False "On vacation" """)
df1 = pd.read_table(DF1, sep='\s+', index_col='id')
df2 = pd.read_table(DF2, sep='\s+', index_col='id')
diff_pd(df1, df2)
输出:
from to
id col
112 score 1.11 1.21
113 isEnrolled True False
Comment On vacation
