我有两个数据帧，它们都具有相同的基本架构。 （4个日期字段，几个字符串字段和4-5个浮点字段）。称他们为df1和df2。
我想做的是基本上得到一个＆＃34;差异＆＃34;两者中的所有行 - 在两个数据帧之间没有共享的行（不在集合交集中）。注意，两个数据帧的长度不必相同。
我尝试使用pandas.merge(how='outer')，但我不确定要传递哪个列作为＆＃39;键＆＃39;因为真的没有，我试过的各种组合都没有用。 df1或df2可能有两个（或更多）相同的行。
在pandas / Python中执行此操作的好方法是什么？
答案 0 :(得分：2)
尝试一下：
diff_df = pd.merge(df1, df2, how='outer', indicator='Exist')
diff_df = diff_df.loc[diff_df['Exist'] != 'both']
您将拥有一个在df1和df2上都不存在的所有行的数据框。
答案 1 :(得分：1)
设置df2.columns = df1.columns
现在，将每列设置为索引：df1 = df1.set_index(df1.columns.tolist())，同样设置为df2。
您现在可以执行df1.index.difference(df2.index)和df2.index.difference(df1.index)，这两个结果就是您的不同列。
答案 2 :(得分：1)
IIUC：
您可以使用pd.Index.symmetric_difference
pd.concat([df1, df2]).loc[
df1.index.symmetric_difference(df2.index)
]
答案 3 :(得分：0)
您可以使用此功能，输出是6个数据帧的有序字典，您可以将其写入excel以进行进一步分析。
在这里：
df1 = pd.DataFrame([['apple', '1'], ['banana', 2], ['coconut',3]], columns=['Fruits','Quantity'])
df2 = pd.DataFrame([['apple', '1'], ['banana', 3], ['durian',4]], columns=['Fruits','Quantity'])
dict1 = diff_func(df1, df2, 'Fruits')
In [10]: dict1['df1_only']:
Out[10]:
Fruits Quantity
1 coconut 3
In [11]: dict1['df2_only']:
Out[11]:
Fruits Quantity
3 durian 4
In [12]: dict1['Diff']:
Out[12]:
Fruits Quantity df1 or df2
0 banana 2 df1
1 banana 3 df2
In [13]: dict1['Merge']:
Out[13]:
Fruits Quantity
0 apple 1
代码如下：
import pandas as pd
from collections import OrderedDict as od
def diff_func(df1, df2, uid, dedupe=True, labels=('df1', 'df2'), drop=[]):
dict_df = {labels[0]: df1, labels[1]: df2}
col1 = df1.columns.values.tolist()
col2 = df2.columns.values.tolist()
# There could be columns known to be different, hence allow user to pass this as a list to be dropped.
if drop:
print ('Ignoring columns {} in comparison.'.format(', '.join(drop)))
col1 = list(filter(lambda x: x not in drop, col1))
col2 = list(filter(lambda x: x not in drop, col2))
df1 = df1[col1]
df2 = df2[col2]
# Step 1 - Check if no. of columns are the same:
len_lr = len(col1), len(col2)
assert len_lr[0]==len_lr[1], \
'Cannot compare frames with different number of columns: {}.'.format(len_lr)
# Step 2a - Check if the set of column headers are the same
# (order doesnt matter)
assert set(col1)==set(col2), \
'Left column headers are different from right column headers.' \
+'\n Left orphans: {}'.format(list(set(col1)-set(col2))) \
+'\n Right orphans: {}'.format(list(set(col2)-set(col1)))
# Step 2b - Check if the column headers are in the same order
if col1 != col2:
print ('[Note] Reordering right Dataframe...')
df2 = df2[col1]
# Step 3 - Check datatype are the same [Order is important]
if set((df1.dtypes == df2.dtypes).tolist()) - {True}:
print ('dtypes are not the same.')
df_dtypes = pd.DataFrame({labels[0]:df1.dtypes,labels[1]:df2.dtypes,'Diff':(df1.dtypes == df2.dtypes)})
df_dtypes = df_dtypes[df_dtypes['Diff']==False][[labels[0],labels[1],'Diff']]
print (df_dtypes)
else:
print ('DataType check: Passed')
# Step 4 - Check for duplicate rows
if dedupe:
for key, df in dict_df.items():
if df.shape[0] != df.drop_duplicates().shape[0]:
print(key + ': Duplicates exists, they will be dropped.')
dict_df[key] = df.drop_duplicates()
# Step 5 - Check for duplicate uids.
if type(uid)==str or type(uid)==list:
print ('Uniqueness check: {}'.format(uid))
for key, df in dict_df.items():
count_uid = df.shape[0]
count_uid_unique = df[uid].drop_duplicates().shape[0]
var = [0,1][count_uid_unique == df.shape[0]] #<-- Round off to the nearest integer if it is 100%
pct = round(100*count_uid_unique/df.shape[0], var)
print ('{}: {} out of {} are unique ({}%).'.format(key, count_uid_unique, count_uid, pct))
# Checks complete, begin merge. '''Remenber to dedupe, provide labels for common_no_match'''
dict_result = od()
df_merge = pd.merge(df1, df2, on=col1, how='inner')
if not df_merge.shape[0]:
print ('Error: Merged DataFrame is empty.')
else:
dict_result[labels[0]] = df1
dict_result[labels[1]] = df2
dict_result['Merge'] = df_merge
if type(uid)==str:
uid = [uid]
if type(uid)==list:
df1_only = df1.append(df_merge).reset_index(drop=True)
df1_only['Duplicated']=df1_only.duplicated(subset=uid, keep=False) #keep=False, marks all duplicates as True
df1_only = df1_only[df1_only['Duplicated']==False]
df2_only = df2.append(df_merge).reset_index(drop=True)
df2_only['Duplicated']=df2_only.duplicated(subset=uid, keep=False)
df2_only = df2_only[df2_only['Duplicated']==False]
label = labels[0]+' or '+labels[1]
df_lc = df1_only.copy()
df_lc[label] = labels[0]
df_rc = df2_only.copy()
df_rc[label] = labels[1]
df_c = df_lc.append(df_rc).reset_index(drop=True)
df_c['Duplicated'] = df_c.duplicated(subset=uid, keep=False)
df_c1 = df_c[df_c['Duplicated']==True]
df_c1 = df_c1.drop('Duplicated', axis=1)
df_uc = df_c[df_c['Duplicated']==False]
df_uc_left = df_uc[df_uc[label]==labels[0]]
df_uc_right = df_uc[df_uc[label]==labels[1]]
dict_result[labels[0]+'_only'] = df_uc_left.drop(['Duplicated', label], axis=1)
dict_result[labels[1]+'_only'] = df_uc_right.drop(['Duplicated', label], axis=1)
dict_result['Diff'] = df_c1.sort_values(uid).reset_index(drop=True)
return dict_result
答案 4 :(得分：-2)
带
left_df.merge(df,left_on=left_df.columns.tolist(),right_on=df.columns.tolist(),how='outer')
您可以获得外部联接结果 类似地，你可以得到内连接结果。然后做一个你想要的差异。