[英]python dataframe boolean values with if statement
我想讓if語句顯示所有重復的REF_INT我試過這個:
(df_picru['REF_INT'].value_counts()==1)
並且它向我顯示了所有值的真或假,但我不想做這樣的事情:
if (df_picru['REF_INT'].value_counts()==1)
print "df_picru['REF_INT']"
In [28]: df_picru['new'] = \
df_picru['REF_INT'].duplicated(keep=False) \
.map({True:'duplicates',False:'unique'})
In [29]: df_picru
Out[29]:
REF_INT new
0 1 unique
1 2 duplicates
2 3 unique
3 8 duplicates
4 8 duplicates
5 2 duplicates
我認為你需要duplicated
布爾掩碼和新列numpy.where
:
mask = df_picru['REF_INT'].duplicated(keep=False)
樣品:
df_picru = pd.DataFrame({'REF_INT':[1,2,3,8,8,2]})
mask = df_picru['REF_INT'].duplicated(keep=False)
print (mask)
0 False
1 True
2 False
3 True
4 True
5 True
Name: REF_INT, dtype: bool
df_picru['new'] = np.where(mask, 'duplicates', 'unique')
print (df_picru)
REF_INT new
0 1 unique
1 2 duplicates
2 3 unique
3 8 duplicates
4 8 duplicates
5 2 duplicates
如果需要檢查至少一個,如果unique
值需要any
轉換boolean mask
- array
到標量True
或False
:
if mask.any():
print ('at least one unique')
at least one unique
使用groupby的另一個解決方案
#groupby REF_INT and then count the occurrence and set as duplicate if count is greater than 1
df_picru.groupby('REF_INT').apply(lambda x: 'Duplicated' if len(x)> 1 else 'Unique')
Out[21]:
REF_INT
1 Unique
2 Duplicated
3 Unique
8 Duplicated
dtype: object
如果你做了一個小改動,value_counts實際上可以工作:
df_picru.REF_INT.value_counts()[lambda x: x>1]
Out[31]:
2 2
8 2
Name: REF_INT, dtype: int64
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