pandas 带字符串的熊猫“diff()”
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Pandas "diff()" with string
提问by guilhermecgs
How can I flag a row in a dataframe every time a column change its string value?
每次列更改其字符串值时，如何标记数据框中的行？
Ex:
前任：
Input
输入
ColumnA ColumnB
1 Blue
2 Blue
3 Red
4 Red
5 Yellow
# diff won't work here with strings.... only works in numerical values
dataframe['changed'] = dataframe['ColumnB'].diff()
ColumnA ColumnB changed
1 Blue 0
2 Blue 0
3 Red 1
4 Red 0
5 Yellow 1
采纳答案by root
I get better performance with neinstead of using the actual !=comparison:
我通过ne而不是使用实际!=比较获得了更好的性能：
df['changed'] = df['ColumnB'].ne(df['ColumnB'].shift().bfill()).astype(int)
Timings
时间安排
Using the following setup to produce a larger dataframe:
使用以下设置生成更大的数据框：
df = pd.concat([df]*10**5, ignore_index=True)
I get the following timings:
我得到以下时间：
%timeit df['ColumnB'].ne(df['ColumnB'].shift().bfill()).astype(int)
10 loops, best of 3: 38.1 ms per loop
%timeit (df.ColumnB != df.ColumnB.shift()).astype(int)
10 loops, best of 3: 77.7 ms per loop
%timeit df['ColumnB'] == df['ColumnB'].shift(1).fillna(df['ColumnB'])
10 loops, best of 3: 99.6 ms per loop
%timeit (df.ColumnB.ne(df.ColumnB.shift())).astype(int)
10 loops, best of 3: 19.3 ms per loop
回答by Kartik
回答by jezrael
For me works compare with shift, then NaNwas replaced 0because before no value:
对我来说，作品与 比较shift，然后NaN被替换，0因为之前没有价值：
df['diff'] = (df.ColumnB != df.ColumnB.shift()).astype(int)
df.ix[0,'diff'] = 0
print (df)
ColumnA ColumnB diff
0 1 Blue 0
1 2 Blue 0
2 3 Red 1
3 4 Red 0
4 5 Yellow 1
Edit by timingsof another answer - fastest is use ne:
按另一个答案的时间编辑- 最快的是使用ne：
df['diff'] = (df.ColumnB.ne(df.ColumnB.shift())).astype(int)
df.ix[0,'diff'] = 0
