pandas 熊猫替换为默认值
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Pandas replace with default value
提问by Vikash B
I have a pandas dataframe I want to replace a certain column conditionally.
我有一个 Pandas 数据框，我想有条件地替换某个列。
eg:
例如：
col
0 Mr
1 Miss
2 Mr
3 Mrs
4 Col.
I want to map them as
我想将它们映射为
{'Mr': 0, 'Mrs': 1, 'Miss': 2}
If there are other titles now available in the dict then I want them to have a default value of 3
如果现在字典中有其他标题可用，那么我希望它们的默认值为 3
The above example becomes
上面的例子变成
col
0 0
1 2
2 0
3 1
4 3
Can I do this with pandas.replace() without using regex ?
我可以在不使用正则表达式的情况下使用 pandas.replace() 执行此操作吗？
回答by jezrael
You can use maprather as replace, because faster, then fillnaby 3and cast to intby astype:
你可以使用map而不是 as replace，因为更快， then fillnaby3和 cast to intby astype：
df['col'] = df.col.map({'Mr': 0, 'Mrs': 1, 'Miss': 2}).fillna(3).astype(int)
print (df)
col
0 0
1 2
2 0
3 1
4 3
Another solution with numpy.whereand condition with isin:
另一个解决方案numpy.where和条件isin：
d = {'Mr': 0, 'Mrs': 1, 'Miss': 2}
df['col'] = np.where(df.col.isin(d.keys()), df.col.map(d), 3).astype(int)
print (df)
col
0 0
1 2
2 0
3 1
4 3
Solution with replace:
解决方案replace：
d = {'Mr': 0, 'Mrs': 1, 'Miss': 2}
df['col'] = np.where(df.col.isin(d.keys()), df.col.replace(d), 3)
print (df)
col
0 0
1 2
2 0
3 1
4 3
Timings:
时间：
df = pd.concat([df]*10000).reset_index(drop=True)
d = {'Mr': 0, 'Mrs': 1, 'Miss': 2}
df['col0'] = df.col.map(d).fillna(3).astype(int)
df['col1'] = np.where(df.col.isin(d.keys()), df.col.replace(d), 3)
df['col2'] = np.where(df.col.isin(d.keys()), df.col.map(d), 3).astype(int)
print (df)
In [447]: %timeit df['col0'] = df.col.map(d).fillna(3).astype(int)
100 loops, best of 3: 4.93 ms per loop
In [448]: %timeit df['col1'] = np.where(df.col.isin(d.keys()), df.col.replace(d), 3)
100 loops, best of 3: 14.3 ms per loop
In [449]: %timeit df['col2'] = np.where(df.col.isin(d.keys()), df.col.map(d), 3).astype(int)
100 loops, best of 3: 7.68 ms per loop
In [450]: %timeit df['col3'] = df.col.map(lambda L: d.get(L, 3))
10 loops, best of 3: 36.2 ms per loop
回答by Paul Mwaniki
To add on the answer by @jezrael: The most straight forward solution is to use a defaultdictinstead of dict. This is especially useful when you want missing values not to be replaced with your default value.
添加@jezrael 的答案：最直接的解决方案是使用defaultdict而不是dict。当您希望不使用默认值替换缺失值时，这尤其有用。
from collections import defaultdict
df['col'] = df.col.map(defaultdict(lambda: 3,Mr= 0, Mrs= 1, Miss= 2),na_action='ignore')
The first argument of defaultdictis a function that return the default value.
defaultdict的第一个参数是一个返回默认值的函数。
