用途factorize：
df['col'] = pd.factorize(df.col)[0]
print (df)
col
0 0
1 1
2 0
3 0
4 1
编辑：
如Jeff评论中所述，那么最好是将column转换为，categorical主要是因为更少的内存使用量：
df['col'] = df['col'].astype("category")
：
有趣的是，在大dfpandas中，速度更快numpy。我不敢相信。
len(df)=500k：
In [29]: %timeit (a(df1))
100 loops, best of 3: 9.27 ms per loop
In [30]: %timeit (a1(df2))
100 loops, best of 3: 9.32 ms per loop
In [31]: %timeit (b(df3))
10 loops, best of 3: 24.6 ms per loop
In [32]: %timeit (b1(df4))
10 loops, best of 3: 24.6 ms per loop
len(df)=5k：
In [38]: %timeit (a(df1))
1000 loops, best of 3: 274 µs per loop
In [39]: %timeit (a1(df2))
The slowest run took 6.71 times longer than the fastest. This Could mean that an intermediate result is being cached.
1000 loops, best of 3: 273 µs per loop
In [40]: %timeit (b(df3))
The slowest run took 5.15 times longer than the fastest. This Could mean that an intermediate result is being cached.
1000 loops, best of 3: 295 µs per loop
In [41]: %timeit (b1(df4))
1000 loops, best of 3: 294 µs per loop
len(df)=5：
In [46]: %timeit (a(df1))
1000 loops, best of 3: 206 µs per loop
In [47]: %timeit (a1(df2))
1000 loops, best of 3: 204 µs per loop
In [48]: %timeit (b(df3))
The slowest run took 6.30 times longer than the fastest. This Could mean that an intermediate result is being cached.
10000 loops, best of 3: 164 µs per loop
In [49]: %timeit (b1(df4))
The slowest run took 6.44 times longer than the fastest. This Could mean that an intermediate result is being cached.
10000 loops, best of 3: 164 µs per loop
：
d = {'col': ["baked","beans","baked","baked","beans"]}
df = pd.DataFrame(data=d)
print (df)
df = pd.concat([df]*100000).reset_index(drop=True)
#test for 5k
#df = pd.concat([df]*1000).reset_index(drop=True)
df1,df2,df3, df4 = df.copy(),df.copy(),df.copy(),df.copy()
def a(df):
df['col'] = pd.factorize(df.col)[0]
return df
def a1(df):
idx,_ = pd.factorize(df.col)
df['col'] = idx
return df
def b(df):
df['col'] = np.unique(df['col'],return_inverse=True)[1]
return df
def b1(df):
_,idx = np.unique(df['col'],return_inverse=True)
df['col'] = idx
return df
print (a(df1))
print (a1(df2))
print (b(df3))
print (b1(df4))
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