I have a Pandas dataframe and I want to find all the unique values in that dataframe...irrespective of row/columns. If I have a 10 x 10 dataframe, and suppose they have 84 unique values, I need to find them - Not the count.
我有一个熊猫dataframe，我想在这个dataframe中找到所有唯一的值。无论行/列。如果我有一个10×10的dataframe，假设它们有84个唯一值，我需要找到它们——而不是计数。
I can create a set and add the values of each rows by iterating over the rows of the dataframe. But, I feel that it may be inefficient (cannot justify that). Is there an efficient way to find it? Is there a predefined function?
我可以通过遍历dataframe的行来创建一个集合并添加每个行的值。但是，我觉得它可能是低效的(不能证明这一点)。有有效的方法找到它吗?有预定义的函数吗?
2 个解决方案
#1
74
In [1]: df = DataFrame(np.random.randint(0,10,size=100).reshape(10,10))
In [2]: df
Out[2]:
0 1 2 3 4 5 6 7 8 9
0 2 2 3 2 6 1 9 9 3 3
1 1 2 5 8 5 2 5 0 6 3
2 0 7 0 7 5 5 9 1 0 3
3 5 3 2 3 7 6 8 3 8 4
4 8 0 2 2 3 9 7 1 2 7
5 3 2 8 5 6 4 3 7 0 8
6 4 2 6 5 3 3 4 5 3 2
7 7 6 0 6 6 7 1 7 5 1
8 7 4 3 1 0 6 9 7 7 3
9 5 3 4 5 2 0 8 6 4 7
In [13]: Series(df.values.ravel()).unique()
Out[13]: array([9, 1, 4, 6, 0, 7, 5, 8, 3, 2])
Numpy unique sorts, so its faster to do it this way (and then sort if you need to)
Numpy唯一排序，所以这样做更快(如果需要的话，还可以排序)
In [14]: df = DataFrame(np.random.randint(0,10,size=10000).reshape(100,100))
In [15]: %timeit Series(df.values.ravel()).unique()
10000 loops, best of 3: 137 ﾵs per loop
In [16]: %timeit np.unique(df.values.ravel())
1000 loops, best of 3: 270 ﾵs per loop
#2
5
Or you can use:
或者你可以使用:
df.stack().unique()
.unique df.stack()()
Then you don't need to worry if you have NaN values, as they are excluded when doing the stacking.
那么，如果您有NaN值，就不需要担心了，因为在进行堆栈时，它们被排除在外。
#1
74
In [1]: df = DataFrame(np.random.randint(0,10,size=100).reshape(10,10))
In [2]: df
Out[2]:
0 1 2 3 4 5 6 7 8 9
0 2 2 3 2 6 1 9 9 3 3
1 1 2 5 8 5 2 5 0 6 3
2 0 7 0 7 5 5 9 1 0 3
3 5 3 2 3 7 6 8 3 8 4
4 8 0 2 2 3 9 7 1 2 7
5 3 2 8 5 6 4 3 7 0 8
6 4 2 6 5 3 3 4 5 3 2
7 7 6 0 6 6 7 1 7 5 1
8 7 4 3 1 0 6 9 7 7 3
9 5 3 4 5 2 0 8 6 4 7
In [13]: Series(df.values.ravel()).unique()
Out[13]: array([9, 1, 4, 6, 0, 7, 5, 8, 3, 2])
Numpy unique sorts, so its faster to do it this way (and then sort if you need to)
Numpy唯一排序，所以这样做更快(如果需要的话，还可以排序)
In [14]: df = DataFrame(np.random.randint(0,10,size=10000).reshape(100,100))
In [15]: %timeit Series(df.values.ravel()).unique()
10000 loops, best of 3: 137 ﾵs per loop
In [16]: %timeit np.unique(df.values.ravel())
1000 loops, best of 3: 270 ﾵs per loop
#2
5
Or you can use:
或者你可以使用:
df.stack().unique()
.unique df.stack()()
Then you don't need to worry if you have NaN values, as they are excluded when doing the stacking.
那么，如果您有NaN值，就不需要担心了，因为在进行堆栈时，它们被排除在外。