问题
Give this dataframe 'x':
col1 col2 col3 col4
0 5 -2 1
-5 2 -1 9
3 -7 3 5
How I could get a list of pairs with the min and max of each column? The result would be:
list = [ [-5 , 3], [-7 , 5], [-2 , 3], [1 , 9] ]
回答1:
You could define a function and call apply
passing the function name, this will create a df with min and max as the index names:
In [203]:
def minMax(x):
return pd.Series(index=['min','max'],data=[x.min(),x.max()])
df.apply(minMax)
Out[203]:
col1 col2 col3 col4
min -5 -7 -2 1
max 3 5 3 9
If you insist on a list of lists we can transpose the df and convert the values to a list:
In [206]:
def minMax(x):
return pd.Series(index=['min','max'],data=[x.min(),x.max()])
df.apply(minMax).T.values.tolist()
Out[206]:
[[-5, 3], [-7, 5], [-2, 3], [1, 9]]
The function itself is not entirely necessary as you can use a lambda instead:
In [209]:
df.apply(lambda x: pd.Series([x.min(), x.max()])).T.values.tolist()
Out[209]:
[[-5, 3], [-7, 5], [-2, 3], [1, 9]]
Note also that you can use describe and loc
to get what you want:
In [212]:
df.describe().loc[['min','max']]
Out[212]:
col1 col2 col3 col4
min -5 -7 -2 1
max 3 5 3 9
回答2:
Pandas introduced the agg method for dataframes which makes this even easier:
df.agg([min, max])
Out[207]:
col1 col2 col3 col4
min -5 -7 -2 1
max 3 49 6 9
That's all. Conversion into a list, if needed, can then be done as described in the accepted answer. As a bonus, this can be used with groupby also (which doesn't work well with apply):
df.groupby(by='col1').agg([min, max])
Out[208]:
col2 col3 col4
min max min max min max
col1
-5 2 2 -1 -1 9 9
0 5 49 -2 6 1 6
3 -7 -7 3 3 5 5
回答3:
>>> df = pd.DataFrame([[0, 5], [-5, 2], [3, -7]])
>>> list = [ [min, max] for min, max in zip(df.min(), df.max()) ]
>>> list
[[-5, 3], [-7, 5]]
Other note: You might find DataFrame.describe
method helpful: http://pandas.pydata.org/pandas-docs/dev/generated/pandas.DataFrame.describe.html
来源:https://stackoverflow.com/questions/29276301/max-and-min-value-for-each-colum-of-one-dataframe