iloc用pandas float64 vs int64设置值

Question

I am basically trying to set a value in a DataFrame using iloc with an index and a column name. Here is the test code:

import pandas as pd

df = pd.DataFrame({'a': [1, 2], 'b': [3, 4]})
print type(df['a'][0])      # numpy.int64
df.iloc[-1]['a'] = 2000
print df                    # value changed to 2000
df['c'] = [3.5, 4.5]
print type(df['a'][0])      # numpy.float64 -> why does this change automatically?
print type(df['c'][0])      # numpy.float64
df.iloc[-1]['c'] = 2000     # yields warning, no value change
print df
df.iloc[-1]['a'] = 4000     # yields warning, no value change
print df

With Int64, I can do it, but not with Float64. Is there an alternative? Or is this a bug? Thanks

Answer 1

Jeff's answer is correct, but not optimal pandas syntax. It is considerably faster to to use .at instead of .loc (and likewise .iat instead of .iloc). If you're using this to get or set a lot of values, the time savings can really add up. The catch is that .at and .iat are meant only for getting and setting individual values, so you can't get or set a range of values or retrieve an entire column or row. Since all you want to do is change one value I would recommend:

df.at[df.index[-1], 'a'] = 4000

If for whatever reason you only have the label-based location for one axis and the integer-based location on the other, you can also use df.get_loc('Label-based name of column') to get the integer-based location of that column or likewise df.index.get_loc('Label-based name of row') to get the integer-based location of that row.

Answer 2

The warning is telling you that what you are doing is unsafe and it is because you have a mixed-type frame. Instead use loc for this. See the docs on why this ia bad idea and may not work (which it doesn't here), http://pandas.pydata.org/pandas-docs/stable/indexing.html#indexing-view-versus-copy

In [12]: df.loc[df.index[-1],'a'] = 4000

In [13]: df
Out[13]: 
      a  b    c
0     1  3  3.5
1  4000  4  4.5

[2 rows x 3 columns]