Python Pandas:根据位置而非索引值替换值 [英] Python pandas: replace values based on location not index value
问题描述
这是我的df:
In[12]: df = pd.DataFrame(data = list("aabbcc"), columns = ["s"], index=range(11,17))
In[13]: df
Out[13]:
s
11 a
12 a
13 b
14 b
15 c
16 c
现在,根据索引值替换值:
Now, replacing values based on index values:
In[14]: df.loc[11, "s"] = 'A'
In[15]: df
Out[15]:
s
11 A
12 a
13 b
14 b
15 c
16 c
In[16]: df.ix[12, "s"] = 'B'
In[17]: df
Out[17]:
s
11 A
12 B
13 b
14 b
15 c
16 c
是否可以基于位置而不是索引值执行相同的操作,类似这样,但是它显示ValueError(ValueError: Can only index by location with a [integer, integer slice (START point is INCLUDED, END point is EXCLUDED), listlike of integers, boolean array]
):
Is it possible to do the same based on position not index value, something like this, but it shows ValueError (ValueError: Can only index by location with a [integer, integer slice (START point is INCLUDED, END point is EXCLUDED), listlike of integers, boolean array]
):
In[18]: df.iloc[1, "s"] = 'b'
而且,如果我尝试这样的事情:
And, if I try something like this:
df.s.iloc[1] = "b"
我收到此警告:
SettingWithCopyWarning:
A value is trying to be set on a copy of a slice from a DataFrame
推荐答案
您可以使用get_loc获取列的位置并将其传递给iloc:
You can use get_loc to get the location of the column and pass that to iloc:
df.iloc[1, df.columns.get_loc('s')] = 'B'
df
Out:
s
11 a
12 B
13 b
14 b
15 c
16 c
反之亦然:
df.loc[df.index[1], 's'] = 'B'
这篇关于Python Pandas:根据位置而非索引值替换值的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!