pandas 如何在熊猫中设置特定的单元格值？
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How set a particular cell value in pandas?
提问by ajayramesh
I am trying to set a value in panda dataframe.
我正在尝试在Pandas数据框中设置一个值。
ZEROS = np.zeros((4,4), dtype=np.int)
df = pd.DataFrame(ZEROS, columns=['A1','B1','C1','D1'])
df.at[2,3] = 32
df
I don't want NaNfor the entire column, the expected output is below:
我不想要NaN整列，预期的输出如下：
Using numpy I am able to set the value like below
使用 numpy 我可以设置如下值
ZEROS[1][3] = 44
output:
输出：
array([[ 0, 0, 0, 0],
[ 0, 0, 0, 44],
[ 0, 0, 0, 0],
[ 0, 0, 0, 0]])
回答by piRSquared
Use pd.DataFrame.iatto reference and/or assign to the ordinal location of a single cell.
使用pd.DataFrame.iat于参考和/或分配给一个小区的序号位置。
ZEROS = np.zeros((4,4), dtype=np.int)
df = pd.DataFrame(ZEROS, columns=['A1','B1','C1','D1'])
df.iat[2,3] = 32
df
A1 B1 C1 D1
0 0 0 0 0
1 0 0 0 0
2 0 0 0 32
3 0 0 0 0
You could also use ilochowever, iloccan alsotake array like input. This makes ilocmore flexible but also requires more overhead. Therefore, if it is only a single cell you want to change... use iat
你也可以使用iloc但是，iloc可也采取类似阵列输入。这使得iloc更灵活，但也需要更多的开销。因此，如果您只想更改单个单元格...使用iat
Also see this post for more information
另请参阅此帖子以获取更多信息
回答by Psidom
Use iloc:
使用iloc：
df.iloc[2,3] = 32
print(df)
# A1 B1 C1 D1
#0 0 0 0 0
#1 0 0 0 0
#2 0 0 0 32
#3 0 0 0 0
Or if you want to modify by index and column name, use loc:
或者，如果要按索引和列名进行修改，请使用loc：
df.loc[2, 'D1'] = 32
