如何在熊猫中用一个值填充一个列？

我在熊猫数据框中有一个连续数字的列。

我想将所有这些值更改为一个简单的字符串，比如“ foo”，结果是

A
foo
foo
foo
foo

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最佳答案

Just select the column and assign like normal:

In [194]:
df['A'] = 'foo'
df


Out[194]:
A
0  foo
1  foo
2  foo
3  foo

Assigning a scalar value will set all the rows to the same scalar value

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The good answer above throws a warning. You can also do:

df.insert(0, 'A', 'foo')

where 0 is the index where the new column will be inserted.

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You could also try pd.Series.replace:

df['A'] = df['A'].replace(df['A'], 'foo')
print(df)

Output:

     A
0  foo
1  foo
2  foo
3  foo

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You can use the method assign:

df = df.assign(A='foo')

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You can also exploit the power of the .loc property by addressing all the rows using : as the argument. Say that your DataFrame is called df:

df.loc[:,'A'] = 'foo'

Resulting in

     A
0  foo
1  foo
2  foo
3  foo

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For this to work without receiving the slice error/warning, you can do this:

df.loc[:]['A'] = 'foo'