df.append() 未附加到 DataFrame

问题描述

我制定了 this question关于添加带索引的行，但我还不清楚当没有索引时如何/为什么会发生这种情况:

columnsList=['A','B','C','D']df8=pd.DataFrame(columns=columnsList)L=['值 aa','值 bb','值 cc','值 dd']s = pd.Series(dict(zip(df8.columns, L)))df8.append(s,ignore_index=True)df8.append(s,ignore_index=True)

我希望这里有一个 2X4 数据帧.但是没有添加任何值，也没有发生错误.

打印(df8.shape)#>>>(0,4)

为什么没有添加系列，为什么没有报错?

<小时>

如果我尝试使用 LOC 添加一行，则会添加一个索引，

df8.loc[df8.index.max() + 1, :] = [4, 5, 6,7]打印(df8)

结果:

 A B C D南 4 5 6 7

我猜LOC和iLOC都不能用来追加没有索引名的行(即Loc添加索引名NaN，当索引号高于数据库的行时不能使用iLoc)

解决方案

DataFrame.append 不是就地操作.从文档中，

<块引用>

DataFrame.append(other, ignore_index=False, verify_integrity=False, sort=None)

将其他行追加到此帧的末尾，返回一个新对象.不在此框架中的列将作为新列添加.

您需要将结果分配回去.

df8 = df8.append([s] * 2, ignore_index=True)df8A B C D0 值 aa 值 bb 值 cc 值 dd1 值 aa 值 bb 值 cc 值 dd

I formulated this question about adding rows WITH index, but it is not yet clear to me how/why this happens when there are no indexes:

columnsList=['A','B','C','D']
df8=pd.DataFrame(columns=columnsList)
L=['value aa','value bb','value cc','value dd']
s = pd.Series(dict(zip(df8.columns, L)))
df8.append(s,ignore_index=True)
df8.append(s,ignore_index=True)

I EXPECT HERE A 2X4 DATAFRAME. nevertheless no values where added, nor an error occurred.

print(df8.shape)
#>>> (0,4)

Why is the series not being added, and why is not given any error?

---

If I try to add a row with LOC, an index is added,

df8.loc[df8.index.max() + 1, :] = [4, 5, 6,7]
print(df8)

result:

     A  B  C  D
NaN  4  5  6  7

I guess neither LOC, nor iLOC could be used to append rows without index name (i.e. Loc adds the index name NaN, and iLoc can not be used when the index number is higher than the rows of the database)

解决方案

DataFrame.append is not an in-place operation. From the docs,

DataFrame.append(other, ignore_index=False, verify_integrity=False, sort=None)

Append rows of other to the end of this frame, returning a new object. Columns not in this frame are added as new columns.

You need to assign the result back.

df8 = df8.append([s] * 2, ignore_index=True)
df8
          A         B         C         D
0  value aa  value bb  value cc  value dd
1  value aa  value bb  value cc  value dd