pandas 熊猫添加行而不是列
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pandas add row instead of column
提问by Pavel M.
I'm new to pandas, but trying to simply add a row
我是Pandas的新手，但试图简单地添加一行
class Security:
def __init__(self):
self.structure = ['timestamp', 'open', 'high', 'low', 'close', 'vol']
self.df = pd.DataFrame(columns=self.structure) # index =
def whats_inside(self):
return self.df
"""
Some skipped code...
"""
def add_data(self, timestamp, open, high, low, close, vol):
data = [timestamp, open, high, low, close, vol]
self.df = self.df.append (data)
sec = Security()
print sec.whats_inside()
sec.add_data ('2015/06/01', '1', '2', '0.5', '1', '100')
print sec.whats_inside()
but the output is:
但输出是：
0 close high low open timestamp vol
0 2015/06/01 NaN NaN NaN NaN NaN NaN
1 1 NaN NaN NaN NaN NaN NaN
2 2 NaN NaN NaN NaN NaN NaN
3 0.5 NaN NaN NaN NaN NaN NaN
4 1 NaN NaN NaN NaN NaN NaN
5 100 NaN NaN NaN NaN NaN NaN
This means, I'm adding a column instead of row. Yes, I've tried to google but still didnt get the point how do make it simple pythonic way.
这意味着，我要添加一列而不是行。是的，我试过谷歌，但仍然没有明白如何使它简单的pythonic方式。
p.s. I know that's simple, but I'm just missing something important.
ps 我知道这很简单，但我只是遗漏了一些重要的东西。
回答by DeepSpace
There are several ways to add a new row. Perhaps the easiest one is (if you want to add the row to the end) is to use loc:
有多种方法可以添加新行。也许最简单的方法是（如果您想将行添加到最后）是使用loc：
df.loc[len(df)] = ['val_a', 'val_b', .... ]
locexpects an index. len(df)will return the number of rows in the dataframe so the new row will be added to the end of the dataframe.
loc期待一个索引。len(df)将返回数据帧中的行数，因此新行将被添加到数据帧的末尾。
'['val_a', 'val_b', .... ]' is a list of values of the row, in the same order of the columns, so the list's length must be equal to the number of columns, otherwise you will get a ValueErrorexception.
An exception for this is that if you want all the columns to have the same values you are allowed to have that value as a single element in the list, for example df.loc[len(df)] = ['aa'].
'['val_a', 'val_b', .... ]' 是行值的列表，列的顺序相同，所以列表的长度必须等于列数，否则会得到一个ValueError例外。一个例外是，如果您希望所有列都具有相同的值，则可以将该值作为列表中的单个元素，例如df.loc[len(df)] = ['aa'].
NOTE:a good idea will be to always use reset_indexbefore using this method because if you ever delete a row or work on a filtered dataframe you are not guaranteed that the rows' indexes will be in sync with the number of rows.
注意：一个好主意是reset_index在使用此方法之前始终使用，因为如果您曾经删除过一行或处理过滤过的数据帧，则不能保证行的索引与行数同步。
回答by yangjie
You should append Series or DataFrame. (Series would be more appropriate in your case)
您应该附加 Series 或 DataFrame。（系列更适合您的情况）
import pandas as pd
from pandas import Series, DataFrame
class Security:
def __init__(self):
self.structure = ['timestamp', 'open', 'high', 'low', 'close', 'vol']
self.df = pd.DataFrame(columns=self.structure) # index =
def whats_inside(self):
return self.df
"""
Some skipped code...
"""
def add_data(self, timestamp, open, high, low, close, vol):
data = [timestamp, open, high, low, close, vol]
# append Series
self.df = self.df.append(pd.Series(data, index=self.structure), ignore_index=True)
# or DataFrame
# self.df = self.df.append(pd.DataFrame([data], columns=self.structure), ignore_index=True)
sec = Security()
print sec.whats_inside()
sec.add_data ('2015/06/01', '1', '2', '0.5', '1', '100')
sec.add_data ('2015/06/02', '1', '2', '0.5', '1', '100')
print sec.whats_inside()
Output:
输出：
timestamp open high low close vol
0 2015/06/01 1 2 0.5 1 100
1 2015/06/02 1 2 0.5 1 100
