pandas 大熊猫在行上迭代作为字典
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pandas iteration over rows as dict
提问by Matina G
Helllo,
你好，
I need to iterate over a pandas dataframe in order to pass each row as argument of a function (actually, class constructor) with **kwargs. This means that each row should behave as a dictionary with keys the column names and values the corresponding ones for each row.
我需要遍历 Pandas 数据框，以便将每一行作为函数（实际上是类构造函数）的参数传递给**kwargs. 这意味着每一行都应该像一个字典，键是列名，值是每行对应的值。
This works, but it performs very badly:
这有效，但它的表现非常糟糕：
import pandas as pd
def myfunc(**kwargs):
try:
area = kwargs.get('length', 0)* kwargs.get('width', 0)
return area
except TypeError:
return 'Error : length and width should be int or float'
df = pd.DataFrame({'length':[1,2,3], 'width':[10, 20, 30]})
for i in range(len(df)):
print myfunc(**df.iloc[i])
Any suggestions on how to make that more performing ? I have tried iterating with tried df.iterrows(),
but I get the following error :
关于如何提高性能的任何建议？我尝试使用 try 进行迭代df.iterrows()，但出现以下错误：
TypeError: myfunc() argument after ** must be a mapping, not tuple
类型错误：** 之后的 myfunc() 参数必须是映射，而不是元组
I have also tried df.itertuples()and df.values, but either I am missing something, or it means that I have to convert each tuple / np.array to a pd.Series or dict , which will also be slow.
My constraint is that the script has to work with python 2.7 and pandas 0.14.1.
我也试过df.itertuples()and df.values，但要么我遗漏了一些东西，要么意味着我必须将每个元组/ np.array 转换为 pd.Series 或 dict ，这也会很慢。我的限制是脚本必须使用 python 2.7 和 pandas 0.14.1。
Thanks in advance for your help!
在此先感谢您的帮助！
采纳答案by stellasia
You can try:
你可以试试：
for k, row in df.iterrows():
myfunc(**row)
Here kis the dataframe index and rowis a dict, so you can access any column with: row["my_column_name"]
这k是数据框索引并且row是一个字典，因此您可以使用以下方式访问任何列：row["my_column_name"]
回答by avloss
one clean option is this one:
一个干净的选择是这个：
for row_dict in df.to_dict(orient="row"):
print(row_dict['column_name']
回答by jpp
Defining a separate function for this will be inefficient, as you are applying row-wise calculations. More efficient would be to calculate a new series, then iterate the series:
为此定义一个单独的函数将是低效的，因为您正在应用逐行计算。更有效的是计算一个新系列，然后迭代该系列：
df = pd.DataFrame({'length':[1,2,3,'test'], 'width':[10, 20, 30,'hello']})
df2 = df.iloc[:].apply(pd.to_numeric, errors='coerce')
error_str = 'Error : length and width should be int or float'
print(*(df2['length'] * df2['width']).fillna(error_str), sep='\n')
10.0
40.0
90.0
Error : length and width should be int or float
