Python 如何组合熊猫中一个低于另一个的数据帧?
当我调用fun(列表_1)时,我会得到4个数据帧,一个低于另一个:Python 如何组合熊猫中一个低于另一个的数据帧?,python,python-2.7,pandas,Python,Python 2.7,Pandas,当我调用fun(列表_1)时,我会得到4个数据帧,一个低于另一个: list_1 =[1, 2, 3, 4 ] def fun(list_1): for each value in list1: # perform some operation and create a new data frame(pandas) for each value in the list # so in total I should get 4 data frames.
list_1 =[1, 2, 3, 4 ]
def fun(list_1):
for each value in list1:
# perform some operation and create a new data frame(pandas) for each value in the list
# so in total I should get 4 data frames.
yield new_data_frame
我的问题是,如果我需要垂直组合熊猫数据帧,该怎么办
data_frame_1
data_frame_2
data_frame_3
data_frame_4
将它们存储在列表中,然后
pd.concat(df\u list,ignore\u index=True)
或
pd.concat([data\u frame\u 1,data\u frame\u 2,data\u frame\u 3,data\u frame\u 4],ignore\u index=True)
os simpler只需将乐趣作为参数传递给:
那太神奇了!!成功了。你能解释一下ignore_index=True在这里做了什么吗?
ignore_index
实际上是对连接的数据帧重新编制索引,否则会得到重复的索引值
data_frame_1
+
data_frame_2
+
data_frame_3
+
data_frame_4
In [8]:
list_1 = [1,2,3,4]
def fun(list_1):
for each_value in list_1:
# perform some operation and create a new data frame(pandas) named "new_data_frame" for each value in the list
yield pd.DataFrame({'a':[each_value]})
pd.concat(fun(list_1), ignore_index=True)
Out[8]:
a
0 1
1 2
2 3
3 4