合并多个大型DataFrame的有效方法
假设我有4个小型DataFrame
df1
,df2
,df3
和df4
import pandas as pdfrom functools import reduce
import numpy as np
df1 = pd.DataFrame([['a', 1, 10], ['a', 2, 20], ['b', 1, 4], ['c', 1, 2], ['e', 2, 10]])
df2 = pd.DataFrame([['a', 1, 15], ['a', 2, 20], ['c', 1, 2]])
df3 = pd.DataFrame([['d', 1, 10], ['e', 2, 20], ['f', 1, 1]])
df4 = pd.DataFrame([['d', 1, 10], ['e', 2, 20], ['f', 1, 15]])
df1.columns = ['name', 'id', 'price']
df2.columns = ['name', 'id', 'price']
df3.columns = ['name', 'id', 'price']
df4.columns = ['name', 'id', 'price']
df1 = df1.rename(columns={'price':'pricepart1'})
df2 = df2.rename(columns={'price':'pricepart2'})
df3 = df3.rename(columns={'price':'pricepart3'})
df4 = df4.rename(columns={'price':'pricepart4'})
上面创建的是4个DataFrame,下面的代码是我想要的。
# Merge dataframesdf = pd.merge(df1, df2, left_on=['name', 'id'], right_on=['name', 'id'], how='outer')
df = pd.merge(df , df3, left_on=['name', 'id'], right_on=['name', 'id'], how='outer')
df = pd.merge(df , df4, left_on=['name', 'id'], right_on=['name', 'id'], how='outer')
# Fill na values with 'missing'
df = df.fillna('missing')
因此,我为4个没有很多行和列的DataFrame实现了这一点。
因此,我通过使用lambda reduce的另一个StackOverflow答案构建了这个解决方案:
from functools import reduceimport pandas as pd
import numpy as np
dfList = []
#To create the 48 DataFrames of size 62245 X 3
for i in range(0, 49):
dfList.append(pd.DataFrame(np.random.randint(0,100,size=(62245, 3)), columns=['name', 'id', 'pricepart' + str(i + 1)]))
#The solution I came up with to extend the solution to more than 3 DataFrames
df_merged = reduce(lambda left, right: pd.merge(left, right, left_on=['name', 'id'], right_on=['name', 'id'], how='outer'), dfList).fillna('missing')
这引起了MemoryError
。
我不知道该怎么做才能阻止内核崩溃。.我已经坚持了两天。.我执行的EXACT
merge操作的一些代码不会导致MemoryError
或产生与您相同的结果结果,将不胜感激。
另外,在主数据帧(不是可再现48个DataFrames中的例子)的3列的类型int64
,int64
和float64
与我宁愿他们留,因为整数和浮子,它代表的这种方式。
编辑:
我不是以迭代方式尝试运行合并操作或使用reduce
lambda函数,而是以2为一组来完成它!另外,我更改了某些列的数据类型,而有些则不需要float64
。所以我把它归结为float16
。它距离很远,但最终仍会抛出MemoryError
。
intermediatedfList = dfListtempdfList = []
#Until I merge all the 48 frames two at a time, till it becomes size 2
while(len(intermediatedfList) != 2):
#If there are even number of DataFrames
if len(intermediatedfList)%2 == 0:
#Go in steps of two
for i in range(0, len(intermediatedfList), 2):
#Merge DataFrame in index i, i + 1
df1 = pd.merge(intermediatedfList[i], intermediatedfList[i + 1], left_on=['name', 'id'], right_on=['name', 'id'], how='outer')
print(df1.info(memory_usage='deep'))
#Append it to this list
tempdfList.append(df1)
#After DataFrames in intermediatedfList merging it two at a time using an auxillary list tempdfList,
#Set intermediatedfList to be equal to tempdfList, so it can continue the while loop.
intermediatedfList = tempdfList
else:
#If there are odd number of DataFrames, keep the first DataFrame out
tempdfList = [intermediatedfList[0]]
#Go in steps of two starting from 1 instead of 0
for i in range(1, len(intermediatedfList), 2):
#Merge DataFrame in index i, i + 1
df1 = pd.merge(intermediatedfList[i], intermediatedfList[i + 1], left_on=['name', 'id'], right_on=['name', 'id'], how='outer')
print(df1.info(memory_usage='deep'))
tempdfList.append(df1)
#After DataFrames in intermediatedfList merging it two at a time using an auxillary list tempdfList,
#Set intermediatedfList to be equal to tempdfList, so it can continue the while loop.
intermediatedfList = tempdfList
有什么我可以优化代码来避免的方法MemoryError
,我什至使用过AWS 192GB
RAM(我现在欠他们7美元,我本来可以给你们一个),这比我得到的要远远得多,而且将MemoryError
28个DataFrame的列表减少到4个后仍然抛出。
回答:
通过使用执行索引对齐的串联,您可能会获得一些好处pd.concat
。希望它应该比外部合并更快,更有效地利用内存。
df_list = [df1, df2, ...]for df in df_list:
df.set_index(['name', 'id'], inplace=True)
df = pd.concat(df_list, axis=1) # join='inner'
df.reset_index(inplace=True)
或者,您可以用concat
迭代代替(第二步)join
:
from functools import reducedf = reduce(lambda x, y: x.join(y), df_list)
这可能会更好,也可能不会更好merge
。
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