Python Pandas 按列值拆分 DataFrame
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Pandas split DataFrame by column value
提问by 146 percent Russian
I have DataFramewith column Sales.
我有DataFrame列Sales。
How can I split it into 2 based on Salesvalue?
如何根据Sales价值将其分成 2 个？
First DataFramewill have data with 'Sales' < sand second with 'Sales' >= s
第一个DataFrame将有数据，'Sales' < s第二个有'Sales' >= s
采纳答案by jezrael
You can use boolean indexing:
您可以使用boolean indexing：
df = pd.DataFrame({'Sales':[10,20,30,40,50], 'A':[3,4,7,6,1]})
print (df)
A Sales
0 3 10
1 4 20
2 7 30
3 6 40
4 1 50
s = 30
df1 = df[df['Sales'] >= s]
print (df1)
A Sales
2 7 30
3 6 40
4 1 50
df2 = df[df['Sales'] < s]
print (df2)
A Sales
0 3 10
1 4 20
It's also possible to invert maskby ~:
也可以mask通过~以下方式反转：
mask = df['Sales'] >= s
df1 = df[mask]
df2 = df[~mask]
print (df1)
A Sales
2 7 30
3 6 40
4 1 50
print (df2)
A Sales
0 3 10
1 4 20
print (mask)
0 False
1 False
2 True
3 True
4 True
Name: Sales, dtype: bool
print (~mask)
0 True
1 True
2 False
3 False
4 False
Name: Sales, dtype: bool
回答by Zero
Using groupbyyou could split into two dataframes like
使用groupby您可以拆分为两个数据帧，例如
In [1047]: df1, df2 = [x for _, x in df.groupby(df['Sales'] < 30)]
In [1048]: df1
Out[1048]:
A Sales
2 7 30
3 6 40
4 1 50
In [1049]: df2
Out[1049]:
A Sales
0 3 10
1 4 20
回答by keryruo
Using "groupby" and list comprehension:
使用“groupby”和列表理解：
Storing all the split dataframe in list variable and accessing each of the seprated dataframe by their index.
将所有拆分的数据帧存储在列表变量中，并通过索引访问每个分离的数据帧。
DF = pd.DataFrame({'chr':["chr3","chr3","chr7","chr6","chr1"],'pos':[10,20,30,40,50],})
ans = [pd.DataFrame(y) for x, y in DF.groupby('chr', as_index=False)]
accessing the separated DF like this:
像这样访问分离的 DF：
ans[0]
ans[1]
ans[len(ans)-1] # this is the last separated DF
accessing the column value of the separated DF like this:
像这样访问分离的 DF 的列值：
ansI_chr=ans[i].chr
