I want to read every nth row of a list of DataFrames and create a new DataFrames by appending all the Nth rows.
Let's say we have the following DataFrames:
>>> df1
A B C D
0 -0.8 -2.8 -0.3 -0.1
1 -0.1 -0.9 0.2 -0.7
2 0.7 -3.3 -1.1 -0.4
>>> df2
A B C D
0 1.4 -0.7 1.5 -1.3
1 1.6 1.4 1.4 0.2
2 -1.4 0.2 -1.7 0.7
>>> df3
A B C D
0 0.3 -0.5 -1.6 -0.8
1 0.2 -0.5 -1.1 1.6
2 -0.3 0.7 -1.0 1.0
I have used the following approach to get the desired df:
df = pd.DataFrame()
df_list = [df1, df2, df3]
for i in range(len(df1)):
for x in df_list:
df = df.append(x.loc[i], ignore_index = True)
Here's the result:
>>> df
A B C D
0 -0.8 -2.8 -0.3 -0.1
1 1.4 -0.7 1.5 -1.3
2 0.3 -0.5 -1.6 -0.8
3 -0.1 -0.9 0.2 -0.7
4 1.6 1.4 1.4 0.2
5 0.2 -0.5 -1.1 1.6
6 0.7 -3.3 -1.1 -0.4
7 -1.4 0.2 -1.7 0.7
8 -0.3 0.7 -1.0 1.0
I was just wondering if there is a pandas way of rewriting this code which would do the same thing (maybe by using .iterrows, pd.concat, pd.join, or pd.merge)?
Cheers
Update Simply appending one df after another is not what I am looking for here.
The code should do:
df.row1 = df1.row1
df.row2 = df2.row1
df.row3 = df3.row1
df.row4 = df1.row2
df.row5 = df2.row2
df.row6 = df3.row2
...
For a single output dataframe, you can concatenate and sort by index:
res = pd.concat([df1, df2, df3]).sort_index().reset_index(drop=True)
A B C D
0 -0.8 -2.8 -0.3 -0.1
1 1.4 -0.7 1.5 -1.3
2 0.3 -0.5 -1.6 -0.8
3 -0.1 -0.9 0.2 -0.7
4 1.6 1.4 1.4 0.2
5 0.2 -0.5 -1.1 1.6
6 0.7 -3.3 -1.1 -0.4
7 -1.4 0.2 -1.7 0.7
8 -0.3 0.7 -1.0 1.0
For a dictionary of dataframes, You can concatenate and then group by index:
res = dict(tuple(pd.concat([df1, df2, df3]).groupby(level=0)))
With the dictionary defined as above, each value represents a row number. For example, res[0] will give the first row from each input dataframe.
There is pd.concat
df=pd.concat([df1,df2,df3]).reset_index(drop=True)
recommended by Jez
df=pd.concat([df1,df2,df3],ignore_index=True)
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