CODEWITHSUNDEEP
pythonpandasnumpydataframe
Ashish Lal
Ashish Lal

Reputation: 21

Is there an efficient way to select multiple rows in a large pandas data frame?

I am working on a large pandas adatframe with about 100 million rows and 2 columns. I want to iterate over the dataframe and efficiently set a third column depending on the values of col1 and col2. This is what I am currently doing -

df[col3] = 0
for idx, row in df.iterrows():
    val1 = row[col1]
    val2 = row[col2]
    df1 = df.loc[(df.col1 == val2) & (df.col2 == val1)]
    if len(df1) > 0:
        df.loc[(df.col1 == val2) & (df.col2 == val1), col3] = 1
Example:
    df = pd.DataFrame({'col1':[0,1,2,3,4,11], 'col2':[10,11,12,4,3,0]})
    >> df.head()
        col1 col2
     0  0   10
     1  1   11
     2  2   12
     3  3   4
     4  4   3
     5  3   10
    I want to add 'col3' such that last 2 rows of the third column are
    1. Think of it as a reverse_edge column which is 1 when for each 
    (val1, val2) in col1, col2 there is a (val2, val1) in col1, col2
        col1    col2    col3
      0 0        10      0
      1 1        11      0
      2 2        12      0
      3 3        4       1
      4 4        3       1
      5 11       0       0

What is the most efficient way to do this computation? It is currently taking me hours to traverse the entire dataframe.

EDIT: Think of each value in col1 and corresponding value in col2 as an edge in a graph (val1 -> val2). I want to know if a reverse edge exists or not (val2 -> val1).

Upvotes: 1

Views: 594

Answers (3)

jezrael
jezrael

Reputation: 862761

Use:

df1 = pd.DataFrame(np.sort(df[['col1', 'col2']], axis=1), index=df.index)
df['col3'] = df1.duplicated(keep=False).astype(int)
print (df)
   col1  col2  col3
0     0    10     0
1     1    11     0
2     2    12     0
3     3     4     1
4     4     3     1

Another solution with merge and compare subsets, compare to 2d arrays, last use np.all for check all True per rows:

df2 = df.merge(df, how='left', left_on='col2', right_on='col1')

df['col3'] = ((df2[['col1_x','col2_x']].values == 
               df2[['col2_y','col1_y']].values).all(axis=1).astype(int))
#pandas 0.24+
#https://stackoverflow.com/a/54508052
#df['col3'] = ((df2[['col1_x','col2_x']].to_numpy() ==
                df2[['col2_y','col1_y']].to_numpy()).all(axis=1).astype(int))
print (df)
   col1  col2  col3
0     0    10     0
1     1    11     0
2     2    12     0
3     3     4     1
4     4     3     1
5    11     0     0

print ((df2[['col1_x','col2_x']].values == df2[['col2_y','col1_y']].values))


[[False False]
 [False  True]
 [False False]
 [ True  True]
 [ True  True]
 [False  True]]

Upvotes: 0

fuglede
fuglede

Reputation: 18201

Along the same lines as @Jondiedoop's answer, you can safe a bit of suffix wrangling and stick to an inner join by merging on both columns at once,

df['col3'] = df.index.isin(df.merge(df, left_on=['col1', 'col2'], right_on=['col2', 'col1'], left_index=True).index).astype(int)

For example:

In [40]: df
Out[40]:
   col1  col2
0     0    10
1     1    11
2     2    12
3     3     4
4     4     3
5    11     0
6     0    10

In [41]: df['col3'] = df.index.isin(df.merge(df, left_on=['col1', 'col2'], right_on=['col2', 'col1'], left_index=True).index).astype(int)

In [42]: df
Out[42]:
   col1  col2  col3
0     0    10     0
1     1    11     0
2     2    12     0
3     3     4     1
4     4     3     1
5    11     0     0
6     0    10     0

An equivalent approach would be:

df['col3'] = 0
df.loc[df.merge(df, left_on=['col1', 'col2'], right_on=['col2', 'col1'], left_index=True).index, 'col3'] = 1

Upvotes: 1

Jondiedoop
Jondiedoop

Reputation: 3353

My solution would be to merge the frame to itself (merging column 2 to column 1) and then checking if the other two columns would be identical: that would mean the reverse also exists:

df2 = df.merge(df, how='left', left_on='col2', right_on='col1')
df['rev_exists'] = (df2['col1_x'] == df2['col2_y']).astype(int)
df
#   col1  col2  rev_exists
#0     0    10           0
#1     1    11           0
#2     2    12           0
#3     3     4           1
#4     4     3           1
#5    11     0           0

Upvotes: 1

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