How to drop a list of rows from Pandas dataframe?

Note that it may be important to use the “inplace” command when you want to do the drop in line.

df.drop(df.index[[1,3]], inplace=True)

Because your original question is not returning anything, this command should be used.
http://pandas.pydata.org/pandas-docs/version/0.17.0/generated/pandas.DataFrame.drop.html

You can also pass to DataFrame.drop the label itself (instead of Series of index labels):

In[17]: df
Out[17]: 
            a         b         c         d         e
one  0.456558 -2.536432  0.216279 -1.305855 -0.121635
two -1.015127 -0.445133  1.867681  2.179392  0.518801

In[18]: df.drop('one')
Out[18]: 
            a         b         c         d         e
two -1.015127 -0.445133  1.867681  2.179392  0.518801

Which is equivalent to:

In[19]: df.drop(df.index[[0]])
Out[19]: 
            a         b         c         d         e
two -1.015127 -0.445133  1.867681  2.179392  0.518801

In a comment to @theodros-zelleke’s answer, @j-jones asked about what to do if the index is not unique. I had to deal with such a situation. What I did was to rename the duplicates in the index before I called drop(), a la:

dropped_indexes = <determine-indexes-to-drop>
df.index = rename_duplicates(df.index)
df.drop(df.index[dropped_indexes], inplace=True)

where rename_duplicates() is a function I defined that went through the elements of index and renamed the duplicates. I used the same renaming pattern as pd.read_csv() uses on columns, i.e., "%s.%d" % (name, count), where name is the name of the row and count is how many times it has occurred previously.

If the DataFrame is huge, and the number of rows to drop is large as well, then simple drop by index df.drop(df.index[]) takes too much time.

In my case, I have a multi-indexed DataFrame of floats with 100M rows x 3 cols, and I need to remove 10k rows from it. The fastest method I found is, quite counterintuitively, to take the remaining rows.

Let indexes_to_drop be an array of positional indexes to drop ([1, 2, 4] in the question).

indexes_to_keep = set(range(df.shape[0])) - set(indexes_to_drop)
df_sliced = df.take(list(indexes_to_keep))

In my case this took 20.5s, while the simple df.drop took 5min 27s and consumed a lot of memory. The resulting DataFrame is the same.

If I want to drop a row which has let’s say index x, I would do the following:

df = df[df.index != x]

If I would want to drop multiple indices (say these indices are in the list unwanted_indices), I would do:

desired_indices = [i for i in len(df.index) if i not in unwanted_indices]
desired_df = df.iloc[desired_indices]

I solved this in a simpler way – just in 2 steps.

1. Make a dataframe with unwanted rows/data.

2. Use the index of this unwanted dataframe to drop the rows from the original dataframe.

Example:
Suppose you have a dataframe df which as many columns including ‘Age’ which is an integer. Now let’s say you want to drop all the rows with ‘Age’ as negative number.

df_age_negative = df[ df['Age'] < 0 ] # Step 1
df = df.drop(df_age_negative.index, axis=0) # Step 2

Hope this is much simpler and helps you.

Here is a bit specific example, I would like to show. Say you have many duplicate entries in some of your rows. If you have string entries you could easily use string methods to find all indexes to drop.

ind_drop = df[df['column_of_strings'].apply(lambda x: x.startswith('Keyword'))].index

And now to drop those rows using their indexes

new_df = df.drop(ind_drop)

Determining the index from the boolean as described above e.g.

df[df['column'].isin(values)].index

can be more memory intensive than determining the index using this method

pd.Index(np.where(df['column'].isin(values))[0])

applied like so

df.drop(pd.Index(np.where(df['column'].isin(values))[0]), inplace = True)

This method is useful when dealing with large dataframes and limited memory.

Use only the Index arg to drop row:-

df.drop(index = 2, inplace = True)

For multiple rows:-

df.drop(index=[1,3], inplace = True)

Consider an example dataframe

df =     
index    column1
0           00
1           10
2           20
3           30

we want to drop 2nd and 3rd index rows.

Approach 1:

df = df.drop(df.index[2,3])
 or 
df.drop(df.index[2,3],inplace=True)
print(df)

df =     
index    column1
0           00
3           30

 #This approach removes the rows as we wanted but the index remains unordered

Approach 2

df.drop(df.index[2,3],inplace=True,ignore_index=True)
print(df)
df =     
index    column1
0           00
1           30
#This approach removes the rows as we wanted and resets the index. 

Look at the following dataframe df

df

   column1  column2  column3
0        1       11       21
1        2       12       22
2        3       13       23
3        4       14       24
4        5       15       25
5        6       16       26
6        7       17       27
7        8       18       28
8        9       19       29
9       10       20       30

Lets drop all the rows which has an odd number in column1

Create a list of all the elements in column1 and keep only those elements that are even numbers (the elements that you dont want to drop)

keep_elements = [x for x in df.column1 if x%2==0]

All the rows with the values [2, 4, 6, 8, 10] in its column1 will be retained or not dropped.

df.set_index('column1',inplace = True)
df.drop(df.index.difference(keep_elements),axis=0,inplace=True)
df.reset_index(inplace=True)

We make the column1 as index and drop all the rows that are not required. Then we reset the index back.
df

   column1  column2  column3
0        2       12       22
1        4       14       24
2        6       16       26
3        8       18       28
4       10       20       30

To drop rows with indices 1, 2, 4 you can use:

df[~df.index.isin([1, 2, 4])]

The tilde operator ~ negates the result of the method isin. Another option is to drop indices:

df.loc[df.index.drop([1, 2, 4])]

As Dennis Golomazov’s answer suggests, using drop to drop rows. You can select to keep rows instead. Let’s say you have a list of row indices to drop called indices_to_drop. You can convert it to a mask as follows:

mask = np.ones(len(df), bool)
mask[indices_to_drop] = False

You can use this index directly:

df_new = df.iloc[mask]

The nice thing about this method is that mask can come from any source: it can be a condition involving many columns, or something else.

The really nice thing is, you really don’t need the index of the original DataFrame at all, so it doesn’t matter if the index is unique or not.

The disadvantage is of course that you can’t do the drop in-place with this method.

This worked for me

# Create a list containing the index numbers you want to remove
index_list = list(range(42766, 42798))
df.drop(df.index[index_list], inplace =True)
df.shape

This should drop all indexes within that created range