如何从熊猫数据框中删除行列表？

I have a dataframe df :

>>> df
                  sales  discount  net_sales    cogs
STK_ID RPT_Date                                     
600141 20060331   2.709       NaN      2.709   2.245
       20060630   6.590       NaN      6.590   5.291
       20060930  10.103       NaN     10.103   7.981
       20061231  15.915       NaN     15.915  12.686
       20070331   3.196       NaN      3.196   2.710
       20070630   7.907       NaN      7.907   6.459

Then I want to drop rows with certain sequence numbers which indicated in a list, suppose here is [1,2,4], then left:

                  sales  discount  net_sales    cogs
STK_ID RPT_Date                                     
600141 20060331   2.709       NaN      2.709   2.245
       20061231  15.915       NaN     15.915  12.686
       20070630   7.907       NaN      7.907   6.459

How or what function can do that ?

Use DataFrame.drop and pass it a Series of index labels:

In [65]: df
Out[65]: 
       one  two
one      1    4
two      2    3
three    3    2
four     4    1


In [66]: df.drop(df.index[[1,3]])
Out[66]: 
       one  two
one      1    4
three    3    2

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

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.

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

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.

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]

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)

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.

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.