Is there a faster way to update dataframe column v

问题:

I am trying to process a dataframe. This includes creating new columns and updating their values based on the values in other columns. More concretely, I have a predefined "source" that I want to classify. This source can fall under three different categories 'source_dtp', 'source_dtot', and 'source_cash'. I want to add three new columns to the dataframe that are comprised of either 1's or 0's based on the original "source" column.

I am currently able to do this, it's just really slow...

Original column sample:

source
_id                     
AV4MdG6Ihowv-SKBN_nB    DTP
AV4Mc2vNhowv-SKBN_Rn    Cash 1
AV4MeisikOpWpLdepWy6    DTP
AV4MeRh6howv-SKBOBOn    Cash 1
AV4Mezwchowv-SKBOB_S    DTOT
AV4MeB7yhowv-SKBOA5b    DTP

Desired output:

source_dtp  source_dtot source_cash
_id         
AV4MdG6Ihowv-SKBN_nB    1.0 0.0 0.0
AV4Mc2vNhowv-SKBN_Rn    0.0 0.0 1.0
AV4MeisikOpWpLdepWy6    1.0 0.0 0.0
AV4MeRh6howv-SKBOBOn    0.0 0.0 1.0
AV4Mezwchowv-SKBOB_S    0.0 1.0 0.0
AV4MeB7yhowv-SKBOA5b    1.0 0.0 0.0

This is my current approach, but it's very slow. I would much prefer a vectorized form of doing this but I don't know how - as the condition is very elaborate.

# For 'source' we will use the following classes:
source_cats = ['source_dtp', 'source_dtot', 'source_cash']
# [0, 0, 0] would imply 'other', hence no need for a fourth category

# add new features to dataframe, initializing to nan
for cat in source_cats:
    data[cat] = np.nan

for row in data.itertuples():
    # create series to hold the result per row e.g. [1, 0, 0] for `cash`
    cat = [0, 0, 0]
    index = row[0]
    # to string as some entries are numerical
    source_type = str(data.loc[index, 'source']).lower()
    if 'dtp' in source_type:
        cat[0] = 1
    if 'dtot' in source_type:
        cat[1] = 1
    if 'cash' in source_type:
        cat[2] = 1
    data.loc[index, source_cats] = cat

I am using itertuples() as it proved faster than interrows().

Is there a faster way of achieving the same functionality as above?

EDIT: This is not just with regards to creating a one hot encoding. It boils down to updating the column values dependent on the value of another column. E.g. if I have a certain location_id I want to update its respective longitude and latitude columns - based on that original id (without iterating in the way that I do above because it's really slow for large datasets).

回答1:

Another way to do this is to use pd.get_dummies on the dataframe. First put '_id' into the index.

source = source.set_index('_id')
df_out = pd.get_dummies(source).reset_index()

print(df_out)

Output:

                    _id  source_Cash 1  source_DTOT  source_DTP
0  AV4MdG6Ihowv-SKBN_nB              0            0           1
1  AV4Mc2vNhowv-SKBN_Rn              1            0           0
2  AV4MeisikOpWpLdepWy6              0            0           1
3  AV4MeRh6howv-SKBOBOn              1            0           0
4  AV4Mezwchowv-SKBOB_S              0            1           0
5  AV4MeB7yhowv-SKBOA5b              0            0           1

回答2:

You can use str.get_dummies to get your OHEncodings.

c = df.source.str.get_dummies().add_prefix('source_').iloc[:, ::-1]
c.columns = c.columns.str.lower().str.split().str[0]
print(c)
   source_dtp  source_dtot  source_cash
0           1            0            0
1           0            0            1
2           1            0            0
3           0            0            1
4           0            1            0
5           1            0            0

Next, concatenate c with _id using pd.concat.

df = pd.concat([df._id, c], 1)
print(df)
                    _id  source_dtp  source_dtot  source_cash
0  AV4MdG6Ihowv-SKBN_nB           1            0            0
1  AV4Mc2vNhowv-SKBN_Rn           0            0            1
2  AV4MeisikOpWpLdepWy6           1            0            0
3  AV4MeRh6howv-SKBOBOn           0            0            1
4  AV4Mezwchowv-SKBOB_S           0            1            0
5  AV4MeB7yhowv-SKBOA5b           1            0            0

---

Improvement! Now slightly smoother, thanks to Scott Boston's set_index - reset_index paradigm:

df = df.set_index('_id')\
      .source.str.get_dummies().iloc[:, ::-1]
df.columns = df.columns.str.lower().str.split().str[0]
df = df.add_prefix('source_').reset_index()

print(df)
                    _id  source_dtp  source_dtot  source_cash
0  AV4MdG6Ihowv-SKBN_nB           1            0            0
1  AV4Mc2vNhowv-SKBN_Rn           0            0            1
2  AV4MeisikOpWpLdepWy6           1            0            0
3  AV4MeRh6howv-SKBOBOn           0            0            1
4  AV4Mezwchowv-SKBOB_S           0            1            0
5  AV4MeB7yhowv-SKBOA5b           1            0            0