CODEWITHSUNDEEP
pythonloopspandasdataframe
user5421875
user5421875

Reputation:

iterate over the rows , and control count i++

I'm looking for method, that iterates over the rows, but apply some method only for every 20th or 30th row values so something like:

UPDATED CODE

for index, row in df.iterrows(), index=+20:
     location= geolocator.reverse("%s, %s" % (row['lat'],row['long']),timeout=None)
      row['location']=location.address
      time.sleep(3)
return df

Actually I try to minimize the number of requests, cause otherwise I have the timeout issue. That's why I tried iterate over the rows, and apply the function of request only for every 20th or 60th row (cause I have 7000 rows) and not to speed the process by applying the time.sleep method

Upvotes: 1

Views: 1241

Answers (3)

EdChum
EdChum

Reputation: 394399

Why not just slice the df using iloc and a step param:

Example:

In [120]:
df = pd.DataFrame({'c':np.random.randn(30)})
df

Out[120]:
           c
0  -0.737805
1   1.158012
2  -0.348384
3   0.044989
4   0.962584
5   2.041479
6   1.376785
7   0.208565
8  -1.535244
9   0.389831
10  0.049862
11 -0.142717
12 -0.794087
13  1.316492
14  0.182952
15  0.850953
16  0.015589
17  0.062692
18 -1.551303
19  0.937899
20  0.583003
21 -0.612411
22  0.762307
23 -0.682298
24 -0.897314
25 -0.101144
26 -0.617573
27 -2.168498
28  0.631021
29 -1.592888

In [121]:
df['c'].iloc[::5] = 0
df

Out[121]:
           c
0   0.000000
1   1.158012
2  -0.348384
3   0.044989
4   0.962584
5   0.000000
6   1.376785
7   0.208565
8  -1.535244
9   0.389831
10  0.000000
11 -0.142717
12 -0.794087
13  1.316492
14  0.182952
15  0.000000
16  0.015589
17  0.062692
18 -1.551303
19  0.937899
20  0.000000
21 -0.612411
22  0.762307
23 -0.682298
24 -0.897314
25  0.000000
26 -0.617573
27 -2.168498
28  0.631021
29 -1.592888

This will be much faster than iterating over every row

So in your case:

df['C'].iloc[::20] = some_function()

should work

Upvotes: 0

TigerhawkT3
TigerhawkT3

Reputation: 49330

Just use enumerate and the modulus operator:

for index, row in enumerate(df.iterrows()):
    if not index%20:
        row['C']=some_function()
return df

I took the return out of the loop so that the loop wouldn't end after one iteration.

Upvotes: 2

lapinkoira
lapinkoira

Reputation: 8998

Try this:

for index, row in enumerate(df):
    if index % 20 == 0:
        # do something

Upvotes: 4

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