pandas for python is neat. I'm trying to replace a list-of-dictionaries with a pandas-dataframe. However, I'm wondering of there's a way to change values row-by-row in a for-loop just as easy?
Here's the non-pandas dict-version:
trialList = [
{'no':1, 'condition':2, 'response':''},
{'no':2, 'condition':1, 'response':''},
{'no':3, 'condition':1, 'response':''}
] # ... and so on
for trial in trialList:
# Do something and collect response
trial['response'] = 'the answer!'
... and now trialList
contains the updated values because trial
refers back to that. Very handy! But the list-of-dicts is very unhandy, especially because I'd like to be able to compute stuff column-wise which pandas excel at.
So given trialList from above, I though I could make it even better by doing something pandas-like:
import pandas as pd
dfTrials = pd.DataFrame(trialList) # makes a nice 3-column dataframe with 3 rows
for trial in dfTrials.iterrows():
# do something and collect response
trials[1]['response'] = 'the answer!'
... but trialList
remains unchanged here. Is there an easy way to update values row-by-row, perhaps equivalent to the dict-version? It is important that it's row-by-row as this is for an experiment where participants are presented with a lot of trials and various data is collected on each single trial.
If you really want row-by-row ops, you could use iterrows
and loc
:
>>> for i, trial in dfTrials.iterrows():
... dfTrials.loc[i, "response"] = "answer {}".format(trial["no"])
...
>>> dfTrials
condition no response
0 2 1 answer 1
1 1 2 answer 2
2 1 3 answer 3
[3 rows x 3 columns]
Better though is when you can vectorize:
>>> dfTrials["response 2"] = dfTrials["condition"] + dfTrials["no"]
>>> dfTrials
condition no response response 2
0 2 1 answer 1 3
1 1 2 answer 2 3
2 1 3 answer 3 4
[3 rows x 4 columns]
And there's always apply
:
>>> def f(row):
... return "c{}n{}".format(row["condition"], row["no"])
...
>>> dfTrials["r3"] = dfTrials.apply(f, axis=1)
>>> dfTrials
condition no response response 2 r3
0 2 1 answer 1 3 c2n1
1 1 2 answer 2 3 c1n2
2 1 3 answer 3 4 c1n3
[3 rows x 5 columns]