python的熊猫很整洁。我正在尝试用pandas-dataframe替换字典列表。但是,我想知道有一种方法可以在for循环中逐行更改值吗?
这是非熊猫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!'
...现在trialList
包含更新后的值,因为trial
引用了这一点。非常便利!但是这些名单是非常不方便的,特别是因为我希望能够以列大小的方式计算大熊猫擅长的东西。
所以从上面给出了trialList,虽然我可以通过做一些类似熊猫的事情来做得更好:
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!'
...但trialList
在这里保持不变。有没有一种简单的方法可以逐行更新值,也许等同于dict-version?重要的是它是逐行的,因为这是一个实验,参与者被提交了大量的试验,每个单独的试验收集了各种数据。
答案 0 :(得分:36)
如果您真的想要逐行操作,可以使用iterrows
和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]
更好的是,当你可以矢量化时:
>>> 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]
总是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]