Python 替换熊猫中的行值
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/30459485/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Replacing row values in pandas
提问by blaz
I would like to replace row values in pandas.
我想替换熊猫中的行值。
In example:
例如：
import pandas as pd
import numpy as np
a = np.array(([100, 100, 101, 101, 102, 102],
np.arange(6)))
pd.DataFrame(a.T)
Result:
结果：
array([[100, 0],
[100, 1],
[101, 2],
[101, 3],
[102, 4],
[102, 5]])
Here, I would like to replace the rows with the values [101, 3]with [200, 10]and the result should therefore be:
在这里，我想用值来代替行[101, 3]用[200, 10]，因此其结果应该是：
array([[100, 0],
[100, 1],
[101, 2],
[200, 10],
[102, 4],
[102, 5]])
Update
更新
In a more general case I would like to replace multiple rows.
在更一般的情况下，我想替换多行。
Therefore the old and new row values are represented by nx2 sized matrices (n is number of row values to replace). In example:
因此，旧的和新的行值由 nx2 大小的矩阵表示（n 是要替换的行值的数量）。例如：
old_vals = np.array(([[101, 3]],
[[100, 0]],
[[102, 5]]))
new_vals = np.array(([[200, 10]],
[[300, 20]],
[[400, 30]]))
And the result is:
结果是：
array([[300, 20],
[100, 1],
[101, 2],
[200, 10],
[102, 4],
[400, 30]])
采纳答案by EdChum
For the single row case:
对于单行情况：
In [35]:
df.loc[(df[0]==101) & (df[1]==3)] = [[200,10]]
df
Out[35]:
0 1
0 100 0
1 100 1
2 101 2
3 200 10
4 102 4
5 102 5
For the multiple row-case the following would work:
对于多行情况，以下将起作用：
In [60]:
a = np.array(([100, 100, 101, 101, 102, 102],
[0,1,3,3,3,4]))
df = pd.DataFrame(a.T)
df
Out[60]:
0 1
0 100 0
1 100 1
2 101 3
3 101 3
4 102 3
5 102 4
In [61]:
df.loc[(df[0]==101) & (df[1]==3)] = 200,10
df
Out[61]:
0 1
0 100 0
1 100 1
2 200 10
3 200 10
4 102 3
5 102 4
For multi-row update like you propose the following would work where the replacement site is a single row, first construct a dict of the old vals to search for and use the new values as the replacement value:
对于像您建议的多行更新，在替换站点是单行的情况下，以下将起作用，首先构造旧 val 的 dict 以搜索并使用新值作为替换值：
In [78]:
old_keys = [(x[0],x[1]) for x in old_vals]
new_valss = [(x[0],x[1]) for x in new_vals]
replace_vals = dict(zip(old_keys, new_vals))
replace_vals
Out[78]:
{(100, 0): array([300, 20]),
(101, 3): array([200, 10]),
(102, 5): array([400, 30])}
We can then iterate over the dict and then set the rows using the same method as my first answer:
然后我们可以迭代字典，然后使用与我的第一个答案相同的方法设置行：
In [93]:
for k,v in replace_vals.items():
df.loc[(df[0] == k[0]) & (df[1] == k[1])] = [[v[0],v[1]]]
df
0 1
0 100 0
0 1
5 102 5
0 1
3 101 3
Out[93]:
0 1
0 300 20
1 100 1
2 101 2
3 200 10
4 102 4
5 400 30
回答by alec_djinn
The simplest way should be this one:
最简单的方法应该是这个：
df.loc[[3],0:1] = 200,10
In this case, 3 is the third row of the data frame while 0 and 1 are the columns.
在这种情况下，3 是数据框的第三行，而 0 和 1 是列。
This code instead, allows you to iterate over each row, check its content and replace it with what you want.
相反，此代码允许您遍历每一行，检查其内容并将其替换为您想要的内容。
target = [101,3]
mod = [200,10]
for index, row in df.iterrows():
if row[0] == target[0] and row[1] == target[1]:
row[0] = mod[0]
row[1] = mod[1]
print(df)
回答by blaz
Another possibility is:
另一种可能是：
import io
a = np.array(([100, 100, 101, 101, 102, 102],
np.arange(6)))
df = pd.DataFrame(a.T)
string = df.to_string(header=False, index=False, index_names=False)
dictionary = {'100 0': '300 20',
'101 3': '200 10',
'102 5': '400 30'}
def replace_all(text, dic):
for i, j in dic.items():
text = text.replace(i, j)
return text
string = replace_all(string, dictionary)
df = pd.read_csv(io.StringIO(string), delim_whitespace=True)
I found this solution better, since when dealing with large amount of data to replace, the time is shorter than by EdChum's solution.
我发现这个解决方案更好，因为在处理大量要替换的数据时，时间比 EdChum 的解决方案短。
