Python 熊猫：如果 A 列中的行包含“x”，则将“y”写入 B 列中的行
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Pandas: if row in column A contains "x", write "y" to row in column B
提问by Winterflags
For pandas, I'm looking for a way to write conditional values to each row in column B, based on substrings for corresponding rows in column A.
对于pandas，我正在寻找一种方法，根据 A 列中相应行的子字符串，将条件值写入 B 列中的每一行。
So if cell in Acontains "BULL", write "Long"to B. Or if cell in Acontains "BEAR", write "Short"to B.
所以，如果在细胞A中包含"BULL"，写"Long"来B。或者，如果细胞中A含有"BEAR"，写"Short"来B。
Desired output:
期望的输出：
A B
"BULL APPLE X5" "Long"
"BEAR APPLE X5" "Short"
"BULL APPLE X5" "Long"
B is initially empty: df = pd.DataFrame([['BULL APPLE X5',''],['BEAR APPLE X5',''],['BULL APPLE X5','']],columns=['A','B'])
B 初始为空： df = pd.DataFrame([['BULL APPLE X5',''],['BEAR APPLE X5',''],['BULL APPLE X5','']],columns=['A','B'])
采纳答案by Padraic Cunningham
Your code would error as you creating the Dataframe incorrectly, just create a single column Athen add Bbased on A:
当您错误地创建数据框时，您的代码会出错，只需创建一个列A然后添加B基于A：
import pandas as pd
df = pd.DataFrame(["BULL","BEAR","BULL"], columns=['A'])
df["B"] = ["Long" if ele == "BULL" else "Short" for ele in df["A"]]
print(df)
A B
0 BULL Long
1 BEAR Short
2 BULL Long
Or do you logic with the data before you create the dataframe:
或者在创建数据框之前对数据进行逻辑处理：
import pandas as pd
data = ["BULL","BEAR","BULL"]
data2 = ["Long" if ele == "BULL" else "Short" for ele in data]
df = pd.DataFrame(list(zip(data, data2)), columns=['A','B'])
print(df)
A B
0 BULL Long
1 BEAR Short
2 BULL Long
For your edit:
对于您的编辑：
df = pd.DataFrame([['BULL APPLE X5',''],['BEAR APPLE X5',''],['BULL APPLE X5','']], columns=['A','B'])
df["B"] = df["A"].map(lambda x: "Long" if "BULL" in x else "Short" if "BEAR" in x else "")
print(df)
A B
0 BULL APPLE X5 Long
1 BEAR APPLE X5 Short
2 BULL APPLE X5 Long
Or just add the column after:
或者只是在后面添加列：
df = pd.DataFrame(['BULL APPLE X5','BEAR APPLE X5','BLL APPLE X5'], columns=['A'])
df["B"] = df["A"].map(lambda x: "Long" if "BULL" in x else "Short" if "BEAR" in x else "")
print(df)
Or using contains:
或使用包含：
df = pd.DataFrame([['BULL APPLE X5',''],['BEAR APPLE X5',''],['BULL APPLE X5','']], columns=['A','B'])
df["B"][df['A'].str.contains("BULL")] = "Long"
df["B"][df['A'].str.contains("BEAR")] = "Short"
print(df)
0 BULL APPLE X5 Long
1 BEAR APPLE X5 Short
2 BULL APPLE X5 Long
回答by Anand S Kumar
Also, for populating the df['B']you can try the below method -
此外，为了填充df['B']您可以尝试以下方法 -
def applyFunc(s):
if s == 'BULL':
return 'Long'
elif s == 'BEAR':
return 'Short'
return ''
df['B'] = df['A'].apply(applyFunc)
df
>>
A B
0 BULL Long
1 BEAR Short
2 BULL Long
What the applyfunction does, is that for each row value of df['A'], it calls the applyFuncfunction with the parameter as the value of that row , and the returned value is put into the same row for df['B'], what really happens behind the scene is a bit different though, the value is not directly put into df['B']but rather a new Seriesis created and at the end, the new Series is assigned to df['B'].
该apply函数的作用是，对于 的每一行值df['A']，它applyFunc以参数作为该行的值调用该函数，并将返回的值放入同一行 for 中df['B']，但幕后真正发生的事情有点不同，该值不是直接放入df['B']，而是Series创建一个新的，最后，新的系列被分配给df['B']。
回答by unutbu
You could use str.extractto search for regex pattern BULL|BEAR, and then use Series.mapto replace those strings with Longor Short:
您可以使用str.extract搜索正则表达式模式BULL|BEAR，然后使用或Series.map替换这些字符串：LongShort
In [50]: df = pd.DataFrame([['BULL APPLE X5',''],['BEAR APPLE X5',''],['BULL APPLE X5','']],columns=['A','B'])
In [51]: df['B'] = df['A'].str.extract(r'(BULL|BEAR)').map({'BULL':'Long', 'BEAR':'Short'})
In [55]: df
Out[55]:
A B
0 BULL APPLE X5 Long
1 BEAR APPLE X5 Short
2 BULL APPLE X5 Long
However, forming the intermediate Series with str.extractis quite slow compared to df['A'].map(lambda x:...). Using IPython's %timeitto time the benchmarks,
但是，与str.extract相比，形成中间系列的速度相当慢df['A'].map(lambda x:...)。使用 IPython%timeit来计时基准，
In [5]: df = pd.concat([df]*10000)
In [6]: %timeit df['A'].str.extract(r'(BULL|BEAR)').map({'BULL':'Long', 'BEAR':'Short'})
10 loops, best of 3: 39.7 ms per loop
In [7]: %timeit df["A"].map(lambda x: "Long" if "BULL" in x else "Short" if "BEAR" in x else "")
100 loops, best of 3: 4.98 ms per loop
The majority of time is spent in str.extract:
大部分时间花在str.extract：
In [8]: %timeit df['A'].str.extract(r'(BULL|BEAR)')
10 loops, best of 3: 37.1 ms per loop
while the call to Series.mapis relatively fast:
虽然调用Series.map相对较快：
In [9]: x = df['A'].str.extract(r'(BULL|BEAR)')
In [10]: %timeit x.map({'BULL':'Long', 'BEAR':'Short'})
1000 loops, best of 3: 1.82 ms per loop
