将 R 条件跨两列转换为 Python
Translating R mutate conditional across two columns to Python
我正在将 R 中的某些内容翻译成 Python 并且无法理解如何跨两列实现条件变异函数以实现 Python.
中的类似内容
df <- df %>%
select(col_1,…,col_n) %>%
mutate(new_col = ifelse(is.na(col_1), NA, 2),
new_col = ifelse(is.na(col_2), new_col, ifelse(col_3 == 1, new_col+1, new_col-1))).
到目前为止,我有以下内容,其中 col_1 是一个字符串,col_2 是一个字符串,col_3 是一个浮点数:
df = df[[col_1,...,col_n]]
df[new_col] = df[col_1].apply(lambda x: np.nan if x is np.nan else 2)
并且不知道如何进行下一次变异。我尝试了以下方法:
1. df[new_col] = df.apply[col_2](lambda x: x if x is ' ' else df[col_3].apply(lambda x: x+1 if x == 1 else x-1)
# This timesout the kernel
2. df[new_col] = df.apply(lambda x: x if x[col_2] == ' ' else x+1 if x[col_3] == 1 else x-1 if x[col_3] != else x)
# This results in an error of unsuporrted operand type(s) for -: 'str' and 'int'
# I also don't think the 'else x' at the end is the correct way to get the same result
有没有一种方法可以通过比嵌套应用的计算成本更低的方法(如果这甚至是一种正确的攻击方法)或某种方法来手动提取必要的信息?
我想“直译”应该是这样的:
df = df[[col_1,...,col_n]]
df['new_col'] = np.where(df['col_1'].isnull(),
np.nan,
2)
df['new_col'] = np.where(df['col_2'].isnull(),
df['new_col'],
np.where(df['col_3'] == 1,
df['new_col']+1,
df['new_col']-1))
你也可以把你的R代码翻译成python with datar
顺利:
>>> from datar.all import NA, f, tibble, select, mutate, if_else, is_na
>>>
>>> df = tibble(
... col_1 = [1, NA, 2, 3],
... col_2 = [4, 5, NA, 7],
... col_3 = [8, 9, 10, 1],
... unk_col = list('abcd')
... )
>>>
>>> df >> select(f.col_1, f.col_2, f.col_3) >> mutate(
... new_col_ = if_else(is_na(f.col_1), NA, 2),
... new_col = if_else(
... is_na(f.col_2),
... f.new_col,
... if_else(f.col_3 == 1, f.new_col+1, f.new_col-1)
... )
... )
col_1 col_2 col_3 new_col
<float64> <float64> <int64> <float64>
0 1.0 4.0 8 1.0
1 NaN 5.0 9 NaN
2 2.0 NaN 10 2.0
3 3.0 7.0 1 3.0
我是 datar
包的作者。有问题欢迎提issue
我正在将 R 中的某些内容翻译成 Python 并且无法理解如何跨两列实现条件变异函数以实现 Python.
中的类似内容df <- df %>%
select(col_1,…,col_n) %>%
mutate(new_col = ifelse(is.na(col_1), NA, 2),
new_col = ifelse(is.na(col_2), new_col, ifelse(col_3 == 1, new_col+1, new_col-1))).
到目前为止,我有以下内容,其中 col_1 是一个字符串,col_2 是一个字符串,col_3 是一个浮点数:
df = df[[col_1,...,col_n]]
df[new_col] = df[col_1].apply(lambda x: np.nan if x is np.nan else 2)
并且不知道如何进行下一次变异。我尝试了以下方法:
1. df[new_col] = df.apply[col_2](lambda x: x if x is ' ' else df[col_3].apply(lambda x: x+1 if x == 1 else x-1)
# This timesout the kernel
2. df[new_col] = df.apply(lambda x: x if x[col_2] == ' ' else x+1 if x[col_3] == 1 else x-1 if x[col_3] != else x)
# This results in an error of unsuporrted operand type(s) for -: 'str' and 'int'
# I also don't think the 'else x' at the end is the correct way to get the same result
有没有一种方法可以通过比嵌套应用的计算成本更低的方法(如果这甚至是一种正确的攻击方法)或某种方法来手动提取必要的信息?
我想“直译”应该是这样的:
df = df[[col_1,...,col_n]]
df['new_col'] = np.where(df['col_1'].isnull(),
np.nan,
2)
df['new_col'] = np.where(df['col_2'].isnull(),
df['new_col'],
np.where(df['col_3'] == 1,
df['new_col']+1,
df['new_col']-1))
你也可以把你的R代码翻译成python with datar
顺利:
>>> from datar.all import NA, f, tibble, select, mutate, if_else, is_na
>>>
>>> df = tibble(
... col_1 = [1, NA, 2, 3],
... col_2 = [4, 5, NA, 7],
... col_3 = [8, 9, 10, 1],
... unk_col = list('abcd')
... )
>>>
>>> df >> select(f.col_1, f.col_2, f.col_3) >> mutate(
... new_col_ = if_else(is_na(f.col_1), NA, 2),
... new_col = if_else(
... is_na(f.col_2),
... f.new_col,
... if_else(f.col_3 == 1, f.new_col+1, f.new_col-1)
... )
... )
col_1 col_2 col_3 new_col
<float64> <float64> <int64> <float64>
0 1.0 4.0 8 1.0
1 NaN 5.0 9 NaN
2 2.0 NaN 10 2.0
3 3.0 7.0 1 3.0
我是 datar
包的作者。有问题欢迎提issue