# height = [70, 90, 100, 120, 140, 160, 180, 200，220，240， 260] # 長(zhǎng)度為 11
# df.shape ? (1000, 11)
# 對(duì)df的每一行的每一個(gè)元素操作，然后再返回多列
#----------返回多列-----------------
df = df.apply(lambda x: pd.Series([math.atan(i[0]/(i[1])*180/math.pi for i in zip(height, x)]).astype(float), axis=1)
.rename(columns=dict(zip(list(range(0, 11)), df.columns)))

#df['slope'] = df_10min.apply(lambda x: math.atan(210 / (x['ws_260'] - x['ws_50'] + 1e-7))*180/math.pi, axis=1)

import pandas as pd
df_tmp = pd.DataFrame([
? ? {"a":"data1", "cnt":100},{"a":"data2", "cnt":200},
])

df_tmp
a?? ?cnt
data1?? ?100
data2?? ?200

方法一：使用apply 的參數(shù)result_type 來處理
def formatrow(row):
? ? a = row["a"] + str(row["cnt"])
? ? b = str(row["cnt"]) + row["a"]
? ? return a, b?
?
df_tmp[["fomat1", "format2"]] = df_tmp.apply(formatrow, axis=1, result_type="expand")
df_tmp
a?? ?cnt?? ?fomat1?? ?format2
data1?? ?100?? ?data1100?? ?100data1
data2?? ?200?? ?data2200?? ?200data2

方法一：使用zip打包返回結(jié)果來處理
df_tmp["fomat1-1"], df_tmp["format2-2"] = zip(*df_tmp.apply(formatrow, axis=1))
df_tmp
a?? ?cnt?? ?fomat1?? ?format2?? ?fomat1-1?? ?format2-2
data1?? ?100?? ?data1100?? ?100data1?? ?data1100?? ?100data1
data2?? ?200?? ?data2200?? ?200data2?? ?data2200?? ?200data2

np.random.seed(1)
df = pd.DataFrame(np.random.randn(4,2), columns=['A', 'B'])
df
>>>
       A    B
0    1.624345    -0.611756
1    -0.528172    -1.072969
2    0.865408    -2.301539
3    1.744812    -0.761207

df['C'] = df.apply(lambda x: '{:.2f}±{:.2f}'.format(x['A'], x['B']), axis=1)

# 對(duì)第 10 行進(jìn)行操作，基于第2、3兩行
df.loc[10] = df.apply(lambda x: '{:.2f}±{:.2f}'.format(x[2], x[3]) )